Chapter 5 | Integration
601
− ∫ e u du .
Next, change the limits of integration:
u = (1) −1 =1 u = (2) −1 = 1 2 .
Notice that now the limits begin with the larger number, meaning we must multiply by −1 and interchange the limits. Thus,
1/2
1/2 1
− ∫
e u du = ∫
e u du
1
= e u |
1/2 1
= e − e 1/2 = e − e .
Evaluate the definite integral using substitution: ⌠ ⌡ 1 2 1 x 3 e 4 x −2 dx .
5.37
Integrals Involving Logarithmic Functions Integrating functions of the form f ( x ) = x −1 result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as f ( x ) = ln x and f ( x ) = log a x , are also included in the rule.
Rule: Integration Formulas Involving Logarithmic Functions The following formulas can be used to evaluate integrals involving logarithmic functions.
(5.22)
∫ x −1 dx = ln| x | + C ∫ ln xdx = x ln x − x + C = x (ln x −1)+ C ∫ log a xdx = x ln a (ln x −1)+ C
Example 5.45 Finding an Antiderivative Involving ln x
3 x −10 .
Find the antiderivative of the function
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