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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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Chapter
Section
BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Evaluate the integral, if it exists.

0 1 y ( y 2 + 1 ) 5 d y

To determine

The value of the integral function.

Explanation

Given information:

The integral function is 01y(y2+1)5dy

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

Calculation:

Consider (y2+1) as u.

u=y2+1 (1)

Differentiate both sides of the Equation.

du=(2y+0)dy=2ydy (2)

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

ydy=12du (3)

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

Substitute 0 for y in Equation (1)

u=0+1=1

Calculate theupper limit value of u using Equation (1).

Substitute 1 for y in Equation (1)

u=12+1=2

Substitute u for (y2+1) and 12du for ydy in the function as shown below,

01y(y2+1)5dy=12

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