Lesson 13. Solving Definite Integrals


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1 Lesson Solving Definite Integrals
2 How to find antiderivatives We have three methods:. Basic formulas. Algebraic simplification. Substitution
3 Basic Formulas If f(x) is then an antiderivative is x n k cos(kx) x n + except if n = n + kx (assuming the variable is x) sin(kx)/k sin(kx) cos(kx)/k e kx e kx /k x ln x a x a x /ln(a)
4 Algebraic Simplification ( x )( x + ) dx = x dx = x x + C " " ( x ) + dx = 9x + 6x + dx = x + x + x + C = x + x + x + C 9 6 x + x x dx = x( + x) dx = + x dx = x + x + C x
5 Integrals by Substitution Start with Let u = g(x). du dx = g (x) so du = g (x) dx du x sin( x ) dx Let u =? u = x du = xdx x sin( x ) dx = sin( x )x dx = sin( u) du = cos( x ) + C " " "
6 Problem Problem Problem Antiderivative Practice " sin(t) e 4 t + 4 t dt z + z dz z z z z z 6t dt 4 + t Use basic formulas: sin(t) e 4t + 4 t dt " = cos(t) 4 e 4t + ln(4) 4 t + C Simplify algebraically first, then integrate. z + z z z dz = + dz = + dz = z + ln z + C 6t t 4 + t 4 + t u Make a substitution: Let u=4+t, so du=tdt. dt = dt = du = ln u + C = ln 4 + t + C
7 Antiderivative Practice Problem 4 y ln(ky) dy Make a substitution: u=ln(ky), so du=dy/y. dy = dy = du = ln u + C = ln ln( ky) + C y ln( ky) ln( ky) y u Problem 5 Find the particular function F(x) such that F'(x) = x and the graph of F(x) passes through (, ). The general antiderivative is x dx = x + C Then to find C, we must have F() = + C = Thus, C = 5/, and our function is F(x) = x + 5
8 Solving Definite Integrals Theorem: (Fundamental Theorem I) Or: If F is an antiderivative for f, then Example We determined using Simpson s Rule : Now use the fundamental theorem: An antiderivative for f(x) = x + 5 is So: x + 5 dx = F()  F() = 76  = 76
9 Example x dx We have to find an antiderivative; evaluate at ; evaluate at ; subtract the results. x dx = 4 x4 = 4 4 " 4 4 = 8 " 6 = 6 5 = This notation means: evaluate the function at and, and subtract the results. Don t need to include + C in our antiderivative, because any antiderivative will work.
10 " sin( x) + x dx ) Examples x " # sin( x) + x dx = $ cos x + = % $ cos + & $ $ cos + ' ( Alternate notation = = ( ) " # = % $ ( $ ) + & $ ( $ () + ) = 4 + ' ( " ds s " ds = ln s s = ln ln = ln
11 Practice Examples 5 e dx = 4 e 9 s ds = 9 s/ ds = s / 9 = 9 ( ) / " ( ) / = 54 " 4 se s + " ds s = " e s + ds = (e s + ln s ) s = e + ln (e + ln ) = e e ln
12 Substitution in Definite Integrals We can use substitution in definite integrals. However, the limits are in terms of the original variable. We get two approaches: Method I: Solve an indefinite integral first Change the limits First solve an indefinite integral to find an antiderivative. Then use that antiderivative to solve the definite integral. Note: Do not say that a definite and an indefinite integral are equal to each other They can t be.
13 Example First: Solve an indefinite integral. t dt t + 4 u = t + 4 du = t dt t t + 4 dt = t t + 4 dt = Pull out the, and put in a. t dt = t + 4 ln t + 4 Here s the antiderivative we just found. ( ) = u du = ln t C t dt becomes du Second: Use the antiderivative to solve the definite integral. Here s an antiderivative ( ln 8 ) " ( ln 5 ) = ln 8 5
14 Example When discussing population growth, we worked backwards to find out what we got from evaluating Let s find an antiderivative using substitution. P (t) " dt = P(t) " P ( t) P( t) du = ln u + C = ln P(t) + C u " dt u = P(t) du = P'(t) dt Of course, P(t) is always nonnegative, so we don t need absolute values bp (t) " P(t) dt b $ P(b) ' So we get: = ln( P(t) ) a a = ln ( P(b) )#ln ( P(a) ) = ln & ) % P(a) (
15 Method II: Convert the Limits We start with x = a and x = b, and a substitution formula u = Just put a and b into the substitution formula and get new limits. Note: You do not have to go back to x then t Example dt Start with the same substitution t + 4 u = t + 4 du = t dt t dt = t dt + 4 t + 4 t becomes u What happens to t =? u = t + 4 = ^ + 4 = 5. becomes du = = = u 8 ( ) 8 8 du ln u 5 5 ln 5 When t =, u = 5. When t =, u = 8. And when t =, u = t + 4 = ^ + 4 = 8.
16 8 ( ) 5 y y dy Example u = 5y du = y dy 4 ( ) u y 5y dy = y( 5 y ) dy = ( 5y ) y dy = u du = 5 ) ) ) ) Note that we can also do this problem without usub try algebraic simplification " = % # & $ 4 ' ( y = : u = 5 y = 8: u = 8 ( y 5y ) dy = y( 5y ) 8 8" % dy = 5 y $ ' y dy = 5 # & 8 5 = 5 y = (8 ( ) ) y dy
17 e 4x dx + e 4x Method I: Firstly compute u = + e 4x du = 4e 4x dx Practice Example e 4x e 4x + e 4x dx 4e 4x dx = + e 4x 4 dx = + e 4x 4 u du = 4 u" du = 4 u " + " + + C e 4x = u + C = + e4x dx = + e 4x + e4x = + e4 " + e = + e4 " + C Method II: u = + e 4x u() = + e 4* =,u() = + e 4* = + e 4 e 4x 4e 4x dx = + e 4x 4 dx = + e 4x 4 +e 4 u du = 4 +e 4 u " du = 4 u " + " + +e 4 = u +e 4 = + e4 "
18 Using Definite Integrals We can now evaluate many of the integrals that we have been able to set up. Example Find area between y = sin(x) and the x axis from x = to x = π, and from x = to x = π. The area from to π is clearly: # sin( x) dx = " cos( x ) = + = The area from to π is more complicated. We note that But this is obviously not the area The area from to π can be found by: ( ) ( ) # # sin( x) dx " sin( x) dx = " cos( x) + cos( x) = 4 " sin( x) dx =
19 Summary Used the fundamental theorem to evaluate definite integrals. Made substitutions in definite integrals By solving an indefinite integral first By changing the limits Used the fundamental theorem to evaluate integrals which come from applications.
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