Chapter 7, Problem 1RE

Finding the Area of a Region In Exercises 1-10, sketch the region bounded by the graphs of the equations and find the area of the region.

$y=6-\frac{1}{2}{x}^2,y=\frac{3}{4}x,x=-2,x=2$

To determine

To graph: The region bounded by the equations $y=6-\frac{1}{2}{x}^2,y=\frac{3}{4}x,x=-2,x=2$ and find the area of the obtained region.

Answer to Problem 1RE

Solution: The area is, $\underset{¯}{\frac{64}{3}}$.

Explanation of Solution

Given: The equations $y=6-\frac{1}{2}{x}^2,y=\frac{3}{4}x,x=-2,x=2$

Formula used:

Area of the region is;

$A=\text{area under parabola}-\text{area under line}$.

Graph: Sketch of the region bounded by the two equations is given as:

Calculation:

Area of the region is;

$A=\text{area under parabola}-\text{area under line}$.

So, Area of the region is,

$\begin{array}{c}A={\displaystyle \underset{-2}{\overset{2}{\int }}\left(6-\frac{1}{2}{x}^2-\left(\frac{3}{4}x\right)\right)dx}\\ A={\left(6x-\frac{x^3}{6}-\frac{3}{8}{x}^2\right)}_{-2}^2\\ A=12-\frac{8}{6}-\frac{3}{8}\left(4\right)-\left(-12-\frac{-8}{6}-\frac{3}{8}\left(4\right)\right)\\ A=12-\frac{8}{6}-\frac{3}{2}+12-\frac{8}{6}+\frac{3}{2}\end{array}$

Hence, on further simplification:

$\begin{array}{c}A=24-\frac{8}{3}\\ A=\frac{64}{3}\end{array}$

Hence, the area of the region is, $\frac{64}{3}$.