Chapter 8.2, Problem 7E

To determine

To Find: The area of the region enclosed by the graphs of the two functions

Answer to Problem 7E

The area of the region enclosed by the graphs of the two functions $\approx 1.67$

Explanation of Solution

Given information:

Two function

  $y=\sin x,y=1-x^2$

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 function

  $y=\sin x,y=1-x^2$

Using a calculator, solve the equation

  $\sin x=1-x^2$

  

The solution are $-1.3973433396385$ and $0.6273756757579$

The area is

  $\begin{array}{l}A={\displaystyle \underset{-1.41}{\overset{0.637}{\int }}\left[1-x^2-\sin x\right]}dx\\ A={\left[x-\frac{x^3}{3}+\cos x\right]}_{-1.41}^{0.637}\\ A=\left[0.637-\frac{{(0.637)}^3}{3}+\cos 0.637+1.41-\frac{{(1.41)}^3}{3}-\cos 1.41\right]\\ A\approx 1.67\end{array}$

The area of the region enclosed by the graphs of the two functions $\approx 1.67$