   Chapter 12.2, Problem 33E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Evaluate the integrals in Problems 7-36. Check your results by differentiation. ∫ x 2 − 4 x x 3 − 6 x 2 + 2 d x

To determine

To calculate: The value of the integral x24xx36x2+2dx.

Explanation

Given Information:

The provided integral is x24xx36x2+2dx

Formula used:

The power formula of integrals:

undu=un+1n+1+C (forn1)

The power rule of differentiation:

ddu(un)=nun1

Calculation:

Consider the provided integral:

x24xx36x2+2dx

Rewrite the integral by multiplying and dividing by 3 as:

133x212xx36x2+2dx

Let u=x36x2+2, then derivative will be,

du=d(x36x2+2)=(3x312x)dx

Substitute du for (3x312x)dx and u for (x36x2+2) in provided integration

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