   Chapter 5, Problem 18RE ### 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 2 y 2 1 + y 3   d y

To determine

The value of the integral function.

Explanation

Given information:

The integral function is 02y21+y3dy

The region lies between x=0 and x=2.

Calculation:

Consider (1+y3) as u.

u=1+y3 (1)

Differentiate both sides of the Equation

du=(0+3y2)dy=3y2dy (2)

Rearrange equation (2) to find the value of y2dy as shown below:

y2dy=13du (3)

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

Substitute 0 for y in Equation (1).

u=1+0=1

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

Substitute 2 for y in Equation (1).

u=1+23=9

Substitute u for (1+y3) and 13du for y2dy in the function as shown below,

02y21+y3dy

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