Chapter 5.2, Problem 64E

To determine

To find : The value of integral.

Answer to Problem 64E

The answer: ${{\displaystyle \int }}_{-1}^2{(\left|x\right|-1)}^{3}dx=-\frac{1}{4}$ .

Explanation of Solution

Given information:

The integration is: ${{\displaystyle \int }}_{-1}^2{(\left|x\right|-1)}^{3}dx$ .

Calculation:

The graph of ${{\displaystyle \int }}_0^1{x}^{3}dx$ (green) and ${{\displaystyle \int }}_{-1}^2{(\left|x\right|-1)}^{3}dx$ (red) are shown below.

Notice that the graph of between

  $\begin{array}{c}x=-1\\ x=2\end{array}$

can be broken up into three sections that all have the same areas and the area of each section has an area equal to that of the green region on the left. The two sections below the $x$ -axis get counted as negative areas and the region above the $x$ -axis is counted as a positive area. Since the area of each section is $\frac{1}{4}$ because the green region has an area of ${{\displaystyle \int }}_0^1{x}^{3}dx=\frac{1}{4}$ then,

  ${{\displaystyle \int }}_{-1}^2{(\left|x\right|-1)}^{3}dx={{\displaystyle \int }}_{-1}^0{(\left|x\right|-1)}^{3}dx+{{\displaystyle \int }}_0^1{(\left|x\right|-1)}^{3}dx+{{\displaystyle \int }}_1^2{(\left|x\right|-1)}^{3}dx$

  $\begin{array}{c}=-\frac{1}{4}+\left(-\frac{1}{4}\right)+\frac{1}{4}\\ =-\frac{1}{4}\end{array}$

  

Therefore, the required integral of the given integration is ${{\displaystyle \int }}_{-1}^2{(\left|x\right|-1)}^{3}dx=-\frac{1}{4}$ .