Chapter 6.2, Problem 55E

To determine

To evaluate: the integral.

Answer to Problem 55E

The integral of the equation is $-\frac{3}{4}$ .

Explanation of Solution

Given information :

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

Calculation :

If $y=f(x)$ is non-negative and capable of integration over closed interval $[a,b],$ then the area under the curve $y=f(x)$ from a to b is the integral of $f$ from a to b.

  ${{\displaystyle \int }}_a^{b}f(x)dx=-(\text{area})$

  

  

Since, graph of $|x{|}^3$ symmetrical about the $y$ -axis its area is equal to twice the area of graph of $x^3$ for $0\le x\le 1$ , because left side of $y$ -axis of graph of $|x{|}^3$ is graph of $x^3$ for $0\le x\le 1$ .

So,

  ${\displaystyle {\int }_0^1\left(x^3-1\right)}dx=\text{area under the}\text{curve}\left(x^3-1\right)$

  $\begin{array}{l}=-\left[\text{area of square - area under the curve}{x}^3\right]\hfill \\ =-\left[1\times 1-{{\displaystyle \int }}_0^1{x}^{3}dx\right]\hfill \\ =-\left[1-\frac{1}{4}\right]\hfill \end{array}$

  ${{\displaystyle \int }}_0^1\left(x^3-1\right)dx=-\frac{3}{4}$