Chapter 8.2, Problem 36E

To determine

To Find:

To find the area of the given regions by subtracting the area of a triangular region from the area of a larger region on or above the x-axis bounded by the curves $y=4-x^2$ and $y=3x$ .

Answer to Problem 36E

The area of the given regions bounded by the curves are $\frac{15}{2}$ .

Explanation of Solution

Given information:

The given function is $y=4-x^2$ and $y=3x$ .in the interval $-2\le x\le 1$ .

The red curve is the graph of $y=4-x^2$ and the blue line is the graph of $y=3x$ .

  

Now, we have to find the area by subtracting the triangle formed by $y=3x$ from the area formed by the entire region.

Therefore, the area of the larger region is

  $\begin{array}{l}a={\displaystyle \underset{-2}{\overset{1}{\int }}\left(4-x^2\right)dx}\\ a={\left[4x-\frac{x^3}{3}\right]}_{-2}^1\\ a=4-\frac{1}{3}-\left(-8+\frac{8}{3}\right)=9\end{array}$

Area of the triangle outside of area that we are trying to find but formed from larger region is

  $\begin{array}{l}b={\displaystyle \underset{0}{\overset{1}{\int }}3xdx}\\ b={\left[\frac{3x^2}{2}\right]}_0^1\\ b=\frac{3}{2}\end{array}$

Subtract the area of triangle from area of larger region,

  $\begin{array}{l}A={\displaystyle \underset{-2}{\overset{1}{\int }}\left(4-x^2\right)dx}-{\displaystyle \underset{0}{\overset{1}{\int }}3xdx}\\ A=9-\frac{3}{2}\\ A=\frac{15}{2}\end{array}$

Therefore, the area of the given regions bounded by the curves are $\frac{15}{2}$ .