   Chapter 5.5, Problem 63E ### 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 definite integral. ∫ 0 13 d x ( 1 + 2 x ) 2 3

To determine

To evaluate: The definite integral.

Explanation

Given:

The definite integral function is 013dx(1+2x)23.

The region lies between x=0 and x=13.

Calculation:

Consider u=1+2x (1)

Differentiate both sides of the Equation (1).

du=2dxdx=12du

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

Substitute 0 for x in Equation (1).

u=1+2(0)=1

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

Substitute 13 for x in Equation (1).

u=1+2(13)=27

The definite integral function is,

013dx(1+2x)23 (2)

Apply lower and upper limits for u in Equation (2).

Substitute u for (1+2x) and (12du) for (cosxdx) in Equation (2).

013dx(1+2x)23=1271u23(12du<

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