Chapter 8, Problem 14RE

To determine

To find the area of the region enclosed by the lines and curves.

Answer to Problem 14RE

The area of the region enclosed by the lines and curves $y={\sec }^{2}x,y=3-\left|x\right|$ is

  $2.10$ square units.

Explanation of Solution

Given:

The equation of lines and curves are

  $y={\sec }^{2}x,y=3-\left|x\right|$

Calculation:

Using graphing calculator to graph the given equations

  $y={\sec }^{2}x,y=3-\left|x\right|$

Graph:

  

From graph it is observed that the graph of $y=3-\left|x\right|$ lies above the graph of $y={\sec }^{2}x$ and intersect at $\left(-0.825,2.17\right)$ and $\left(0.825,2.17\right)$

The area of region is

  $A={\displaystyle \underset{-0.825}{\overset{0.825}{\int }}\left(3-\left|x\right|-{\sec }^{2}x\right)dx}=2{\displaystyle \underset{0}{\overset{0.825}{\int }}\left(3-x-{\sec }^{2}x\right)dx=2{\left|3x-\frac{x^2}{2}-\tan x\right|}_0^{0.825}=2.10}$

Therefore,

the area of the region enclosed by the lines and curves $y={\sec }^{2}x,y=3-\left|x\right|$ is

  $2.10$ square units.