   Chapter 5, Problem 28RE ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
1 views

# Applying the General Power Rule In Exercises 21–32, find the indefinite integral. Check your result by differentiating. ∫ ( 2 x 4 − 16 x ) 3 ( x 3 − 2 )   d x

To determine

To calculate: The indefinite integral (2x416x)3(x32)dx.

Explanation

Given Information:

The provided indefinite integral is (2x416x)3(x32)dx.

Formula used:

The power rule of integrals:

undu=un+1n+1+C (n1)

The power rule of differentiation:

ddxun=nun1+C

Calculation:

Consider the indefinite integral:

(2x416x)3(x32)dx

Let u=2x416x, then derivative will be,

du=d(2x416x)=(8x316)dx

Rewrite integration as,

(2x416x)3(x32)dx=18(2x416x)3(8x316)dx

Substitute du for (8x316)dx and u for 2x416x in provided integration

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Convert the expressions in Exercises 8596 radical form. 34(1x)5/2

Finite Mathematics and Applied Calculus (MindTap Course List)

#### A sample with a mean of M = 8 has X = 56. How many scores are in the sample?

Essentials of Statistics for The Behavioral Sciences (MindTap Course List)

#### In Exercises 21-24, find the distance between the given points. 22. (1, 0) and (4, 4)

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

#### The general solution to (for x, y > 0) is: a) y = ln x + C b) c) y = ln(ln x + C) d)

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