   Chapter 5, Problem 21RE ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Evaluate the integral, if it exists. ∫ 0 1 v 2 cos ( v 3 )   d v

To determine

The value of the integral function.

Explanation

Given information:

The integral function is 01v2cos(v3)dv

The region lies between x=0 and x=1.

Calculation:

Consider v3 as u.

u=v3 (1)

Differentiate on both side of the above equation.

du=3v2dv (2)

Rearrange equation (2) to find the value of dv as shown below:

v2dv=13du (3)

Calculate the lower limit value of u using Equation (1).

Substitute 0 for v in Equation (1)

u=0

Calculate the upper limit value of u using Equation (1).

Substitute 1 for v in Equation (1)

u=13=1

Substitute u for v3 and 13du for v2dv in

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 