Chapter 8.2, Problem 48E

To determine

To Find: The area of the region enclosed between the graphs

Answer to Problem 48E

The area of the region enclosed between the graphs $\approx 0.4303$

Explanation of Solution

Given information:

Two graphs

  $y=\sin x,y=x^3$

Concept used: If $f(x)$ and $g(x)$ are continuous with $f(x)\ge g(x)$ throughout $[a,b]$ , then the area between the curves $y=f(x)$ and $y=g(x)$ from a to b is the integral of $[f(x)-g(x)]$ from a to b

  $A={\displaystyle \underset{a}{\overset{b}{\int }}[f(x)-g(x)]}dx$

Calculation:

Two graphs

  $y=\sin x,y=x^3$

Using a calculator, solve the equation

  $\sin x=x^3$

  

The solution are $-0.9286$ and $0.9286$

On the interval $\left[-0.9286,0\right]$ , $x^3\ge \sin x$ and on interval $\left[\text{0,0}\text{.9286}\right]$ , $\sin x\ge x^3$

Therefore, the area is

  $\begin{array}{l}A={\displaystyle \underset{-0.9286}{\overset{0}{\int }}\left[x^3-\sin x\right]}dx+{\displaystyle \underset{0}{\overset{0.9286}{\int }}\left[\sin x-x^3\right]dx}\\ A={\left[\frac{x^4}{4}+\cos x\right]}_{-0.9286}^0+{\left[-\cos x-\frac{x^4}{4}\right]}_0^{0.9286}\\ A\approx 0.4303\end{array}$

The area of the region enclosed between the graphs $\approx 0.4303$