MATH 222 Second Semester Calculus. Spring 2013

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1 MATH Second Semester Calculus Spring 3 Typeset:March 4, 3

2 Math nd Semester Calculus Lecture notes version. (Spring 3) is is a self contained set of lecture notes for Math. e notes were wri en by Sigurd Angenent, as part ot the MIU calculus project. Some problems were contributed by A.Miller. e L A TEX files, as well as the I and O files which were used to produce these notes are available at the following web site ey are meant to be freely available for non-commercial use, in the sense that free so ware is free. More precisely: Copyright (c) Sigurd B. Angenent. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version. or any later version published by the Free So ware Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled GNU Free Documentation License.

3 Contents Chapter I. Methods of Integration 7. Definite and indefinite integrals 7. Problems 9 3. First trick: using the double angle formulas 9 4. Problems 5. Integration by Parts 6. Reduction Formulas 5 7. Problems 8 8. Partial Fraction Expansion 9. Problems 5. Substitutions for integrals containing the expression ax + bx + c 6. Rational substitution for integrals containing x a or a + x 3. Simplifying ax + bx + c by completing the square Problems Chapter summary Mixed Integration Problems 36 Chapter II. Proper and Improper Integrals 4. Typical examples of improper integrals 4. Summary: how to compute an improper integral More examples Problems Estimating improper integrals Problems 54 Chapter III. First order differential Equations 57. What is a Differential Equation? 57. Two basic examples First Order Separable Equations Problems 6 5. First Order Linear Equations 6 6. Problems Direction Fields Euler s method Problems 67. Applications of Differential Equations 67. Problems 7 Chapter IV. Taylor s Formula 73. Taylor Polynomials 73. Examples Some special Taylor polynomials Problems The Remainder Term 8 6. Lagrange s Formula for the Remainder Term 8 7. Problems The limit as x, keeping n fixed 84 3

4 4 CONTENTS 9. Problems 9. Differentiating and Integrating Taylor polynomials 9. Problems on Integrals and Taylor Expansions 94. Proof of Theorem Proof of Lagrange s formula for the remainder 95 Chapter V. Sequences and Series 97. Introduction 97. Sequences Problems on Limits of Sequences 4. Series 5. Convergence of Taylor Series 5 6. Problems on Convergence of Taylor Series 8 7. Leibniz formulas for ln and π/ Problems 9 Chapter VI. Vectors. Introduction to vectors. Geometric description of vectors 3 3. Parametric equations for lines and planes 6 4. Vector Bases 8 5. Dot Product 9 6. Cross Product 7 7. A few applications of the cross product 3 8. Notation Problems Computing and drawing vectors 35. Problems Parametric equations for a line 37. Problems Orthogonal decomposition of one vector with respect to another 38. Problems The dot product Problems The cross product 39 Chapter VII. Answers, Solutions, and Hints 4 Chapter VIII. GNU Free Documentation License 59

5 CONTENTS 5 f(x) = df (x) (n + )x n = n+ f(x) = F (x) + C x n = xn+ n + + C n d ln x = x = ln x + C x absolute values!! e x = dex sin x = d cos x cos x = d sin x d ln cos x tan x = e x = e x + C sin x = cos x + C cos x = sin x + C tan x = ln cos x + C absolute values!! + x = d arctan x = d arcsin x x f(x) + g(x) = cf(x) = df (x) + G(x) d cf (x) F dg = df G df G = arctan x + C + x = arcsin x + C x {f(x) + g(x)} = F (x) + G(x) + C cf(x) = cf (x) + C F G = F G F G To find derivatives and integrals involving a x instead of e x use a = e ln a, and thus a x = e x ln a, to rewrite all exponentials as e.... The following integral is also useful, but not as important as the ones above: cos x = + sin x ln + C for cos x. sin x Table. The list of the standard integrals everyone should know

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7 CHAPTER I Methods of Integration e basic question that this chapter addresses is how to compute integrals, i.e. Given a function y = f(x) how do we find a function y = F (x) whose derivative is F (x) = f(x)? e simplest solution to this problem is to look it up on the Internet. Any integral that we compute in this chapter can be found by typing it into the following web page: Other similar websites exist, and more extensive so ware packages are available. It is therefore natural to ask why should we learn how to do these integrals? e question has at least two answers. First, there are certain basic integrals that show up frequently and that are relatively easy to do (once we know the trick), but that are not included in a first semester calculus course for lack of time. Knowing these integrals is useful in the same way that knowing things like + 3 = 5 saves us from a lot of unnecessary calculator use. f(t) Input signal f(t) Electronic Circuit Output signal t g(t) = e τ t f(τ) dτ g(t) t t e second reason is that we o en are not really interested in specific integrals, but in general facts about integrals. For example, the output g(t) of an electric circuit (or mechanical system, or a biochemical system, etc.) is o en given by some integral involving the input f(t). e methods of integration that we will see in this chapter give us the tools we need to understand why some integral gives the right answer to a given electric circuits problem, no ma er what the input f(t) is.. Definite and indefinite integrals We recall some facts about integration from first semester calculus... Definition. A function y = F (x) is called an antiderivative of another function y = f(x) if F (x) = f(x) for all x. For instance, F (x) = x is an antiderivative of f(x) = x, and so is G(x) = x +. e Fundamental eorem of Calculus states that if a function y = f(x) is continuous on an interval a x b, then there always exists an antiderivative F (x) of f, 7

8 8 I. METHODS OF INTEGRATION Indefinite integral Definite integral f(x) is a function of x. b a f(x) is a number. By definition f(x) is any function F (x) whose derivative is f(x). If F (x) is an antiderivative of f(x), then so is F (x) + C. Therefore f(x) = F (x) + C; an indefinite integral contains a constant ( +C ). x is not a dummy variable, for example, x = x + C and tdt = t + C are functions of different variables, so they are not equal. (See Problem.) b a f(x) was defined in terms of Riemann sums and can be interpreted as area under the graph of y = f(x) when f(x). b a f(x) is one uniquely defined number; an indefinite integral does not contain an arbitrary constant. x is a dummy variable, for example, x =, and tdt =, so x = tdt. Whether we use x or t the integral makes no difference. Table. Important differences between definite and indefinite integrals and one has () b a f(x) = F (b) F (a). For example, if f(x) = x, then F (x) = x is an antiderivative for f(x), and thus b a x = F (b) F (a) = b a. e best way of computing an integral is o en to find an antiderivative F of the given function f, and then to use the Fundamental eorem (). How to go about finding an antiderivative F for some given function f is the subject of this chapter. e following notation is commonly used for antiderivatives: () F (x) = f(x). e integral that appears here does not have the integration bounds a and b. It is called an indefinite integral, as opposed to the integral in () which is called a definite integral. It is important to distinguish between the two kinds of integrals. Table lists the main differences.

9 3. FIRST TRICK: USING THE DOUBLE ANGLE FORMULAS 9. Problems. Compute the following integrals: (a) A = x, (b) B = t dt, 3. Which of the following inequalities are true? (a) 4 ( x ) > (c) C = x dt, (d) I = xt dt, (e) J = xt.. One of the following three integrals is not the same as the other two: A = B = C = x, t dt, x dt. Which one? Explain your answer. (b) 4 ( x )dt > (c) ( + x ) > 4. One of the following statements is correct. Which one, and why? (a) x t dt = 3 x3. (b) t dt = 3 x3. (c) t dt = 3 x3 + C. 3. First tri : using the double angle formulas e first method of integration we see in this chapter uses trigonometric identities to rewrite functions in a form that is easier to integrate. is particular trick is useful in certain integrals involving trigonometric functions and while these integrals show up frequently, the double angle trick is not a general method for integration. 3.. e double angle formulas. e simplest of the trigonometric identities are the double angle formulas. ese can be used to simplify integrals containing either sin x or cos x. Recall that cos α sin α = cos α and cos α + sin α =, Adding these two equations gives while subtracting them gives cos α = (cos α + ) sin α = ( cos α). ese are the two double angle formulas that we will use.

10 I. METHODS OF INTEGRATION 3... Example. e following integral shows up in many contexts, so it is worth knowing: cos x = ( + cos x) = {x + } sin x + C = x + sin x + C. 4 Since sin x = sin x cos x this result can also be wri en as cos x = x + sin x cos x + C A more complicated example. If we need to find I = cos 4 x then we can use the double angle trick once to rewrite cos x as ( + cos x), which results in I = cos 4 x = { ( + cos x)} ( = + cos x + cos x ). 4 e first two terms are easily integrated, and now that we know the double angle trick we also can do the third term. We find cos x = Going back to the integral I we get ( ) x + cos 4x = + sin 4x + C. 8 I = 4 ( + cos x + cos x ) = x sin x + (x sin 4x) + C = 3x sin x + sin 4x + C Example without the double angle trick. e integral J = cos 3 x looks very much like the two previous examples, but there is very different trick that will give us the answer. Namely, substitute u = sin x. en du = cos x, and cos x =

11 4. PROBLEMS sin x = u, so that J = = = cos x cos x ( sin x) cos x ( u ) du = u 3 u3 + C = sin x 3 sin3 x + C. In summary, the double angle formulas are useful for certain integrals involving powers of sin( ) or cos( ), but not all. In addition to the double angle identities there are other trigonometric identities that can be used to find certain integrals. See the exercises. 4. Problems Compute the following integrals using the double angle formulas if necessary:. ( + sin θ) dθ.. (cos θ + sin θ) dθ. 3. Find sin x cos x (hint: use the other double angle formula sin α = sin α cos α.) 4. cos 5 θ dθ 5. Find ( sin θ + cos θ) dθ The double angle formulas are special cases of the following trig identities: sin A sin B = cos(a B) cos(a + B) cos A cos B = cos(a B) + cos(a + B) sin A cos B = sin(a + B) + sin(a B) Use these identities to compute the following integrals. 6. sin x sin x 7. π sin 3x sin x (sin ) 8. θ cos 3θ dθ. 9.. π/ π ( sin θ + sin 4θ ) dθ. sin kx sin mx where k and m are constant positive integers. Simplify your answer! (careful: a er working out your solution, check if you didn t divide by zero anywhere.). Let a be a positive constant and I a = π/ (a) Find I a if a. sin(aθ) cos(θ) dθ. (b) Find I a if a =. (Don t divide by zero.). The input signal for a given electronic circuit is a function of time V in (t). The output signal is given by V out(t) = t sin(t s)v in(s)ds. Find V out (t) if V in (t) = sin(at) where a > is some constant. 3. The alternating electric voltage coming out of a socket in any American living room is said to be Volts and 5Herz (or 6, depending on where you are). This means that the voltage is a function of time of the form V (t) = A sin(π t T ) where T = 5 sec is how long one oscillation takes (if the frequency is 5 Herz, then

12 I. METHODS OF INTEGRATION there are 5 oscillations per second), and A is the amplitude (the largest voltage during any oscillation). V(t) T T 3T 4T 5T 6T V RMS A=amplitude t The Volts that is specified is not the amplitude A of the oscillation, but instead it refers to the Root Mean Square of the voltage. By definition the R.M.S. of the oscillating voltage V (t) is T = V (t) T dt. (it is the square root of the mean of the square of V (t)). Compute the amplitude A. 5. Integration by Parts While the double angle trick is just that, a (useful) trick, the method of integration by parts is very general and appears in many different forms. It is the integration counterpart of the product rule for differentiation. 5.. e product rule and integration by parts. Recall that the product rule says that df (x)g(x) df (x) = G(x) + F (x)dg(x) and therefore, a er rearranging terms, F (x) dg(x) df (x)g(x) df (x) = G(x). If we integrate both sides we get the formula for integration by parts F (x) dg(x) df (x) = F (x)g(x) G(x). Note that the effect of integration by parts is to integrate one part of the function (G (x) got replaced by G(x)) and to differentiate the other part (F (x) got replaced by F (x)). For any given integral there are many ways of choosing F and G, and it not always easy to see what the best choice is. 5.. An Example Integrating by parts once. Consider the problem of finding I = xe x. We can use integration by parts as follows: }{{} x }{{} e x = }{{} x }{{} e x }{{} e x }{{} = xe x e x + C. F (x) G (x) F (x) G(x) G(x) F (x) Observe that in this example e x was easy to integrate, while the factor x becomes an easier function when you differentiate it. is is the usual state of affairs when integration by parts works: differentiating one of the factors (F (x)) should simplify the integral, while integrating the other (G (x)) should not complicate things (too much).

13 5. INTEGRATION BY PARTS Another example. What is x sin x? d( cos x) Since sin x = we can integrate by parts }{{} x sin }{{} x = }{{} x ( cos x) }{{}}{{} F (x) G (x) F (x) G(x) F (x) ( cos x) = x cos x + sin x + C. }{{} G(x) 5.4. Example Repeated Integration by Parts. Let s try to compute I = x e x by integrating by parts. Since e x = d ex (3) x }{{} e x }{{} F (x) G (x) one has = x ex e x x = x e x e x x. To do the integral on the le we have to integrate by parts again: e x x = ex }{{} x }{{} ex }{{}. = F (x) }{{} xex e x = xex 4 ex +C. F (x) G(x) G(x) Combining this with (3) we get x e x = x e x xex + 4 ex C (Be careful with all the minus signs that appear when integrating by parts.) 5.5. Another example of repeated integration by parts. e same procedure as in the previous example will work whenever we have to integrate P (x)e ax where P (x) is any polynomial, and a is a constant. Every time we integrate by parts, we get this P (x) }{{}}{{} e ax = P (x) eax e ax a a P (x) F (x) G (x) = a P (x)eax P (x)e ax. a We have replaced the integral P (x)e ax with the integral P (x)e ax. is is the same kind of integral, but it is a li le easier since the degree of the derivative P (x) is less than the degree of P (x).

14 4 I. METHODS OF INTEGRATION 5.6. Example sometimes the factor G (x) is invisible. Here is how we can get the antiderivative of ln x by integrating by parts: ln x = ln x }{{} }{{} F (x) G (x) = ln x x x x = x ln x = x ln x x + C. We can do P (x) ln x in the same way if P (x) is any polynomial. For instance, to compute (z + z) ln z dz we integrate by parts: (z + z) }{{}}{{} ln z G (z) F (z) dz = ( ( 3 z3 + z) ln z 3 z3 + z) z dz = ( ( 3 z3 + z) ln z 3 z + z) dz = ( 3 z3 + z) ln z 9 z3 4 z + C An example where we get the original integral ba. It can happen that a er integrating by parts a few times the integral we get is the same as the one we started with. When this happens we have found an equation for the integral, which we can then try to solve. e standard example in which this happens is the integral I = e x sin x. We integrate by parts twice: }{{} e x sin }{{ x } = e x sin x }{{} e x } cos {{ x } F (x) G(x) F (x) G (x) = e x sin x e x cos x = e x sin x e x cos x e x sin x = e x sin x e x cos x 4 e x sin x. Note that the last integral here is exactly I again. erefore the integral I satisfies We solve this equation for I, with result I = e x sin x e x cos x 4I. 5I = e x sin x e x cos x = I = 5( e x sin x e x cos x ).

15 6. REDUCTION FORMULAS 5 Since I is an indefinite integral we still have to add the arbitrary constant: I = 5( e x sin x e x cos x ) + C. 6. Reduction Formulas We have seen that we can compute integrals by integrating by parts, and that we sometimes have to integrate by parts more than once to get the answer. ere are integrals where we have to integrate by parts not once, not twice, but n-times before the answer shows up. To do such integrals it is useful to carefully describe what happens each time we integrate by parts before we do the actual integrations. e formula that describes what happens a er one partial integration is called a reduction formula. All this is best explained by an example. 6.. First example of a reduction formula. Consider the integral I n = x n e ax, (n =,,, 3,...) or, in other words, consider all the integrals I = e ax, I = xe ax, I = x e ax, I 3 = x 3 e ax,... and so on. We will consider all these integrals at the same time. Integration by parts in I n gives us I n = }{{} x n }{{} e ax F (x) G (x) = x n a eax nx n a eax = a xn e ax n x n e ax. a We haven t computed the integral, and in fact the integral that we still have to do is of the same kind as the one we started with (integral of x n e ax instead of x n e ax ). What we have derived is the following reduction formula I n = a xn e ax n a I n, which holds for all n. For n = we do not need the reduction formula to find the integral. We have I = e ax = a eax + C. When n the reduction formula tells us that we have to compute I n if we want to find I n. e point of a reduction formula is that the same formula also applies to I n, and I n, etc., so that a er repeated application of the formula we end up with I, i.e., an integral we know.

16 6 I. METHODS OF INTEGRATION For example, if we want to compute x 3 e ax we use the reduction formula three times: I 3 = a x3 e ax 3 a I = a x3 e ax 3 { a a x e ax } a I = a x3 e ax 3 { a a x e ax ( a a xeax )} a I Insert the known integral I = a eax + C and simplify the other terms and we get x 3 e ax = a x3 e ax 3 a x e ax + 6 a 3 xeax 6 a 4 eax + C. 6.. Reduction formula requiring two partial integrations. Consider S n = x n sin x. en for n one has S n = x n cos x + n x n cos x = x n cos x + nx n sin x n(n ) x n sin x. us we find the reduction formula S n = x n cos x + nx n sin x n(n )S n. Each time we use this reduction, the exponent n drops by, so in the end we get either S or S, depending on whether we started with an odd or even n. ese two integrals are S = sin x = cos x + C S = x sin x = x cos x + sin x + C. (Integrate by parts once to find S.) As an example of how to use the reduction formulas for S n let s try to compute S 4 : x 4 sin x = S 4 = x 4 cos x + 4x 3 sin x 4 3S = x 4 cos x + 4x 3 sin x 4 3 { x cos x + x sin x S } At this point we use S = sin x = cos x + C, and we combine like terms. is results in x 4 sin x = x 4 cos x + 4x 3 sin x 4 3 { x cos x + x sin x ( cos x) } + C = ( x 4 + x 4 ) cos x + ( 4x 3 + 4x ) sin x + C.

17 6. REDUCTION FORMULAS A reduction formula where you have to solve for I n. We try to compute I n = (sin x) n by a reduction formula. Integrating by parts we get I n = (sin x) n sin x = (sin x) n cos x ( cos x)(n )(sin x) n cos x = (sin x) n cos x + (n ) (sin x) n cos x. We now use cos x = sin x, which gives {sin I n = (sin x) n cos x + (n ) n x sin n x } = (sin x) n cos x + (n )I n (n )I n. We can think of this as an equation for I n, which, when we solve it tells us ni n = (sin x) n cos x + (n )I n and thus implies (4) I n = n sinn x cos x + n n I n. Since we know the integrals I = (sin x) = = x + C and I = sin x = cos x + C the reduction formula (4) allows us to calculate I n for any n A reduction formula that will come in handy later. In the next section we will see how the integral of any rational function can be transformed into integrals of easier functions, the most difficult of which turns out to be I n = ( + x ) n. When n = this is a standard integral, namely I = = arctan x + C. + x When n > integration by parts gives us a reduction formula. Here s the computation: I n = ( + x ) n x = ( + x ) n x ( n) ( + x ) n x x = ( + x ) n + n x ( + x ) n+

18 8 I. METHODS OF INTEGRATION Apply to get x ( + x ) n+ = ( + x ) ( + x ) n+ = ( + x ) n ( + x ) n+ x { } ( + x = ) n+ ( + x ) n ( + x ) n+ Our integration by parts therefore told us that x I n = ( + x ) n + n( ) I n I n+, which we can solve for I n+. We find the reduction formula I n+ = x n ( + x ) n + n n I n. = I n I n+. As an example of how we can use it, we start with I = arctan x + C, and conclude that ( + x ) = I = I + = x ( + x ) + I = x + x + arctan x + C. Apply the reduction formula again, now with n =, and we get ( + x ) 3 = I 3 = I + = x ( + x ) + I { } = x 4 ( + x ) + 3 x 4 + x + arctan x = x 4 ( + x ) + 3 x 8 + x arctan x + C. 7. Problems. Evaluate x n ln x where n.. Assume a and b are constants, and compute e ax sin bx. [Hint: Integrate by parts twice; you can assume that b.] 3. Evaluate e ax cos bx where a, b. 4. Prove the formula x n e x = x n e x n and use it to evaluate x e x. x n e x 5. Use 6.3 to evaluate sin x. Show that the answer is the same as the answer you get using the half angle formula. 6. Use the reduction formula in 6.3 to compute π/ sin 4 x.

19 7. In this problem you ll look at the numbers A n = π sin n x. (a) Check that A = π and A =. (b) Use the reduction formula in 6.3 to compute A 5, A 6, and A 7. (c) Explain why A n < A n is true for all n =,, 3, 4,... (Hint: Interpret the integrals A n as area under the graph, and check that (sin x) n (sin x) n for all x.) (d) Based on your values for A 5, A 6, and A 7 find two fractions a and b such that a < π < b. 8. Prove the formula cos n x = n sin x cosn x + n cos n x, n for n, and use it to evaluate π/4 9. Prove the formula x m (ln x) n = cos 4 x. x m+ (ln x) n m + n x m (ln x) n, m + for m, and use it to evaluate the following integrals:. ln x. (ln x) 7. PROBLEMS 9. x 3 (ln x) 3. Evaluate x ln x by another method. [Hint: the solution is short!] 4. For any integer n > derive the formula tan n x = tann x n tan n x Using this, find π/4 tan 5 x. Use the reduction formula from example 6.4 to compute these integrals: 5. ( + x ) 3 6. ( + x ) 4 x 7. ( + x ) [Hint: x/( + x ) n is 4 easy.] x ( + x ) ( 49 + x ) 3.. The reduction formula from example 6.4 is valid for all n. In particular, n does not have to be an integer, and it does not have to be positive. Find a relation between + x and n =. + x. Apply integration by parts to x Let u = x du = and v = x. x x = ( x )(x) by se ing and dv =. This gives us, x x Simplifying x = + x and subtracting the integral from both sides gives us =. How can this be?

20 I. METHODS OF INTEGRATION 8. Partial Fraction Expansion By definition, a rational function is a quotient (a ratio) of polynomials, f(x) = P (x) Q(x) = p nx n + p n x n + + p x + p q d x d + q d x d + + q x + q. Such rational functions can always be integrated, and the trick that allows you to do this is called a partial fraction expansion. e whole procedure consists of several steps that are explained in this section. e procedure itself has nothing to do with integration: it s just a way of rewriting rational functions. It is in fact useful in other situations, such as finding Taylor expansions (see Chapter IV) and computing inverse Laplace transforms (see M 39.) 8.. Reduce to a proper rational function. A proper rational function is a rational function P (x)/q(x) where the degree of P (x) is strictly less than the degree of Q(x). e method of partial fractions only applies to proper rational functions. Fortunately there s an additional trick for dealing with rational functions that are not proper. If P /Q isn t proper, i.e. if degree(p ) degree(q), then you divide P by Q, with result P (x) R(x) = S(x) + Q(x) Q(x) where S(x) is the quotient, and R(x) is the remainder a er division. In practice you would do a long division to find S(x) and R(x). 8.. Example. Consider the rational function f(x) = x3 x + x. Here the numerator has degree 3 which is more than the degree of the denominator (which is ). e function f(x) is therefore not a proper rational function. To apply the method of partial fractions we must first do a division with remainder. One has so that x x x 3 x+ x 3 x f(x) = x3 x + x = S(x) x+ = R(x) = x + x + x When we integrate we get x 3 { x + x = x + x + x = x + } x + x. e rational function that we still have to integrate, namely x+ x, is proper: its numerator has lower degree than its denominator.

21 8. PARTIAL FRACTION EXPANSION 8.3. Partial Fraction Expansion: e Easy Case. To compute the partial fraction expansion of a proper rational function P (x)/q(x) you must factor the denominator Q(x). Factoring the denominator is a problem as difficult as finding all of its roots; in Math we shall only do problems where the denominator is already factored into linear and quadratic factors, or where this factorization is easy to find. In the easiest partial fractions problems, all the roots of Q(x) are real numbers and distinct, so the denominator is factored into distinct linear factors, say P (x) Q(x) = P (x) (x a )(x a ) (x a n ). To integrate this function we find constants A, A,..., A n so that P (x) Q(x) = A + A + + A n. (#) x a x a x a n en the integral is P (x) Q(x) = A ln x a + A ln x a + + A n ln x a n + C. One way to find the coefficients A i in (#) is called the method of equating coefficients. In this method we multiply both sides of (#) with Q(x) = (x a ) (x a n ). e result is a polynomial of degree n on both sides. Equating the coefficients of these polynomial gives a system of n linear equations for A,, A n. You get the A i by solving that system of equations. Another much faster way to find the coefficients A i is the Heaviside tri ¹. Multiply equation (#) by x a i and then plug in² x = a i. On the right you are le with A i so A i = P (x)(x a i) Q(x) = x=ai P (a i ) (a i a ) (a i a i )(a i a i+ ) (a i a n ) Previous Example continued. To integrate x + x e partial fraction expansion of x + x (5) Multiply with (x )(x + ) to get x = (x )(x + ). then is x + x = x + (x )(x + ) = A x + B x +. x + = A(x + ) + B(x ) = (A + B)x + (A B). we factor the denominator, e functions of x on the le and right are equal only if the coefficient of x and the constant term are equal. In other words we must have A + B = and A B =. ¹ Named a er O H, a physicist and electrical engineer in the late 9 th and early th century. ² More properly, you should take the limit x a i. e problem here is that equation (#) has x a i in the denominator, so that it does not hold for x = a i. erefore you cannot set x equal to a i in any equation derived from (#). But you can take the limit x a i, which in practice is just as good.

22 I. METHODS OF INTEGRATION ese are two linear equations for two unknowns A and B, which we now proceed to solve. Adding both equations gives A =, so that A = ; from the first equation one then finds B = A = 3. So x + x = / x 3/ x +. Instead, we could also use the Heaviside trick: multiply (5) with x to get x + x + = A + B x x + Take the limit x and you find + + = A, i.e. A =. Similarly, a er multiplying (5) with x + one gets and le ing x you find x + x = Ax + x + B, B = ( ) + ( ) = 3, as before. Either way, the integral is now easily found, namely, x 3 x + x = x x + + x = x + { / x 3/ x + } = x + ln x 3 ln x + + C Partial Fraction Expansion: e General Case. When the denominator Q(x) contains repeated factors or quadratic factors (or both) the partial fraction decomposition is more complicated. In the most general case the denominator Q(x) can be factored in the form (6) Q(x) = (x a ) k (x a n ) kn (x + b x + c ) l (x + b m x + c m ) lm Here we assume that the factors x a,, x a n are all different, and we also assume that the factors x + b x + c,, x + b m x + c m are all different. It is a theorem from advanced algebra that you can always write the rational function P (x)/q(x) as a sum of terms like this (7) P (x) Q(x) = + A (x a i ) k + + Bx + C (x + b j x + c j ) l + How did this sum come about? For each linear factor (x a) k in the denominator (6) you get terms A x a + A (x a) + + A k (x a) k in the decomposition. ere are as many terms as the exponent of the linear factor that generated them.

23 8. PARTIAL FRACTION EXPANSION 3 For each quadratic factor (x + bx + c) l you get terms B x + C x + bx + c + B x + C (x + bx + c) + + B mx + C m (x + bx + c) l. Again, there are as many terms as the exponent l with which the quadratic factor appears in the denominator (6). In general, you find the constants A..., B... and C... by the method of equating coefficients. Unfortunately, in the presence of quadratic factors or repeated linear factors the Heaviside trick does not give the whole answer; we really have to use the method of equating coefficients. e workings of this method are best explained in an example Example. Find the partial fraction decomposition of and compute f(x) = x + x (x + ) I = x + x (x + ). e degree of the denominator x (x + ) is four, so our partial fraction decomposition must also contain four undetermined constants. e expansion should be of the form x + x (x + ) = A x + B x + Cx + D x +. To find the coefficients A, B, C, D we multiply both sides with x ( + x ), x + = Ax(x + ) + B(x + ) + x (Cx + D) x + = (A + C)x 3 + (B + D)x + Ax + B x 3 + x + x + = (A + C)x 3 + (B + D)x + Ax + B Comparing terms with the same power of x we find that A + C =, B + D =, A =, B =. ese are four equations for four unknowns. Fortunately for us they are not very difficult in this example. We find A =, B =, C = A =, and D = B =, whence e integral is therefore f(x) = x + x (x + ) = x x +. I = x + x (x + ) = arctan x + C. x

24 4 I. METHODS OF INTEGRATION 8.7. A complicated example. Find the integral x + 3 x (x + )(x + ) 3. e procedure is exactly the same as in the previous example. We have to expand the integrand in partial fractions: (8) x + 3 x (x + )(x + ) 3 = A x + A x + A 3 x + + B x + C x + + B x + C (x + ) + B 3x + C 3 (x + ) 3. Note that the degree of the denominator x (x + )(x + ) 3 is = 9, and also that the partial fraction decomposition has nine undetermined constants A, A, A 3, B, C, B, C, B 3, C 3. A er multiplying both sides of (8) with the denominator x (x + )(x + ) 3, expanding everything, and then equating coefficients of powers of x on both sides, we get a system of nine linear equations in these nine unknowns. e final step in finding the partial fraction decomposition is to solve those linear equations. A computer program like Maple or Mathematica can do this easily, but it is a lot of work to do it by hand A er the partial fraction decomposition. Once we have the partial fraction decomposition (8) we still have to integrate the terms that appeared. e first three terms are of the form A(x a) p and they are easy to integrate: A = A ln x a + C x a and A (x a) p = A ( p)(x a) p + C if p >. e next, fourth term in (8) can be wri en as B x + C x + = B x x + + C x + = B ln(x + ) + C arctan x + K, where K is the integration constant (normally +C but there are so many other C s in this problem that we chose a different le er, just for this once.) While these integrals are already not very simple, the integrals Bx + C (x with p > + bx + c) p which can appear are particularly unpleasant. If we really must compute one of these, then we should first complete the square in the denominator so that the integral takes the form Ax + B ((x + b) + a ) p. A er the change of variables u = x + b and factoring out constants we are le with the integrals du u du (u + a ) p and (u + a ) p.

25 9. PROBLEMS 5 e reduction formula from example 6.4 then allows us to compute this integral. An alternative approach is to use complex numbers. If we allow complex numbers then the quadratic factors x + bx + c can be factored, and our partial fraction expansion only contains terms of the form A/(x a) p, although A and a can now be complex numbers. e integrals are then easy, but the answer has complex numbers in it, and rewriting the answer in terms of real numbers again can be quite involved. In this course we will avoid complex numbers and therefore we will not explain this any further. 9. Problems. Express each of the following rational functions as a polynomial plus a proper rational function. (See 8. for definitions.) (a) x 3 x 3 4 (b) x3 + x x 3 4 (c) x3 x x 5 x 3 4 (d) x3 x. Compute the following integrals by completing the square: (a) x + 6x + 8, (b) (c) x + 6x +, 5x + x Use the method of equating coefficients to find numbers A, B, C such that x + 3 x(x + )(x ) = A x + B x + + and then evaluate the integral x + 3 x(x + )(x ). C x 4. Do the previous problem using the Heaviside trick. 5. Find the integral x + 3 x (x ). 6. Simplicio had to integrate 4x (x 3)(x + ). He set 4x (x 3)(x + ) = A x 3 + B x +. Using the Heaviside trick he then found A = 4x x 3 =, x= and B = 4x x + = 9, x=3 which leads him to conclude that 4x (x 3)(x + ) = x x +. To double check he now sets x = which leads to What went wrong? = 3 + 9???? Evaluate the following integrals: x 4 x + x 3 x 4 + x 5 x x 5 x 4 x 3 x x + x 3x + x + x 3x +

26 6 I. METHODS OF INTEGRATION e 3x e 4x e x + e x e x e x + e x + + e x x(x + ) x(x + ) x (x ) (x )(x )(x 3) x + (x )(x )(x 3) x 3 + (x )(x )(x 3) 4. (a) Compute positive number. where h is a x(x h) (b) What happens to your answer to (a) when h? (c) Compute x.. Substitutions for integrals containing the expression ax + bx + c e main method for finding antiderivatives that we saw in Math is the method of substitution. is method will only let us compute an integral if we happen to guess the right substitution, and guessing the right substitution is o en not easy. If the integral contains the square root of a linear or quadratic function, then there are a number of substitutions that are known to help. Integrals with ax + b: substitute ax + b = u with u >. See.. Integrals with ax + bx + c: first complete the square to reduce the integral to one containing one of the following three forms u, u, u +. en, depending on which of these three cases presents itself, you choose an appropriate substitution. ere are several options: a trigonometric substitution; this works well in some cases, but o en you end up with an integral containing trigonometric functions that is still not easy (see. and.4.). use hyperbolic functions; the hyperbolic sine and hyperbolic cosine sometimes let you handle cases where trig substitutions do not help. ( a rational substitution (see ) using the two functions U(t) = ) t+t and V (t) = ( ) t t... Integrals involving ax + b. If an integral contains the square root of a linear function, i.e. ax + b then you can remove this square root by substituting u = ax + b.... Example. To compute I = x x + 3 we substitute u = x + 3. en x = (u 3) so that = u du,

27 . SUBSTITUTIONS FOR INTEGRALS CONTAINING THE EXPRESSION ax + bx + c 7 and hence I = (u 3) }{{}}{{} u u }{{} du x x+3 (u 4 3u ) du = = { 5 u5 u 3} + C. To write the antiderivative in terms of the original variable you substitute u = x + 3 again, which leads to I = (x + 3)5/ (x + 3)3/ + C. A comment: se ing u = ax + b is usually the best choice, but sometimes other choices also work. You, the reader, might want to try this same example substituting v = x + 3 instead of the substitution we used above. You should of course get the same answer.... Another example. Compute I = + + x. Again we substitute u = x +, or, u = x +. We get I = + u = x + so u du = + x u du = + u ( ) = du + u = u ln( + u) + C = x + ln ( + x + ) + C. A rational function: we know what to do. Note that u = x + is positive, so that + x + >, and so that we do not need absolute value signs in ln( + u)... Integrals containing x. If an integral contains the expression x then this expression can be removed at the expense of introducing trigonometric functions. Sometimes (but not always) the resulting integral is easier. e substitution that removes the x is x = sin θ.... Example. To compute note that I = so we have an integral involving x. ( x ) 3/ ( x ) = 3/ ( x ) x,

28 8 I. METHODS OF INTEGRATION θ x x We set x = sin θ, and thus = cos θ dθ. We get cos θ dθ I = ( sin θ). 3/ Use sin θ = cos θ and you get ( sin θ) 3/ = ( cos θ ) 3/ = cos θ 3. We were forced to include the absolute values here because of the possibility that cos θ might be negative. However it turns out that cos θ > in our situation since, in the original integral I the variable x must lie between and +: hence, if we set x = sin θ, then we may assume that π < θ < π. For those θ one has cos θ >, and therefore we can write ( sin θ) 3/ = cos 3 θ. A er substitution our integral thus becomes cos θ dθ I = cos 3 θ = dθ cos = tan θ + C. θ To express the antiderivative in terms of the original variable we use x = sin θ x x = cos θ. x = sin θ = tan θ =. x e final result is I = ( x ) = x + C. 3/ x... Example: sometimes you don t have to do a trig substitution. e following integral is very similar to the one from the previous example: x Ĩ = ( ). x 3/ e only difference is an extra x in the numerator. To compute this integral you can substitute u = x, in which case du = x. us we find x ( ) = x 3/ du u = 3/ u / u 3/ du = ( /) + C = + C u = + C. x.3. Integrals containing a x. If an integral contains the expression a x for some positive number a, then this can be removed by substituting either x = a sin θ or x = a cos θ. Since in the integral we must have a < x < a, we only need values of θ in the interval ( π, π ). us we substitute x = a sin θ, π < θ < π. For these values of θ we have cos θ >, and hence a x = a cos θ.

29 . SUBSTITUTIONS FOR INTEGRALS CONTAINING THE EXPRESSION ax + bx + c Example. To find J = 9 x we substitute x = 3 sin θ. Since θ ranges between π and + π we have cos θ > and thus 9 x = 9 9 sin θ = 3 sin θ = 3 cos θ = 3 cos θ = 3 cos θ. We also have = 3 cos θ dθ, which then leads to J = 3 cos θ 3 cos θ dθ = 9 cos θ dθ. is example shows that the integral we get a er a trigonometric substitution is not always easy and may still require more tricks to be computed. For this particular integral we use the double angle trick. Just as in 3 we find J = 9 cos θ dθ = 9 θ + ( sin θ) + C. 3 θ 9 x x x = 3 sin θ 9 x = 3 cos θ. e last step is to undo the substitution x = 3 sin θ. ere are several strategies: one approach is to get rid of the double angles again and write all trigonometric expressions in terms of sin θ and cos θ. Since θ ranges between π and + π we have x = 3 sin θ θ = arcsin x 3, To substitute θ = arcsin( ) in sin θ we need a double angle formula, sin θ = sin θ cos θ = x 9 x 3 = 3 9 x 9 x. We get 9 x = 9 θ + 9 sin θ cos θ + C. = 9 arcsin x 3 + x 9 x + C..4. Integrals containing x a or a + x. ere are trigonometric substitutions that will remove either x a or a + x from an integral. In both cases they come from the identities (9) ( ) = tan θ + or cos θ ( ) = tan θ. cos θ You can remember these identities either by drawing a right triangle with angle θ and with base of length, or else by dividing both sides of the equations a cos θ θ a tan θ a x = a tan θ a + x = a/ cos θ y = a/ cos θ y a = a tan θ = sin θ + cos θ or cos θ = sin θ by cos θ.

30 3 I. METHODS OF INTEGRATION.4.. Example turn the integral 4 x 4 into a trigonometric integral. Since x 4 = x we substitute x = cos θ, which then leads to x 4 = 4 tan θ = tan θ. In this last step we have to be careful with the sign of the square root: since < x < 4 in our integral, we can assume that < θ < π and thus that tan θ >. erefore tan θ = tan θ instead of tan θ. e substitution x = cos θ also implies that = sin θ cos θ dθ. We finally also consider the integration bounds: and erefore we have 4 π/3 x 4 = x = = = = cos θ = = θ =, cos θ x = 4 = cos θ = 4 = cos θ = = θ = π 3. tan θ sin θ π/3 cos θ dθ = 4 sin θ cos 3 θ dθ. is integral is still not easy: it can be done by integration by parts, and you have to know the antiderivative of / cos θ.. Rational substitution for integrals containing x a or a + x.. e functions U(t) and V (t). Instead of using a trigonometric substitution one can also use the following identity to get rid of either x a or x + a. e y = x 4 Area =? 4 Figure. What is the area of the shaded region under the hyperbola? We first try to compute it using a trigonometric substitution (.4.), and then using a rational substitution involving the U and V functions (..). The answer turns out to be 4 3 ln ( + 3 ).

31 . RATIONAL SUBSTITUTION FOR INTEGRALS CONTAINING x a OR a + x 3 U(t) = ( t + ) t t/ V (t) = ( t ) t t U = ( t + ) t V = ( t ) t t = U + V t = U V U V = Figure. The functions U(t) and V (t) identity is a relation between two functions U and V of a new variable t defined by ( () U(t) = t + ) (, V (t) = t t ). t ese satisfy () U = V +, which one can verify by direct substitution of the definitions () of U(t) and V (t). To undo the substitution it is useful to note that if U and V are given by (), then () t = U + V, t = U V.... Example.4. again. Here we compute the integral A = 4 x 4 using the rational substitution (). Since the integral contains the expression x 4 = x we substitute x = U(t). Using U = + V we then have x 4 = 4U(t) 4 = U(t) = V (t). When we use the substitution x = au(t) we should always assume that t. Under that assumption we have V (t) (see Figure ) and therefore x 4 = V (t). To

32 3 I. METHODS OF INTEGRATION summarize, we have (3) x = U(t), x 4 = V (t). We can now do the indefinite integral: x 4 = V (t) }{{} x 4 = = U (t) dt }{{} ( t t ) ( t t + t 3 ) dt = t ln t t + C ( ) t dt To finish the computation we still have to convert back to the original x variable, and substitute the integration bounds. e most straightforward approach is to substitute t = U + V, and then remember the relations (3) between U, V, and x. Using these relations the middle term in the integral we just found becomes { x ( ln t = ln(u + V ) = ln + x) }. We can save ourselves some work by taking the other two terms together and factoring them as follows (4) t t = ( t ( ) ) t = ( t + )( t ) t t = x ( x = x x 4. ) a b = (a + b)(a b) t + t = x ( ) t t = ( x ) So we find x 4 = x x ( x 4 ln{ + x) } + C. Hence, substituting the integration bounds x = and x = 4, we get 4 A = x 4 [ x x ( = x 4 ln{ + x) }] x=4 x= = 4 ( ) 6 4 ln + 3 = 4 3 ln ( + 3 ). ( the terms with x = vanish )

33 . SIMPLIFYING ax + bx + c BY COMPLETING THE SQUARE An example with + x. ere are several ways to compute I = + x and unfortunately none of them are very simple. e simplest solution is to avoid finding the integral and look it up in a table, such as Table. But how were the integrals in that table found? One approach is to use the same pair of functions U(t) and V (t) from (). Since U = + V the substitution x = V (t) allows us to take the square root of + x, namely, x = V (t) = + x = U(t). Also, = V (t)dt = ( + t ) dt, and thus we have I = + x }{{} =U(t) }{{} dv (t) ( )( ) = t + + dt t t = (t + 4 t + ) dt t 3 = {t 4 + ln t } + C t = ( t ) + 8 t ln t + C. At this point we have done the integral, but we should still rewrite the result in terms of the original variable x. We could use the same algebra as in (4), but this is not the only possible approach. Instead we could also use the relations (), i.e. t = U + V and t = U V ese imply and conclude t =(U + V ) =U + UV + V t =(U V ) =U UV + V t t = = 4UV I = + x = 8 ( t t ) + ln t + C = UV + ln(u + V ) + C = x + x + ln( x + + x ) + C.. Simplifying ax + bx + c by completing the square Any integral involving an expression of the form ax + bx + c can be reduced by means of a substitution to one containing one of the three forms u, u, or u +. We can achieve this reduction by completing the square of the quadratic expression under the square root. Once the more complicated square root ax + bx + c

34 34 I. METHODS OF INTEGRATION has been simplified to ±u ±, we can use either a trigonometric substitution, or the rational substitution from the previous section. In some cases the end result is one of the integrals listed in Table : du u = arcsin u u du = u u + arcsin u du + u = ln( u + + u ) + u du = u + u + ln( u + + u ) du u = ln( u + u ) u du = u u ln( u + u ) (all integrals +C ) Table. Useful integrals. Except for the first one these should not be memorized. Here are three examples. e problems have more examples... Example. Compute I = 6x x. Notice that since this integral contains a square root the variable x may not be allowed to have all values. In fact, the quantity 6x x = x(6 x) under the square root has to be positive so x must lie between x = and x = 6. We now complete the square: 6x x = ( x 6x ) = ( x 6x ) = [ (x 3) 9 ] At this point we decide to substitute [ (x 3) = 9 9 [( x 3 = 9 3 u = x 3, 3 ] ) ]. which leads to 6x x = 9 ( u ) = 9 ( u ) = 3 u, x = 3u + 3, = 3 du. Applying this change of variable to the integral we get = 6x x 3du 3 u = du x 3 = arcsin u + C = arcsin + C. u 3

35 3. PROBLEMS 35.. Example. Compute I = 4x + 8x + 8. We again complete the square in the quadratic expression under the square root: 4x + 8x + 8 = 4 ( x + x + ) = 4 { (x + ) + }. us we substitute u = x +, which implies du =, a er which we find 4x I = + 8x + 8 = u (x + ) + = + du. is last integral is in table, so we have I = u u + + ln ( u + u + ) + C = (x + ) { (x + ) + + ln x + + } (x + ) + + C..3. Example. Compute: We first complete the square I = x 4x 5. x 4x 5 = x 4x = (x ) 9 u a form { ( x ) } = 9 3 u form is prompts us to substitute u = x, 3 du = 3, i.e. = 3 du. We get { ( x ) } I = 9 = 3 3 u u 3 du = 9 du. Using the integrals in Table and then undoing the substitution we find I = x 4x 5 = 9 u u 9 ln( u + u ) + C = 9 x ( x ) { 9 x ( 3 3 ln x ) } + + C 3 3 { = (x ) (x ) 9 9 ln 3 x + } (x ) 9 + C = (x ) x 4x ln{ x + x 4x + 5 } 9 ln 3 + C = (x ) x 4x ln{ x + x 4x + 5 } + C 3. Problems Evaluate these integrals:

36 36 I. METHODS OF INTEGRATION In any of these integrals, a is a positive constant... x 4 x 3. + x 4. x x x 4x 4 / / 3/ x 4 x x + x + a x x + x a + x 3x + 6x + 6 3x + 6x + 5 a 3 a x +, x + a. 4. Chapter summary ere are several methods for finding the antiderivative of a function. Each of these methods allow us to transform a given integral into another, hopefully simpler, integral. Here are the methods that were presented in this chapter, in the order in which they appeared: () Double angle formulas and other trig identities: some integrals can be simplified by using a trigonometric identity. is is not a general method, and only works for certain very specific integrals. Since these integrals do come up with some frequency it is worth knowing the double angle trick and its variations. () Integration by parts: a very general formula; repeated integration by parts is done using reduction formulas. (3) Partial Fraction Decomposition: a method that allows us to integrate any rational function. (4) Trigonometric and Rational Substitution: a specific group of substitutions that can be used to simplify integrals containing the expression ax + bx + c. 5. Mixed Integration Problems One of the challenges in integrating a function is to recognize which of the methods we know will be most useful so here is an unsorted list of integrals for practice. Evaluate these integrals:.. a a x sin x x cos x x x + x + 7 x x + x /3 /4 4 3 x x x x x x x 4 (x 36) / x 4 x 36 x 4 36 x

37 ... x + x 4 x x + 3 x 4 x (x 3) / 3. e x (x + cos(x)) 4. (e x + ln(x)) (x + 5) x + 5x x x + x + 7 x x + x + 7 x x + x + 7 x 9. x + x + 7 3x + x. x 3 x 4. x e x (x ) 3 4 (x ) 3 (x + ) 6 x 4x x + x + 3 x ln x 7. x ln(x + ) e 3 e e x ln x x(ln x) 3 3. arctan( x) 5. MIXED INTEGRATION PROBLEMS 37 hint: x + 5 = /u 3. x(cos x) 3. π + cos(6w) dw Hint: + cos α = sin α sin(x) 34. Find x(x )(x )(x 3) and (x 3 + ) x(x )(x )(x 3) 35. Compute x 3 + x + x + (Hint: to factor the denominator begin with +x+x +x 3 = (+x)+x (+x) =...) 36. [Group Problem] You don t always have to find the antiderivative to find a definite integral. This problem gives you two examples of how you can avoid finding the antiderivative. (a) To find I = π/ sin x sin x + cos x you use the substitution u = π/ x. The new integral you get must of course be equal to the integral I you started with, so if you add the old and new integrals you get I. If you actually do this you will see that the sum of the old and new integrals is very easy to compute. (b) Use your answer from (a) to compute x + x. (c) Use the same trick to find π/ sin x 37. The Astroid. Draw the curve whose equation is x 3 + y 3 = a 3,

38 38 I. METHODS OF INTEGRATION where a is a positive constant. The curve you get is called the Astroid. Compute the area bounded by this curve. 38. The Bow-Tie Graph. Draw the curve given by the equation y = x 4 x 6. Compute the area bounded by this curve. 39. The Fan-Tailed Fish. Draw the curve given by the equation ( ) y = x x. + x Find the area enclosed by the loop. (H : Rationalize the denominator of the integrand.) 4. Find the area of the region bounded by the curves x =, y =, y = x ln x 4. Find the volume of the solid of revolution obtained by rotating around the x axis the region bounded by the lines x = 5, x =, y =, and the curve x y = x How to find the integral of f(x) = cos x. Note that cos x = cos x cos x = cos x sin x, and apply the substitution s = sin x followed by a partial fraction decomposition to compute cos x. Calculus bloopers As you ll see, the following computations can t be right; but where did they go wrong? 43. Here is a failed computation of the area of this region: y y = x- Clearly the combined area of the two triangles should be. Now let s try to get this answer by integration. Consider x. Let f(x) = x so that { x if x f(x) = x if x < Define F (x) = { x x if x x x if x < Then since F is an antiderivative of f we have by the Fundamental Theorem of Calculus: x = f(x) = F () F () ( ) = =. ) ( But this integral cannot be zero, f(x) is positive except at one point. How can this be? 44. It turns out that the area enclosed by a circle is zero?! According to the ancient formula for the area of a disc, the area of the following half-disc is π/. y = (-x ) / - y We can also compute this area by means of an integral, namely Area = x Substitute u = x so: u = x, x = u = ( u), = ( )( u) ( ) du. Hence x = x x u ( )( u) ( ) du.

39 5. MIXED INTEGRATION PROBLEMS 39 Now take the definite integral from x = to x = and note that u = when x = and u = also when x =, so x = ( ) u ( u) ( ) du = The last being zero since ( anything ) =. But the integral on the le is equal to half the area of the unit disc. Therefore half a disc has zero area, and a whole disc should have twice as much area: still zero! How can this be?

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41 CHAPTER II Proper and Improper Integrals All the definite integrals that we have seen so far were of the form I = b a f(x), where a and b are finite numbers, and where the integrand (the function f(x)) is nice on the interval a x b, i.e. the function f(x) does not become infinite anywhere in the interval. ere are many situations where one would like to compute an integral that fails one of these conditions; i.e. integrals where a or b is not finite, or where the integrand f(x) becomes infinite somewhere in the interval a x b (usually at an endpoint). Such integrals are called improper integrals. If we think of integrals as areas of regions in the plane, then improper integrals usually refer to areas of infinitely large regions so that some care must be taken in interpreting them. e formal definition of the integral as a limit of Riemann sums cannot be used since it assumes both that the integration bounds a and b are finite, and that the integrand f(x) is bounded. Improper integrals have to be defined on a case by case basis. e next section shows the usual ways in which this is done.. Typical examples of improper integrals.. Integral on an unbounded interval. Consider the integral A = x 3. is integral has a new feature that we have not dealt with before, namely, one of the integration bounds is rather than a finite number. e interpretation of this integral is that it is the area of the region under the graph of y = /x 3, with < x <. y = x 3 A = x 3 Because the integral goes all the way to x = the region whose area it represents stretches infinitely far to the right. Could such an infinitely wide region still have a finite area? And if it is, can we compute it? To compute the integral I that has the in its integration bounds, we first replace the integral by one that is more familiar, namely A M = M 4 x 3,

42 4 II. PROPER AND IMPROPER INTEGRALS where M > is some finite number. is integral represents the area of a finite region, namely all points between the graph and the x-axis, and with x M. y = x 3 A M = M x 3 M We know how to compute this integral: A M = M [ x 3 = ] M x = M +. e area we find depends on M. e larger we choose M, the larger the region is and the larger the area should be. If we let M then the region under the graph between x = and x = M will expand and eventually fill up the whole region between graph and x-axis, and to the right of x =. us the area should be A = lim A M = M M lim M x 3 = [ lim M M + ] =. We conclude that the infinitely large region between the graph of y = /x 3 and the x-axis that lies to the right of the line x = has finite area, and that this area is exactly!.. Second example on an unbounded interval. e following integral is very similar to the one we just did: A = e integral represents the area of the region that lies to the right of the line x =, and is caught between the x-axis and the hyperbola y = /x. A M M x. A y=/x As in the previous example the region extends infinitely far to the right while at the same time becoming narrower and narrower. To see what its area is we again look at the truncated region that contains only those points between the graph and the x-axis, and for which x M. is area is A M = M x = [ ln x ] M = ln M ln = ln M. e area of the whole region with x < is the limit A = lim A M = lim ln M = +. M M So we see that the area under the hyperbola is not finite!

43 . TYPICAL EXAMPLES OF IMPROPER INTEGRALS An improper integral on a finite interval. In this third example we consider the integral I =. x e integration bounds for this integral are and so they are finite, but the integrand becomes infinite at one end of the integration interval: lim x x = +. e region whose area the integral I represents does not extend infinitely far to the le or the right, but in this example it extends infinitely far upward. To compute this area we again truncate the region by looking at all points with x a for some constant a <, and compute I a = a e integral I is then the limit of I a, i.e. x = lim a We see that the area is finite. = [ arcsin x ] a = arcsin a. x a x = lim a arcsin a = arcsin = π..4. A doubly improper integral. Let us try to compute I = + x. is example has a new feature, namely, both integration limits are infinite. To compute this integral we replace them by finite numbers, i.e. we compute + x = lim a = lim a = lim lim b lim b b a + x arctan(b) arctan(a) arctan b lim arctan a b a = π ( π ) = π. A different way of ge ing the same example is to replace and in the integral by a and a and then let a. e only difference with our previous approach is that we now use one variable (a) instead of two (a and b). e computation goes as follows: + x = lim a = lim a a a + x ( ) arctan(a) arctan( a) = π ( π ) = π. In this example we got the same answer using either approach. is is not always the case, as the next example shows. y = x a

44 44 II. PROPER AND IMPROPER INTEGRALS Another doubly improper integral. Suppose we try to compute the integral I = x. e shorter approach where we replace both ± by ±a would be x = lim a a a a x = lim a ( a) =. On the other hand, if we take the longer approach, where we replace and by two different constants, then we get this x = lim lim a b = lim a lim b = lim b b b a [ x ] lim a x At this point we see that both limits lim b b / = and lim a a / = do not exist. e result we therefore get is x =. Since is not a number we find that the improper integral does not exist. We conclude that for some improper integrals different ways of computing them can give different results. is means that we have to be more precise and specify which definition of the integral we use. e next section lists the definitions that are commonly used. a. x is the area above minus the area below the axis.. Summary: how to compute an improper integral.. How to compute an improper integral on an unbounded interval. By definition the improper integral is given by a f(x) (5) a f(x) = is is how the integral in. was computed. M lim f(x). M a.. How to compute an improper integral of an unbounded function. If the integration interval is bounded, but if the integrand becomes infinite at one of the endpoints, say at x = a, then we define (6) b a f(x) = lim s a b s f(x).

45 3. MORE EXAMPLES Doubly improper integrals. If the integration interval is the whole real line, i.e. if we need to compute the integral I = f(x), then we must replace both integration bound by finite numbers and then let those finite numbers go to ±. us we would first have to compute an antiderivative of f, F (x) = f(x). e Fundamental eorem of Calculus then implies I a,b = b a f(x) = F (b) F (a) and we set I = lim F (b) lim F (a). b a Note that according to this definition we have to compute both limits separately. e example in Section.5 shows that it really is necessary to do this, and that computing can give different results. In general, if an integral f(x) = lim F (a) F ( a) a b a f(x) is improper at both ends we must replace them by c, d with a < c < d < b and compute the limit For instance, b a lim f(x) = lim c a lim d b d d = lim lim x c d c 3. More examples c f(x). x Area under an exponential. Let a be some positive constant and consider the graph of y = e ax for x >. How much area is there between the graph and the x-axis with x >? (See Figure for the cases a = and a =.) e answer is given by the y = e x y = e x Figure. What is the area under the graph of y = e x? What fraction of the region under the graph of y = e x lies under the graph of y = e x?

46 46 II. PROPER AND IMPROPER INTEGRALS improper integral A = = lim M = lim M e ax M = a lim M = a. e ax [ a e ax] M ( ) e am a > so lim M e am = We see that the area under the graph is finite, and that it is given by /a. In particular the area under the graph of y = e x is exactly, while the area under the graph of y = e x is exactly half that (i.e. /). 3.. Improper integrals involving x p. e examples in. and. are special cases of the following integral I = x p, where p > is some constant. We already know that the integral is if p = 3 (.), and also that the integral is infinite (does not exist) if p = (.). We can compute it in the same way as before, (7) I = lim M = lim M M = p lim M x p [ p x p] M ( ) M p. assume p At this point we have to find out what happens to M p as M. is depends on the sign of the exponent p. If this exponent is positive, then M p is a positive power of M and therefore becomes infinite as M. On the other hand, if the exponent is negative then M p is a negative power of M so that it goes to zero as M. To summarize: If < p < then p > so that lim M M p = ; if p > then p <, and lim M p = M If we apply this to (7) then we find that if < p then and if p > then lim M =. M p x p =, x p = p.

47 4. PROBLEMS 47 Q Q R P R P O Figure. Geometric construction of the Versiera. The figure shows a circle of diameter that touches the x-axis exactly at the origin. For any point Q on the line y = we can define a new point P by intersecting the line segment OQ with the circle, and requiring P to have the same x-coordinate as Q and the same y-coordinate as R. The Versiera is the curve traced out by the point P as we lets the point Q slide along the line y =. See problem. 4. Problems Compute the following improper integrals and draw the region whose area they represent: / 3 7 ( + x). (x ) 3 3 x. / x. xe x. x x + x. x x 3 + x. x. x x. x + x. (suggestion: x = u ) ( x + x ).. The graph of the function y = /(+x ) is called the Versiera (see Figure ). Compute the area of the shaded region between the Versiera and the circle in Figure. 3. e x. How would you define this integral (one of the integration bounds is rather than + )? 4. (a) e x sin πx where a, b are positive constants. (You have done the integral before see Ch I, 7, Problem.) (b) As in the previous problems, draw the region whose area the integral from (a) represents. Note that the function in the integral for this problem is not positive everywhere. How does this affect your answer? Can you tell from your drawing if the integral is positive? 5. (A way to visualize that = ) x (a) Show that for any a > one has a a x = ln ; in particular, this integral is the same for all a >. (b) Compute the area under the graph of y = /x between x = and x = n is n ln by spli ing the region into: the part where x, the part where x 4, the part where 4 x 8,. the part where n x n.

48 48 II. PROPER AND IMPROPER INTEGRALS a f(x) g(x) a y = g(x) y = f(x) a Figure 3. Comparing improper integrals. Here f and g are positive functions that satisfy f(x) g(x) for all x a. If g(x) is finite, then so is f(x). Conversely, if a a f(x) is not finite, then g(x) cannot be finite either. a a y = g(x) y = f(x) a y = f(x) Figure 4. In this figure f(x) is again positive, but g(x) is bounded by f(x) g(x) f(x). The same conclusion still holds, namely, if f(x) exists, then so does g(x). a a (c) Explain how the answer to (b) implies that the integral /x does not exist. 6. The area under the graph of y = /x with x < is infinite. Compute the volume of the funnel-shaped solid you get by revolving this region around the x-axis. Is the volume of this funnel finite or infinite? 5. Estimating improper integrals Sometimes it is just not possible to compute an improper integral because we simply cannot find an antiderivative for the integrand. When this happens we can still try to estimate the integral by comparing it with easier integrals, and even if we cannot compute the integral we can still try to answer the question does the integral exist, i.e. Does lim M M a f(x) exist? In this section we will see a number of examples of improper integrals that much easier to estimate than to compute. roughout there are three principles that we shall use:

49 5. ESTIMATING IMPROPER INTEGRALS 49 Integral of a positive function. If the function f(x) satisfies f(x) for all x a then either the integral f(x) exists, or else it is infinite, by which we mean a lim M M a f(x) =. If the function f(x) is not always positive then the above limit can fail to exist by oscillating without ever becoming infinite. For an example see 5..; see also 5. for further explanation. Comparison with easier integrals. If y = f(x) and y = g(x) are functions defined for x a, and if g(x) f(x) for all x a, then g(x) f(x). In particular, a a a f(x) exists = g(x) does not exist = a a a g(x) exists, f(x) does not exist. Read 5. for more details and examples. Only the tail ma ers. If f(x) is a continuous function for all x a, then for any b a we have a f(x) exists Furthermore, for any b a we have (8) a f(x) = b a b f(x) + f(x) exists. b f(x). is is further explained in eorem 5.3 and the examples following it. 5.. Improper integrals of positive functions. Suppose that the function f(x) is defined for all x a, and that f(x) for all x. To see if the improper integral f(x) a exists we have to figure out what happens to I M = M a f(x) as M. Since we are assuming that f(x) the integral I M represents the area under the graph of f up to x = M. As we let M increase this region expands, and thus the integral I M increases. So, as M there are two possibilities: either I M remains finite and converges to a finite number, or I M becomes infinitely large. e following theorem summarizes this useful observation: 5... eorem. If f(x) for all x a then either the integral I = exists, (i.e. I is a finite number), or else I = a a f(x) f(x) =, i.e. lim M M a f(x) =.

50 5 II. PROPER AND IMPROPER INTEGRALS 5... Example - integral to infinity of the cosine. To illustrate what eorem 5.. says, let s consider an improper integral of a function that is not always positive. For instance, consider I = cos x. e function in the integral is f(x) = cos x, and this function is clearly not always positive. When we try to compute this integral we get M [ ] M I = lim cos x = lim sin x = lim sin M. M M x= M is limit does not exist as sin M oscillates up and down between and + as M. On the other hand, since sin M stays between and +, we cannot say that lim M M cos x = +. eorem 5.. tells us that if you integrate a positive function then this kind of oscillatory behavior cannot occur. 5.. Comparison eorem for Improper Integrals. Suppose f(x) and g(x) are functions that are defined for a x <, and suppose that g(x) f(x) for all x a, i.e. If the improper integral a and one has f(x) g(x) f(x) for all x a. f(x) exists then the improper integral g(x) also exists, a a g(x) f(x). is theorem is used in two ways: it can be used to verify that some improper integral exists without actually computing the integral, and it can also be used to estimate an improper integral Example. Consider the improper integral I = a + x 3. e function in the integral is a rational function so in principle we know how to compute the integral. It turns out the computation is not very easy. If we don t really need to know the exact value of the integral, but only want a rough estimate of the integral, then we could compare the integral with an easier integral. To decide which simpler integral we should use as comparison, we reason as follows. Since only the tail ma ers, we consider the integrand +x for large x. When x is very 3 large x 3 will be much larger than, so that we may guess that we can ignore the in the denominator + x 3 : (9) + x 3 x 3 as x. is suggests that we may be able to compare the given integral with the integral x 3. We know from our very first example in this chapter (.) that this last integral is finite (we found that it is ). erefore we can guess that our integral I also is finite.

51 5. ESTIMATING IMPROPER INTEGRALS 5 Now let s try to use the Comparison eorem 5. to get certainty by proving that the integral I does indeed exist. We want to show that the integral +x exists, so we 3 choose g(x) = + x 3, and thus f(x) = x 3. We can compare these functions as follows: ( ) divide both sides first it follows from: x 3 + x 3 for all x by x 3 and then by + x 3 that: + x 3 x 3 for x is tells us that + x 3 x 3 =. erefore we have found that the integral I does indeed exist, and that it is no more than. We can go beyond this and try to find a lower bound (instead of saying that I is no more than we try to say that it is at least as large as some other number.) Here is one way of doing that: + x 3 x 3 + x 3 for all x = + x 3 x 3 for all x = x 3 + x 3 for x ( ) divide both sides first by x 3 and then by + x 3 is implies that x 3 + x 3. e first integral here is half the integral we computed in., so we get In summary, we have found that Second example. Does the integral I = + x 3. + x 3. x x + exist? Since the function we are integrating is positive, we could also ask is the integral finite? As in the previous example we look at the behavior of the integrand as x. For large x we can assume that x is much larger than x, so that it would be reasonable to ignore the in the denominator x +. If we do that than we find that x x + x x = x.

52 5 II. PROPER AND IMPROPER INTEGRALS If this were correct, then we would end up comparing the integral I with the simpler integral x. We know this la er integral is not finite (it was our second example, see.) and therefore we guess that the integral I probably also is not finite. To give a sound argument we will use the Comparison eorem. Our goal is to show that the integral I = f(x), with f(x) = x + x is not finite. To do this we have to find a function g(x) such that g(x) is smaller than f(x) (so that f will be larger than g(x)), g(x) is easy to integrate, and the integral of g(x) is not finite. e first and last point together imply that I = f(x) g(x) =, which is what we are trying to show. To complete the reasoning we have to find the easy-to-integrate function g(x). Based on what we have done above our first guess would be g(x) = x, but this does not work, since x x + < x x = x. So with this choice of g(x) we get g(x) > f(x) instead of g(x) < f(x). One way to simplify f(x) and get a smaller function is to remember that by increasing the denominator in a fraction you decrease the fraction. us, for x > we have f(x) = x x + > So we let g(x) = x. en we find I = =. x x + x = x x + x x x = x e Tail eorem. If y = f(x) is a continuous function for x a, and if b > a then a f(x) exists if and only if b f(x) exists Moreover, if these integrals exist, then (8) holds: a f(x) = b a f(x) + b f(x). P. For any finite M one has M b M f(x) = f(x) + f(x) a a b e theorem follows by taking the limit for M on both sides of this equation.

53 5. ESTIMATING IMPROPER INTEGRALS 53 e following two examples show how one can use this fact Example. Does the integral I = x 3 + exist? e integrand here is the same as in 5.. where we found that x 3 + <, and in particular is finite. Since the function f(x) = /(x 3 + ) is continuous for x we may conclude that I = x 3 + = x x 3 + so that I is finite. If we want an upper estimate for the integral we can use the fact that we already know that the integral from to is not more than and estimate the integral from x = to x =. For x we have x 3 + = x 3 + is implies x 3 + =, and hence x 3 + = x x = e area under the bell curve. e bell curve which plays a central role in probability and statistics, is the graph of the function n(x) = ae x /b, where a and b are positive constants whose values will not be important in this example. In fact, to simplify this example we will choose them to be a = and b = so that we are dealing with the function n(x) = e x. e question now is, is the area under the bell curve finite, and can we estimate it? In terms of improper integrals, we want to know if the integral exists. A = e x A = e x y = e x y = e x Figure 5. Estimating the area under the bell curve.

54 54 II. PROPER AND IMPROPER INTEGRALS n(x) = n( x) so the bell curve is symmetric We write the integral as the sum of two integrals e x = e x + e x Since the function n(x) = e x is even these two integrals are equal, so if we can show that one of them exists, the other will also exist. When x is large x is much larger than x so that e x will be smaller than e x ; since e x is a function we know how to integrate it seems like a good idea to compare the bell curve with the graph of e x. To make the comparison (and to use the Comparison eorem) we first check if e x e x really is true. We find that this is true if x, namely: x = x x = x x = e x e x. We can therefore use e x to estimate the integral of e x for x : e x e x = lim [ e x] M ( = lim e M + e ) = e. M M Between x = and x = we have e x, so and therefore e x = e x e x + =, e x + e. Since the bell curve is symmetric, we get the same estimates for the integral over < x : e x + e. In the end we have shown that the area under the bell curve is finite, and that it is bounded by A = e x + e. With quite a bit more work, and using an indirect argument one can actually compute the integral (without finding an antiderivative of e x ). e true value turns out to be A = e x = π. For comparison, π and e 6. Problems Which of the following inequalities are true for all x > a (you get to choose a):. x > x? Answer: True. If x > then x > x and therefore <. So the inequality is true if x x x > a where we choose a = x < x? x x + x > x? x x3 > x /?

55 6. PROBLEMS x 3x + (x x) 3 < x 4? sin x Consider the function F (x) = x. (a) Compute lim F (x) and lim F (x). x x Which of the following inequalities are true for all x with < x < a (you get to choose a again, as long as a > ): 3 6. x < 3 x? Answer: False. If < x < then x < x and therefore >, and thus 3 > 3. x x x x So, more precisely, the inequality is false if < x < a where we choose a = x x + x < x? x x < x? x + x < x? 4 x + x < x? x x < 4 x? For each of the following improper integrals draw the graph of the integrand, and decide if the integral exists. If it does try to find an estimate for how large the integral is π u ( u + ) du u 3 ( u + ) du u 4 ( u + ) du sin x x sin(x ) x (what happens at x =?) 7. [Group Problem] (Fresnel integrals from optics.) For background on Fresnel integrals read the Wikipedia article on the subject. (b) Show that F (x) = cos x (c) Show that the integral sin x ex- x ists. (d) Show that cos x = sin x x. sin x x. (e) True or False: if for some function f the improper integral f(x) exists then it must be true that lim x f(x) =? 8. [Group Problem] Suppose p(x) = e ax, where a > is a constant. (a) Show that p(x) <. (hint: you can compute the integral.) (b) Compute xp(x) p(x) by integrating by parts in the integral in the numerator. (c) Compute x p(x) p(x) by integrating by parts twice in the integral in the numerator. 9. [Group Problem] Suppose p(x) = e x /σ where σ > is a constant. (σ is the Greek lower case s, pronounced as sigma. ) (a) Show that p(x) <. (hint: there is no antiderivative of e x /σ, so don t look for it. Instead follow example 5.3..) (b) Compute xp(x) p(x).

56 56 II. PROPER AND IMPROPER INTEGRALS Hint: there is an easy antiderivative for the integral in the numerator. Do that one first. (c) Compute x p(x) p(x) by integrating by parts in the integral in the numerator.. [Group Problem] The area under the bell curve is -factorial In this problem we look at a new function F (n). For any number n we would like to define F (n) = t n e t dt. The integral that defines F (n) is improper so it is not automatically clear that F (n) exists. (a) Show that the integral exists if n. (b) Compute F (), F (), and F (). (c) Use integration by parts to show that F (n + ) = (n + )F (n) Holds for any n. Use this to Compute F (). (d) If you set n = in the relation F (n + ) = (n + )F (n) then you get F () = F ( ) =. This should contradict the value for F () you found in (b). What is wrong? (e) Show that the integral for F ( ) exists, and use a substitution to show that F ( ) = e u du. How does this justify the statement at the beginning of the problem?

57 CHAPTER III First order differential Equations. What is a Differential Equation? A differential equation is an equation involving an unknown function and its derivatives. A general differential equation can contain second derivatives and higher derivatives of the unknown function, but in this course we will only consider differential equations that contain first order derivatives. So the most general differential equation we will look at is of the form dy () = f(x, y). Here f(x, y) is any expression that involves both quantities x and y. For instance, if f(x, y) = x + y then the differential equation represented by () is dy = x + y. In this equation x is a variable, while y is a function of x. Differential equations appear in many science and engineering problems, as we will see in the section on applications. But first let s think of a differential equation as a purely mathematical question: which functions y = y(x) satisfy the equation ()? It turns out that there is no general method that will always give us the answer to this question, but there are methods that work for certain special kinds of differential equations. To explain all this, this chapter is divided into the following parts: Some basic examples that give us a clue as to what the solution to a general differential equation will look like. Two special kinds of differential equation ( separable and linear ), and how to solve them. How to visualize the solution of a differential equation (using direction fields ) and how to compute the solution with a computer using Euler s method. Applications: a number of examples of how differential equations come up and what their solutions mean.. Two basic examples.. Equations where the RHS does not contain y. Which functions y = y(x) satisfy dy () = sin x? is is a differential equation of the form () where the function f that describes the Right Hand Side is given by f(x, y) = sin x. In this example the function f does not depend on the unknown function y. Because of this the differential equation really asks 57

58 58 III. FIRST ORDER DIFFERENTIAL EQUATIONS which functions of x have sin x as derivative? In other words, which functions are the antiderivative of sin x? We know the answer, namely y = sin x = cos x + C where C is an arbitrary constant. is is the solution to the differential equation (). is example shows us that there is not just one solution, but that there are many solutions. e expression that describes all solutions to the differential equation () is called the general solution. It contains an unknown constant C that is allowed to have arbitrary values. To give meaning to the constant C we can observe that when x = we have So the constant C is nothing but y() = cos + C = + C. C = y() +. For instance, the solution of () that also satisfies y() = 4 has C = 4 + = 5, and thus is given by y(x) = cos x + 5. We have found that there are many solutions to the differential equation () (because of the undetermined constant C), but as soon as we prescribe the value of the solution for one value of x, such as x =, then there is exactly one solution (because we can compute the constant C.).. e exponential growth example. Which functions equal their own derivative, i.e. which functions satisfy dy = y? Everyone knows at least one example, namely y = e x. But there are more solutions: the function y = also is its own derivative. From the section on exponential growth in math we know all solutions to dy = y. ey are given by y(x) = Ce x, where C can be an arbitrary number. If we know the solution y(x) for some value of x, such as x =, then we can find C by se ing x = : y() = C. Again we see that instead of there being one solution, the general solution contains an arbitrary constant C..3. Summary. e two examples that we have just seen show us that for certain differential equations there are many solutions, the formula for the general solution contains an undetermined constant C, the undetermined constant C becomes determined once we specify the value of the solution y at one particular value of x. It turns out that these features are found in almost all differential equations of the form (). In the next two sections we will see methods for computing the general solution to two frequently occurring kinds of differential equation, the separable equations, and the linear equations.

59 3. FIRST ORDER SEPARABLE EQUATIONS First Order Separable Equations By definition a separable differential equation is a diffeq of the form () y dy (x) = F (x)g(y(x)), or = F (x)g(y). us the function f(x, y) on the right hand side in () has the special form For example, the differential equation f(x, y) = F (x) G(y). dy = sin(x)( + y ) is separable, and one has F (x) = sin x and G(y) = + y. On the other hand, the differential equation dy = x + y is not separable. 3.. Solution method for separable equations. To solve this equation divide by G(y(x)) to get (3) dy G(y(x)) = F (x). Next find a function H(y) whose derivative with respect to y is (4) H (y) = ( ) dy solution: H(y) = G(y) G(y). en the chain rule implies that the le hand side in (3) can be wri en as dy G(y(x)) = H (y(x)) dy = dh(y(x)). us (3) is equivalent with dh(y(x)) = F (x). In words: H(y(x)) is an antiderivative of F (x), which means we can find H(y(x)) by integrating F (x): (5) H(y(x)) = F (x) + C. Once we have found the integral of F (x) this gives us y(x) in implicit form: the equation (5) gives us y(x) as an implicit function of x. To get y(x) itself we must solve the equation (5) for y(x). A quick way of organizing the calculation goes like this: To solve dy = F (x)g(y) we first separate the variables, dy = F (x), G(y) and then integrate, dy G(y) = F (x).

60 6 III. FIRST ORDER DIFFERENTIAL EQUATIONS e result is an implicit equation for the solution y with one undetermined integration constant. Determining the constant. e solution we get from the above procedure contains an arbitrary constant C. If the value of the solution is specified at some given x, i.e. if y(x ) is known then we can express C in terms of y(x ) by using (5). 3.. A snag: We have to divide by G(y) which is problematic when G(y) =. is has as consequence that in addition to the solutions we found with the above procedure, there are at least a few more solutions: the zeroes of G(y) (see Example 3.4 below). In addition to the zeroes of G(y) there sometimes can be more solutions, as we will see in Example. on Leaky Bucket Dating Example. We solve dz dt = ( + z ) cos t. Separate variables and integrate dz + z = cos t dt, to get arctan z = sin t + C. Finally solve for z and we find the general solution z(t) = tan ( sin(t) + C ) Example: the snag in action. If we apply the method to y (x) = y, we get y(x) = e x+c. No ma er how we choose C we never get the function y(x) =, even though y(x) = satisfies the equation. is is because here G(y) = y, and G(y) vanishes for y =. 4. Problems For each of the following differential equations - find the general solution, - indicate which, if any, solutions were lost while separating variables, - find the solution that satisfies the indicated initial values dy = xy, y() =. dy + x cos y =, y() = π 3. dy + + x =, y() = A. + y 4. y dy + x3 =, y() = A dy + y =, y() = A. dy + + y =, y() = A. dy + x =, y() =. y 8. [Group Problem] Let P (t) be the size of a colony of bacteria in a Petri dish in some experiment. Assume that the size of the colony changes according to the so-called logistic equation: dp dt = P ( P ), 5

61 5. FIRST ORDER LINEAR EQUATIONS 6 Assume also that in the beginning of the experiment the population size is P =. (a) Find the general solution to the differential equation. (b) Find the solution that satisfies the given initial conditions. (c) How long does it take the population to reach size P = 5? (d) How long does it take the population to reach size P = 9? (e) What value does P have when dp is dt the largest (hint: you do not need to solve the differential equation this question has a very short answer.) (6) 5. First Order Linear Equations Differential equations of the form equation are called first order linear. dy + a(x)y = k(x) 5.. e Integrating Factor. Linear equations can always be solved by multiplying both sides of the equation with a specially chosen function called the integrating factor. It is defined by (7) A(x) = a(x), m(x) = e A(x). Here m(x) is the integrating factor. It looks like we just pulled this definition of A(x) and m(x) out of a hat. e example in 5. shows another way of finding the integrating factor, but for now let s go on with these two functions. Multiply the equation (6) by the integrating factor m(x) to get m(x) dy + a(x)m(x)y = m(x)k(x). By the chain rule the integrating factor satisfies erefore one has dm(x) = d ea(x) dm(x)y = A (x) }{{} =a(x) e A(x) = a(x)m(x). }{{} =m(x) = m(x) dy + a(x)m(x)y { dy } = m(x) + a(x)y = m(x)k(x). Integrating and then dividing by the integrating factor gives the solution y = ( ) m(x)k(x) + C. m(x) In this derivation we have to divide by m(x), but since m(x) = e A(x) and since exponentials never vanish we know that m(x), so we can always divide by m(x).

62 6 III. FIRST ORDER DIFFERENTIAL EQUATIONS 5.. An example. Find the general solution to the differential equation en find the solution that satisfies dy = y + x. (8) y() =. Solution. We first write the equation in the standard linear form dy (9) y = x, and then multiply with the integrating factor m(x). We could of course memorize the formulas (7) that lead to the integrating factor, but a safer approach is to remember the following procedure, which will always give us the integrating factor. Assuming that m(x) is as yet unknown we multiply the differential equation (9) with m, (3) m(x) dy m(x)y = m(x)x. If m(x) is such that (3) m(x) = dm(x), then equation (3) implies m(x) dy + dm(x) y = m(x)x. e expression on the le is exactly what comes out of the product rule this is the point of multiplying with m(x) and then insisting on (3). So, if m(x) satisfies (3), then the differential equation for y is equivalent with dm(x)y = m(x)x. We can integrate this equation, m(x)y = m(x)x, and thus find the solution (3) y(x) = m(x) m(x)x. All we have to do is find the integrating factor m. is factor can be any function that satisfies (3). Equation (3) is a differential equation for m, but it is separable, and we can easily solve it: dm = m dm = ln m = x + C. m Since we only need one integrating factor m we are not interested in finding all solutions of (3), and therefore we can choose the constant C. e simplest choice is C =, which leads to ln m = x m = e x m = ±e x. Again, we only need one integrating factor, so we may choose the ± sign: the simplest choice for m here is m(x) = e x.

63 6. PROBLEMS 63 With this choice of integrating factor we can now complete the calculation that led to (3). e solution to the differential equation is y(x) = m(x)x m(x) = e x e x x (integrate by parts) = e x{ } e x x e x + C = x + Ce x. is is the general solution. To find the solution that satisfies not just the differential equation, but also the initial condition (8), i.e. y() =, we compute y() for the general solution, y() = + Ce = 3 + Ce. e requirement y() = then tells us that C = 3e. e solution of the differential equation that satisfies the prescribed initial condition is therefore y(x) = x + 3e x. 6. Problems. In example 5. we needed a function m(x) that satisfies (3). The function m(x) = satisfies that equation. Why did we not choose m(x) =?. Why can t we simplify the computation at the end in example 5. by canceling the two factors m(x) as follows: y(x) = m(x) = x = x + C? m(x) x For each of the following differential equations - specify the differential equation that the integrating factor satisfies, - find one integrating factor, - find the general solution, - find the solution that satisfies the specified initial conditions. In these problems K and N are constants. 3. dy = y + x, y() =. 4. dy = y + x, y() = dy + y + ex =. dy (cos x)y = esin x, y() = A. dy = y + e x, y() =. dy = y tan x +, y() =. dy = y tan x +, y() =.. cos x dy = N y y() =.. x dy = y + x, y() = dy = xy + x3, y() =. dy = y + sin x, y() =. dy = Ky + sin x, y() =. dy + x y =, y() = 5. dy + ( + 3x )y =, y() =.

64 64 III. FIRST ORDER DIFFERENTIAL EQUATIONS 7. Direction Fields We can visualize a differential equation by drawing the corresponding direction field. Consider a differential equation dy = f(x, y) where f is some given function. e differential equation tells us that if we know a point (x, y ) on the graph of a solution y = y(x) of the differential equation, then we also know the slope of the graph at that point. We can draw the tangent line to the graph of the solution: Slope m = f(x, y ) (x, y ) A solution Tangent line to the solution at (x, y ) If we have not yet solved the differential equation then we don t know any points on the solution. In this case we can sample the xy-plane, compute f(x, y) at all our sample points, and draw the tangent lines a solution would have if it passed through one of the sample points. In Figure this is done for the differential equation dy = y + sin πx. e direction field on the le in Figure gives us an idea of what the solutions should look like. Whenever a solution passes through one of the sample points the slope of its graph is given by the line segment drawn at that sample point. It is clear from a quick look at Figure that drawing a direction field involves computing f(x, y) (i.e. y + sin πx in our example) for many, many points (x, y). is kind of repetitive computation is be er Figure. Direction field for dy = y + sin πx on the le, and the same direction field with solutions to the differential equation.

65 8. EULER S METHOD 65 done by a computer, and an internet search quickly leads to a number of websites that produce direction fields for our favorite differential equation. In particular the ODE page at the Virtual Math Museum of UC Irvine, draws both direction fields and approximations to solutions found using Euler s method, to which we now turn. 8. Euler s method 8.. e idea behind the method. Consider again the differential equation (), dy = f(x, y). Suppose we know one point on a solution, i.e. suppose we know that a given solution to this equation satisfies y = y when x = x, i.e. (33) y(x ) = y. e differential equation then tells us what the derivative of the solution is at x = x, namely, y (x ) = f(x, y ). e definition a derivative says that so that we have y (x ) = lim h y(x + h) y(x ) h y(x + h) y(x ) (34) lim = f(x, y ). h h Keep in mind that the right hand side is what we get by substituting the x and y values that we know for the solution in f. So if we know x and y then we can also compute f(x, y ). If we don t know the solution then we cannot compute the le hand side in (34), but, following Euler, we can make an approximation. If instead of le ing h we choose a small number h >, then we may assume that y(x + h) y(x ) (35) f(x, y ). h Here means approximately equal, which is a vaguely defined concept. It means that the difference between the two quantities in (35) is small and we will not worry too much about the error in (35) (such issues are addressed in more advanced courses on Numerical Analysis; e.g. Math 54 at UW Madison). In the approximate equation (35) all quantities are known except y(x + h). A er solving (35) for y(x + h) we find (36) y(x + h) y + hf(x, y ). Euler s idea (see Figure ) was to forget that this is an approximation, and declare that we now know a new point on the graph of the solution, namely (37) x = x + h, y = y + hf(x, y ). Assuming that our solution satisfies y(x ) = y exactly (rather than just approximately), we can apply the same procedure and get another new point on the graph of our solution: x = x + h, y = y + hf(x, y ).

66 66 III. FIRST ORDER DIFFERENTIAL EQUATIONS y tangent y solution? x x = x + h Figure. Approximating a solution with Euler s method. On the le : one step of Euler s method. Knowing one point (x, y ) on the graph does not immediately tell us what the solution is, but it does tell us what the tangent to the graph at (x, y ) is. We can then guess that the point on the tangent with x = x + h and y = y + f(x, y )h almost lies on the graph. On the right: repeating the procedure gives us a sequence of points that approximate the solution. Reducing the step size h should give be er approximations. By repeating this procedure we can generate a whole list of points (x, y ), (x, y ), (x, y ), (x 3, y 3 ), etc that lie approximately on the graph of the solution passing through the given initial point (x, y ). 8.. Setting up the computation. If we are given x start and y(x start ) = y start, and if we want to find y(x end ) for some x end > x start, then we can decide to approximate y(x end ) by applying n steps of Euler s method. Since each step advances x by an amount h, application of Euler s method will advance x by nh a er n steps. us we want nh = x end x start, i.e. h = x end x start. n Starting with the known value of y at x start we then apply Euler s method n times by filling out a table of the following form: x k y k m k = f(x k, y k ) y k+ = y k + m k h x = x start y start m y x = x + h y m y x = x + h y m y 3 x 3 = x + 3h y 3 m 3 y 4.. x n = x + (n )h y n m n y n x n = x end y n e procedure is very mechanical and repetitive, and is best programmed on a computer.

67 . APPLICATIONS OF DIFFERENTIAL EQUATIONS 67 Once the table has been computed the values in the first two columns can be used to graph the approximation to the real solution. 9. Problems. Let y(t) be the solution of dy dt = t y, y(.) =.. We want to compute y(3.). (a) Find an exact formula for y(3.) by solving the equation (the equation is both separable and linear, so we have at least two methods for finding the solution.) (b) If we use Euler s method with step size h =., then how many steps do we have to take to approximate y(3.)? Compute the approximation with h =.. (c) Find the approximations if h =., h = /3, and h =.5. Organize the computations following the table in 8.. (d) Compare the results from the computations in (b) and (c) with the true value of y(3.) from part (a).. The function y(x) = e x is the solution to dy = y, y() =. (a) Approximate e = y() by using Euler s method first with step size h =, then with h = /, and then with h = /3. What are the approximations you find for y() in each case? (b) Look for a pa ern in the answers to (a). (c) Find a formula for the result of applying Euler s method n times with step size h = n. 3. Use Euler s method to approximate the solution to dy = y, y() =, and, in particular to find y(). Use various step sizes. How small do you have to make the step size before the answer seems reasonable? Explain.. Applications of Differential Equations Differential equations are very o en used to describe how some object or system changes or evolves in time. If the object or system is simple enough then its state is completely determined by one number (say y) that changes with time t. A differential equation for the system tells us how the system changes in time, by specifying the rate of change of the state y(t) of the system. is rate of change can depend on time, and it can depend on the state of the system, i.e. on y(t). is dependence can be expressed as an equation of the form (38) dy dt = f(y, t). e function f describes the evolutionary law of our system (synonyms: evolutionary law, dynamical law, evolution equation for y )... Example: carbon dating. Suppose we have a fossil, and we want to know how old it is. All living things contain carbon, which naturally occurs in two isotopes, C 4 (unstable) and C (stable). A long as the living thing is alive it eats & breaths, and its ratio of C to C 4 is kept constant. Once the thing dies the isotope C 4 decays into C at a steady rate that is proportional to the amount of C 4 it contains.

68 68 III. FIRST ORDER DIFFERENTIAL EQUATIONS Let y(t) be the ratio of C 4 to C at time t. e law of radioactive decay says that there is a constant k > such that dy(t) = ky(t). dt Solve this differential equation (it is both separable and first order linear: either method works) to find the general solution y(t; C) = Ce kt. A er some lab work it is found that the current C 4 /C ratio of our fossil is y now. us we have y now = Ce ktnow = C = y now e ktnow. erefore our fossil s C 4 /C ratio at any other time t is/was y(t) = y now e k(t now t). is allows us to compute the time at which the fossil died. At this time the C 4 /C ratio must have been the common value in all living things, which can be measured let s call it y life. en at the time t demise when our fossil became a fossil we would have had y(t demise ) = y life. Hence the age of the fossil would be given by y life = y(t demise ) = y now e k(t now t demise ) = t now t demise = k ln x life x now.. Example: dating a leaky bu et. A bucket is filled with water. ere is a hole in the bo om of the bucket so the water streams out at a certain rate. h(t) the height of water in the bucket A area of cross section of bucket a area of hole in the bucket v velocity with which water goes through the hole. area = A v h(t) We have the following facts to work with: e amount (volume) of water in the bucket is A h(t); e rate at which water is leaving the bucket is a v(t); Hence dah(t) = av(t). dt In fluid mechanics it is shown that the velocity of the water as it passes through the hole only depends on the height h(t) of the water, and that, for some constant K, v(t) = Kh(t). e last two equations together give a differential equation for h(t), namely, dh(t) dt = a A Kh(t).

69 . APPLICATIONS OF DIFFERENTIAL EQUATIONS 69 h(t) t Figure 3. Several solutions h(t; C) of the Leaking Bucket Equation (39). Note how they all have the same values when t. To make things a bit easier we assume that the constants are such that a A K =. en h(t) satisfies (39) h (t) = h(t). is equation is separable, and when we solve it we get dh h = = h(t) = t + C. is formula can t be valid for all values of t, for if we take t > C, the RHS becomes negative and can t be equal to the square root in the LHS. But when t C we do get a solution, h(t; C) = (C t). is solution describes a bucket that is losing water until at time C it is empty. Motivated by the physical interpretation of our solution it is natural to assume that the bucket stays empty when t > C, so that the solution with integration constant C is given by { (C t) when t C (4) h(t) = for t > C. e snag appears again. (See 3. and 3.4.) Note that we had to divide by h to find the solution. is is not allowed when h =. It turns out that h(t) = is a solution to the differential equation. e solution h(t) = satisfies h() =, and our experience with differential equations so far would lead us to believe that this should therefore be the only solution that satisfies h() =. However, every solution from (4) with C also satisfies h() =. is problem is therefore different from the differential equations we have dealt with up to now. Namely, prescribing the value y() = does not single out one solution of the differential equation (39)..3. Heat transfer. We all know that heat flows from warm to cold: if we put a cold spoon in a hot cup of coffee, then heat will flow from the coffee to the spoon. How fast does the spoon warm up?

70 7 III. FIRST ORDER DIFFERENTIAL EQUATIONS in out According to physics the rate of change of the spoon s temperature is proportional to the difference in temperature between the coffee and the spoon. So, if T c and T s are the temperature of the coffee and the spoon, respectively, then dt s (4) = K ( ) T s T c. dt Here K is a constant that depends on the shape and material of the spoon, how much of the spoon is in the coffee, etc., but not on the temperatures T s and T c. If we assume that the spoon is small, then whatever small amount of heat it extracts from the coffee will not change the coffee temperature. Under that assumption we may assume that T c is a constant and the differential equation (4) is both separable and linear so that we have two methods for solving it. If the coffee itself is also cooling or warming up then T c will depend on time and the equation (4) becomes dt s (4) dt + KT s = KT c (t). If we know T c (t) then this is still a linear differential equation for T s and we can solve it..4. Mixing problems. Consider a container containing water and vinegar. If water and vinegar are flowing in and out of the container, then the concentration of vinegar in the container will change with time according to some differential equation. Which differential equation describes the vinegar content of the container depends on the precise details of the set-up. As an example, let us assume that the container has fixed volume V = liters. is means that the amount of liquid that flows in during any time interval must equal the amount of liquid flowing out of the container (the liquid is incompressible, i.e. its density is fixed.) Suppose furthermore that a mixture of 5% vinegar and 95% water flows into the container at 4 liters per minute. And suppose also that the liquid in the container is thoroughly mixed, so that the liquid that flows out of the container has the same vinegar/water mixture as the entire container. Problem: Let D(t) be the fraction of the liquid in the container that is vinegar. How does D(t) change with time? Solution: Instead of tracking the concentration D(t) of vinegar we will look at the total amount of vinegar in the container. is amount is D(t)V. To find the differential equation we consider how much vinegar flows in and out of the container during a short time interval of length t (i.e. between time t and t + t): in: e liquid volume flowing into the tank during a time t is 4 t liters. Since the vinegar concentration of the in-flowing liquid is 5%, this means that 5% 4 t = t liters of vinegar enter the container. out: Since 4 t liters of liquid enter the container, the same amount of liquid must also leave the container. e liquid that leaves is the well-stirred mixture, and thus it contains D(t) 4 t liters vinegar. In total we see that the change in vinegar content during a time t is (43) ( vinegar content ) = t 4D(t) t. To find the change in concentration we divide by the volume of the container D = t 4D(t) t = t 5 D(t) 5 t.

71 . PROBLEMS 7 We find the rate of change by dividing by t and le ing t : dd (44) dt = 5 5 D. is equation is again both linear and separable so we have two methods for solving it.. Problems. Read Example. on Leaky bucket dating again. In that example we assumed that a A K =. (a) Solve diffeq for h(t) without assuming a A K =. Abbreviate C = a A K. (b) If in an experiment one found that the bucket empties in seconds a er being filled to height cm, then how much is the constant C?. A population of bacteria grows at a rate proportional to its size. Write and solve a differential equation that expresses this. If there are bacteria a er one hour and bacteria a er two hours, how many bacteria are there a er three hours? 3. Rabbits in Madison have a birth rate of 5% per year and a death rate (from old age) of % per year. Each year rabbits get run over and 7 rabbits move in from Sun Prairie. (a) Write a differential equation that describes Madison s rabbit population at time t. (b) If there were, rabbits in Madison in 99, how many are there in 994? 4. [Group Problem] Consider a cup of soup that is cooling by exchanging heat with the room. If the temperature of the soup is T s (t), and if the temperature of the room is T r, then Newton s cooling law claims that dt s = K(T s T r) dt for some constant K >. (a) What units does K have, if we measure temperature in degrees Fahrenheit, and time in seconds? (b) Find the general solution to Newton s cooling law. What units does the arbitrary constant in your solution have? What is the limit of the temperature as t? (c) The soup starts at 8 o F, and sits in a room whose temperature is 75 F. In five minutes its temperature has dropped to 5 F. Find the cooling constant K. When will its temperature be 9 F? 5. [Group Problem] A lake gets heated by day when the sun is out, and loses heat at night. It also exchanges heat with the Earth. The Earth s temperature is T e, and the lake s temperature is T l (t). These effects together can be modeled by a differential equation for T l (t) dt l dt = K(T l T e) + S sin(πt). Here the first term on the right represents Newton s cooling law, while the second term accounts for the heating and cooling by radiation: if sin(πt) > then it s day and the sun is heating the lake, if sin(πt) < then it s night and the lake is radiating heat into the cold sky. Time t is measured in days. Temperatures are in degrees Celsius. (a) Assuming K = S =, and T e = find the general solution to the differential equation. (b) Draw the direction field for the differential equation when K = S = and T e =. (First try to do this by hand. Then use a computer.) (c) Does lim t T l (t) exist? Consider the separate terms in the general solution you