Chapter V.G The Definite Integral and Area: Two Views

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1 Chapter V.G The Definite Integral and Area: Two Views In is section we will look at e application of e definite integral to e problem of finding e area of a region in e plane. In particular e regions in our discussion are bounded by eier lines or e graphs of functions. There are two ways to approach is problem as in many oer applications of e definite integral. We will consider e area problem in some detail because it is typical. Our first approach for finding an area will center on e approximation of e area as a sum of areas of rectangles at are determined by e given region. This follows our earlier treatment of e definite integral as a limit of sums: What's good for one is good for all and e more e merrier. These estimating sums will have a limiting value which by definition is a definite integral. A second contrasting approach to e same problem follows e eme at information about a rate of change (a differential equation) can help determine e value of a variable which might oerwise be considered constant. In is approach we'll consider e area as a function of a single variable, much like in e derivative form of e fundamental eorem of calculus. The key here is to estimate e rate at which e area is changing as a function of one variable. The area of a rectangle forms e basis for an estimate of e change in area of e region. 1. Area as a Limit of Sums Our initial work on e definition of e definite integral used an interpretation of e Euler sums as e area of rectangles which approximated e area of a region enclosed by e graph of a non-negative function, e X -axis, and lines X = a and X = b. The next example generalizes is situation. Keep in mind e definition of e definite integral, namely: This definition allows us to recognize e limit of estimates of is form as definite integrals. The advantage of seeing is form is at e definite integral in many cases can be evaluated exactly by using e Fundamental Theorem of Calculus. This connection will be more clear after you read e next example. Example V.G.1. In is example we will a) express e area of is region as e difference of two areas; b) estimate e area of e region in e first quadrant enclosed by e Y-axis, e graph of g(x)=x and f(x)=-x ; and c) express e area of is region as a definite integral, explaining how is integral is related to e difference in part a) and e estimate in part b). Discussion: a) First graph e region as described and note at e curves intersect in e first quadrant when x = -x, i.e., when x = 1, so e x values for is region are between 0 and 1. See Figure ***.The region can be considered as being obtained from e region benea e graph of Y = - X and above e X-axis between e Y - axis and e line X = 1, wi area A1, after cutting away e region benea e graph of Figure 1 Figure Chapter V.G 6/15/14 1

2 Y= X and above e X-axis between e Y-axis and e line X = 1, wi area A. The area can erefore be described as e difference of e areas of e two regions, i.e., A1-A. Each of ese areas can be computed from a definite integral, and. So e area we want to find is 5/3-1/3 = 4/3. b) For a first estimate we cut e interval in half and estimate e area by e areas of two rectangles wi base 1/ and altitudes determined by left hand endpoints, 0 and 1/. Wi ese heights of =-0 and 7/4-1/4 = 3/, e estimate for e area is 1/+ 3/ 1/ = 7/4. See Figure ***. Of course a finer partition of e interval and a correspondingly larger number of rectangles would give a better estimate of e area. In general if we let x=1/n and x = k/n, en e estimate for e area is given by k. Larger values for N will give a better estimate for e area as Figure *** illustrates. Wi N = 10 we find an estimate for e area of Notice at from e figure it should clear at any of e estimates obtained using ese sums will be an overestimate of e actual area. c)the areas in part a) were expressed as integrals, and Figure 3 Figure 4 while e area in question was e difference A1-A.. This area can be expressed also as a single integral by using e linearity property of e definite integral,. Notice at is is precisely e definite integral at e sum in part b) estimates by e definition of e integral. Thus we have two approaches to finding e area: i) recognize e region as being composed of regions for which e areas are computable and ii) estimate e area wi sums. Wi each approach e area can be found by evaluating a single definite integral using e Fundamental Theorem of Calculus,. Generalization. The work in e last example can be generalized wiout much difficulty if we follow e estimation approach but continue to assume at for all x in [a,b], g(x) f(x). In is situation e region will be contained between e graphs of y=f(x) and y=g(x) and e lines X = a and X = b wi a < b and where f and g are continuous functions on [a,b]. We partition e interval [a,b] into N pieces of leng wi x = a + k x. Above e k subinterval we consider a rectangle wi its height h k determined by e difference in e function values at e left hand endpoint of e interval, i.e.,. The sum of areas of e individual rectangles, given by k, is our estimate for e area of e region. As N, ese sums give better estimates for e area of e region, but ey also approach (by definition) a definite integral. Therefore e area of e region must be given by Chapter V.G 6/15/14

3 . Comment. In many situations e functions whose graphs bound a region change over e controlling domain interval. It is not uncommon in fact for two function's graphs to cross so at for a part of e interval f(x) g(x) while for anoer part g(x) f(x). In ese situations e region needs to be broken down for analysis into regions over intervals where e upper and lower curves do not change. Finding e values for x where ese changes happen, for example, where f(x) = g(x), is crucial to is investigation. For convenience it is possible to denote e result in situations where only two functions, f and g, are involved wiout ese complications by noting at in eier case. However, in practice is notation does not actually simplify e computation since to evaluate an integral wi an absolute value in e integrand requires furer analysis of e integrand. Summary: The area of a region between two continuous functions f and g and e lines X=a and X=b is given by. The next example follows e eme of estimation to find an area. It differs from e previous example primarily because e estimating rectangles are controlled by e use of e vertical axis, giving an alternative for regions where e boundary curves are more easily described wi e horizontal (X) component of e points on e curves being deter function of eir vertical (Y) component. We continue to follow e principles of What's Good For One Is Good For All and The More The Merrier! Example V.G.. Consider e region bounded by e curve Y = X and e line wi equation Y = X -. a) Estimate e area of is region. b) Express e area of is region as a definite integral. c) Use part b to find e area of e given region. Solution. a) First we graph e region under discussion and note at e curves intersect when Y = Y +, i.e., e points on e curve where Y =-1 and Y = so e Y values for is region are between -1 and. As a first estimate we cut is interval of Y values [-1,] on e Y-axis into ree equal pieces of leng 1 and estimate e area by e areas of ree horizontal "rectangles" wi wid 1 and lengs of 0, and 3-1 =, giving e estimate of (0 + + ) 1 = 4. Of course a finer partition of e interval and a correspondingly larger number of horizontal rectangles would give a better estimate of e area. In general we let y =3/N and y k = -1+ k y. We use e horizontal segment connecting e points on e curves wi Y coordinates y to determine e side of e k rectangle. This gives e k rectangle dimensions by y so en e estimate for e area of e k rectangle is given by to an estimate for e total area given by e sum Chapter V.G 6/15/14 3 or k-1. This leads. As you might expect larger values for N will give a better estimate for e area. [Draw a figure at illustrates is situation wi rectangles.] For example wi N=15, we find an estimate of 4.48.

4 b) Here e pattern of estimating an area wi sums shows its payoff. You should be able to recognize at e sum in part a) estimates e definite integral:. Since ese sums can estimate only one number as eir limit when N, it must be at e area is e number described by. c) From part b) e area can be found by evaluating e definite integral using e Evaluation Form of e Fundamental Theorem of Calculus, Comment: The generalization of is example is to regions at have eir boundaries determined by curves described wi e X-coordinate of a point on e curve determined as a function of its corresponding Y-coordinate. In e example we had X=f(Y)=+Y and X=g(Y)=Y and an interval of Y values [c,d] which controlled e boundary curves values. The result for regions of is type is at e area can be expressed as.. Area as a Solution to a Differential Equation In our first treatment of e Fundamental Theorem of Calculus we saw at an area function was e solution to a differential equation. We will find e same kind of analysis useful in finding e area between two curves as e next example illustrates by resolving our first example wi a quite different approach using e derivative and e fundamental eorem of calculus for differential equations. Example V.G.3. a) Let A be e function determined by e statement at A(t) denotes e area of e region in e first quadrant enclosed by e Y - axis, e graph of g(x)=x, f(x)=-x, and e line X = t when t is between 0 and 1. Estimate e instantaneous rate of change of A(t) at t where t is in e interval [0,1]. b) Find e instantaneous rate of change of e area A of is region as a function of t. c) Explain how e differential equation in part b) shows e area in question must be. Solution. a) An estimate of e rate of change at t is given by. But is approximately e area of a rectangle wi altitude g(t)-f(t) and base t, so at Chapter V.G 6/15/14 4

5 . Draw a figure at illustrates is for e functions f and g. b) From part a) we have. c) The initial condition for A(t) is A(0)=0. By e Differential Equation form of e Fundamental Theorem of Calculus, e solution to e differential equation A'(t)=f(t)-g(t)=-t wi A(0)=0 is.therefore. Exercises. In each of e following problems a) sketch a graph of e region described, b) use four rectangles to estimate e area of e region, c) express e area of e region using e definite integral, and d) find e exact area of e region described The region bounded by e graph of e equations Y = X and Y = X.. The region bounded by e graph of e equations Y = X and Y = X The region bounded by e graph of e equations Y = X and Y = X. 4. One section of e region bounded by e graph of e equations Y=sin(X) and Y=cos(X). 1/ 5. The region bounded by e graph of Y = X and e graph of Y = X. 6. The region bounded by e graph of e equations Y = X and Y = - X. In each of e following problems a) sketch a graph of e region described, b) use four horizontal rectangles to estimate e area of e region, c) express e area of e region using e definite integral wi y as e controlling variable, and d) find e exact area of e region described. 7. The region bounded by e graph of e equations,y = X, Y = -1 and Y =. 8. The region bounded by e graph of e equations and Y = X, Y = - and Y =. 9. The region bounded by e graph of e equations Y = X and. 10. The region bounded by e graph of e equations X=sin(Y), X=cos(Y), Y=0 and Y=. 11. The region bounded by e graph of and e graph of. 1. The region bounded by e graph of e equations Y= 1/x, Y= 1/x, and e line X=. Chapter V.G 6/15/14 5

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