MATH 12002 - CALCULUS I 4.5: Integration by Substitution Definite - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §4.5: Integration by Substitution — Definite Integrals

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Substitution and Definite Integrals

By the Fundamental Theorem of Calculus, if F is an antiderivative for f , then b

a

f (x) dx = F(b) − F(a). Hence one option for computing a definite integral requiring substitution is to find an antiderivative (i.e., indefinite integral) F(x) by substitution, and then compute F(b) − F(a). This involves some unnecessary steps, however: Computation of the indefinite integral

• f (x) dx must be done

separately. “Back substitution” after integrating is necessary to return the integral to a function of x before evaluating. The antiderivative must be put back in the definite integral computation to evaluate. There is an easier way!

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Substitution and Definite Integrals

When we write b

a f (x) dx, the differential dx tells us that the variable is x

and we are integrating from x = a to x = b. The entire integral, including the limits of integration, is in terms of x. When we make a change of variable, we must change the limits of integration to correspond to the new variable. That is, for b

a f (g(x))g′(x) dx, we have

u = g(x) du = g′(x) dx x = b ⇒ u = g(b) x = a ⇒ u = g(a) and then b

a

f (g(x))g′(x) dx = g(b)

g(a)

f (u) du. The integral can be computed as on the right without back substituting.

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Examples

1 Evaluate the integral

2

1 x2(x3 + 1)4 dx.

2

1

x2(x3 + 1)4 dx = 2

1

x2(x3 + 1)4 dx u = x3 + 1 du = 3x2 dx

1 3du = x2 dx

x = 2 ⇒ u = 23 + 1 = 9 x = 1 ⇒ u = 13 + 1 = 2 =

1 3

9

2

u4 du = 1

3 · 1 5u5

9

2

= 1 15(95 − 25) = 59049 − 32 15 = 59017 15 .

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Examples

2 Evaluate the integral

π/2 cos10 x sin x dx. π/2 cos10 x sin x dx = π/2 (cos x)10 sin x dx u = cos x du = − sin x dx (−1)du = sin x dx x = π/2 ⇒ u = cos(π/2) = 0 x = 0 ⇒ u = cos 0 = 1 = −

1

u10 du = − 1 11u11

• 1

= − 1 11(011 − 111) = − 1 11(−1) = 1 11.

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Notes

Compute the integral in terms of u. Do not back substitute. If you do back substitute, change limits back to x-values. In each step, the limits of integration must correspond to the variable used in that step. The following is incorrect: π/2 cos10 x sin x dx = − π/2 u10 du = − 1

11u11

π/2 = − 1

11(cos x)11

π/2 = − 1

11(cos(π/2) − cos 0)

= − 1

11(0 − 1) = 1 11.

LIE LIE

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