Calculus Volume 1

590

Chapter 5 | Integration

⎞ ⎠ |

⎛ ⎝ u 6 6

3

⎛ ⎝ 1 6

⎞ ⎠

3

1 6 ∫

u 5 du =

1

1

⎡ ⎣ (3) 6 −(1) 6 ⎤ ⎦

= 1

36

= 182 9 .

Use substitution to evaluate the definite integral ⌠ ⌡ −1 0 y ⎛

5.29

5

⎝ 2 y 2 −3 ⎞ ⎠

dy .

Example 5.35 Using Substitution with an Exponential Function

Use substitution to evaluate ∫ 0 1

2 +3

xe 4 x

dx .

Solution Let u =4 x 3 +3. Then, du =8 xdx . To adjust the limits of integration, we note that when x =0, u =3, and when x =1, u =7. So our substitution gives ∫ 0 1 xe 4 x 2 +3 dx = 1 8 ∫ 3 7 e u du = 1 8 e u | 3 7

= e 7 − e 3 8 ≈134.568.

⎛ ⎝ π 2

x 3 ⎞

Use substitution to evaluate ⌠ ⌡ 0 1

5.30

x 2 cos

⎠ dx .

Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for u after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in Example 5.36 . Example 5.36

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