Chapter 6.2, Problem 52E

To determine

To evaluate: the integral.

Answer to Problem 52E

The integral of the equation is $-\frac{1}{4}$ .

Explanation of Solution

Given information :

The fact:

  ${{\displaystyle \int }}_0^1{x}^{3}dx=\frac{1}{4}$

The integral:

  ${{\displaystyle \int }}_{-1}^2{(|x|-1)}^{3}dx$

Calculation :

For any integral function,

  ${{\displaystyle \int }}_a^{b}f(x)dx$ =(area above the x-axis)−(area below the x-axis)

  

  

Since, observe the graph of ${(|x|-1)}^3$ , area under the curve consist three regions and each region is equal to area under the curve $x^3$ for $0\le x\le 1$ .

So,

  ${{\displaystyle \int }}_{-1}^2{(|x|-1)}^{3}dx=\text{area under the curve}{(|x|-1)}^3$

  $\begin{array}{l}=\left(areaabovethex-axis\right)–\left(areabelowthex-axis\right)\hfill \\ =\left(1\times areaunderthecurvex³\right)–\left(2\times areaunderthecurvex³\right)\hfill \end{array}$

  $=1{{\displaystyle \int }}_0^1{x}^{3}dx-2{{\displaystyle \int }}_0^1{x}^{3}dx$

  $\begin{array}{cc}{{\displaystyle \int }}_{-1}^2{(|x|-1)}^{3}dx& =1\times \frac{1}{4}-2\times \frac{1}{4}\\ & =-\frac{1}{4}\end{array}$