 # Integrals of Functions Involving Absolute Values In Exercises 45–52, use integration by parts to evaluate the given integral using the following integral formulas where necessary. (You have seen some of these before; all can be checked by differentiating.) Integral Formula Shortcut Version ∫ | x | x d x = | x | + C Because d d x | x | = | x | x ∫ | a x + b | a x + b d x = 1 a | a x + b | + C ∫ | x | d x = 1 2 x | x | + C ∫ | a x + b | d x = 1 2 a ( a x + b ) | a x + b | + C ∫ x | x | d x = 1 3 x 2 | x | + C ∫ ( a x + b ) | a x + b | d x = 1 3 a ( a x + b ) 2 | a x + b | + C ∫ x 2 | x | d x = 1 4 x 3 | x | + C ∫ ( a x + b ) 2 | a x + b | d x = 1 4 a ( a x + b ) 3 | a x + b | + C ∫ 2 x | x − 3 | x − 3 d x ### Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
Publisher: Cengage Learning
ISBN: 9781337274203 ### Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
Publisher: Cengage Learning
ISBN: 9781337274203

#### Solutions

Chapter
Section
Chapter 14.1, Problem 47E
Textbook Problem

## Integrals of Functions Involving Absolute Values In Exercises 45–52, use integration by parts to evaluate the given integral using the following integral formulas where necessary. (You have seen some of these before; all can be checked by differentiating.) Integral Formula Shortcut Version ∫ | x | x d x = | x | + C  Because  d d x | x | = | x | x ∫ | a x + b | a x + b d x = 1 a | a x + b | + C ∫ | x | d x = 1 2 x | x | + C ∫ | a x + b | d x = 1 2 a ( a x + b ) | a x + b | + C ∫ x | x | d x = 1 3 x 2 | x | + C ∫ ( a x + b ) | a x + b | d x = 1 3 a ( a x + b ) 2 | a x + b | + C ∫ x 2 | x | d x = 1 4 x 3 | x | + C ∫ ( a x + b ) 2 | a x + b | d x = 1 4 a ( a x + b ) 3 | a x + b | + C ∫ 2 x | x − 3 | x − 3 d x

Expert Solution
To determine

To calculate: The value of integral 2x|x3|x3dx by the use of integration by parts.

### Explanation of Solution

Given Information:

The given integral is:

2x|x3|x3dx

Formula used:

Integration by parts:

If u and v are continuous functions of x and u has a continuous derivative, then

uvdx=uI(v)D(u)I(v)dx

Where D(u)=ddxu and I(v)=vdx

The shortcut version of Integration:

|ax+b|ax+bdx=1a|ax+b|+C

And,

|ax+b|dx=12a(ax+b)|ax+b|+C

Where a and b are any constants.

Calculation:

Consider the provided integral:

2x|x3|x3dx

Consider the formula:

uvdx=uI(v)D(u)I(v)dx

Substitute 2x for u and |x3|x3 for v in above equation to get:

2x|x3|x3dx=2xI(|x3|x3)D(2x)I(|x3|x3)dx

Since, D(u)=ddx(u)

Substitute 2x for u in above equation to get:

D(x)=ddx(2x)=2

Since, I(v)=v</

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