   Chapter 6, Problem 8TYS ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# In Exercises 8–10, use integration by parts or the integration table in Appendix C to evaluate the definite integral. ∫ 0 1 ln ( 3 − 2 x ) d x

To determine

To calculate: The value of definite integral 01ln(32x)dx.

Explanation

Given Information:

The integral is provided as:

01ln(32x)dx

Formula used:

The formula 41 for integral lnudu is:

lnudu=u[1+lnu]+C

General power differentiation Rule:

ddx[un]=nun1dudx

Calculation:

Consider u=32x.

Differentiate the considered function with respect to x using power rule of differentiation.

du=(02)dx=2dx

Consider the provided integral:

01ln(32x)dx

Multiply and divide by 2.

01ln(32x)dx=1201[ln(32x)](2dx)

Substitute u for 32x, and du for 2dx.

01ln(32x)dx=1201lnudu

Use the formula 41 and solve the above integral as:

01ln(32x)dx=12[u[1+lnu]]01

Substitute 32x for u

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