Chapter 5 | Integration
585
Returning to the problem we looked at originally, we let u = x 2 −3 and then du =2 xdx . Rewrite the integral in terms of u :
⌠ ⌡
3
⎛ ⎝ x 2 −3
⎞ ⎠
= ∫ u 3 du .
(2 xdx ) ⏟ du
u
Using the power rule for integrals, we have
⌠ ⌡ u
4 4 +
3 du = u
C .
Substitute the original expression for x back into the solution:
⎞ ⎠ 4 4 +
⎛ ⎝ x 2 −3
u 4
C =
C .
4 +
We can generalize the procedure in the following Problem-Solving Strategy.
Problem-Solving Strategy: Integration by Substitution 1. Look carefully at the integrand and select an expression g ( x ) within the integrand to set equal to u . Let’s select g ( x ). such that g ′( x ) is also part of the integrand. 2. Substitute u = g ( x ) and du = g ′( x ) dx . into the integral. 3. We should now be able to evaluate the integral with respect to u . If the integral can’t be evaluated we need to go back and select a different expression to use as u . 4. Evaluate the integral in terms of u . 5. Write the result in terms of x and the expression g ( x ).
Example 5.30 Using Substitution to Find an Antiderivative Use substitution to find the antiderivative ⌠ ⌡ 6 x ⎛ ⎝ 3 x 2 +4 ⎞ ⎠ 4
dx .
Solution The first step is to choose an expression for u . We choose u =3 x 2 +4 because then du =6 xdx , and we already have du in the integrand. Write the integral in terms of u : ⌠ ⌡ 6 x ⎛ ⎝ 3 x 2 +4 ⎞ ⎠ 4 dx = ∫ u 4 du . Remember that du is the derivative of the expression chosen for u , regardless of what is inside the integrand. Now we can evaluate the integral with respect to u :
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