586
Chapter 5 | Integration
∫ u 4 du = u 5 5
+ C
5
⎛ ⎝ 3 x 2 +4 ⎞ ⎠
=
+ C .
5
Analysis We can check our answer by taking the derivative of the result of integration. We should obtain the integrand. Picking a value for C of 1, we let y = 1 5 ⎛ ⎝ 3 x 2 +4 ⎞ ⎠ 5 +1. We have
5
⎛ ⎝ 3 x 2 +4 ⎞ ⎠
y = 1 5
+1,
so
⎛ ⎝ 1 5
⎞ ⎠ 5
4
⎛ ⎝ 3 x 2 +4 ⎞ ⎠
y ′ =
6 x
4
⎛ ⎝ 3 x 2 +4 ⎞ ⎠
=6 x . This is exactly the expression we started with inside the integrand.
Use substitution to find the antiderivative ⌠ ⌡ 3 x 2 ⎛
2
5.25
⎞ ⎠
⎝ x 3 −3
dx .
Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we are substituting. Example 5.31 Using Substitution with Alteration Use substitution to find ∫ z z 2 −5 dz . 1/2 dz . Let u = z 2 −5 and du =2 zdz . Now we have a problem because du =2 zdz and the original expression has only zdz . We have to alter our expression for du or the integral in u will be twice as large as it should be. If we multiply both sides of the du equation by 1 2 . we can solve this problem. Thus, u = z 2 −5 du =2 zdz Solution Rewrite the integral as ⌠ ⌡ z ⎛ ⎝ z 2 −5 ⎞ ⎠
1 2
(2 z ) dz = zdz .
du = 1 2
This OpenStax book is available for free at http://cnx.org/content/col11964/1.12
Made with FlippingBook - professional solution for displaying marketing and sales documents online