Chapter 8.2, Problem 32E

To determine

To Find:

To find the area of the given regions enclosed by the lines $y={\sec }^{2}x$ and $y={\tan }^{2}x$ .

Answer to Problem 32E

The area of the given regions enclosed by the lines is $\frac{\pi }{2}$ .

Explanation of Solution

Given information:

The given function is $y={\sec }^{2}x$ and $y={\tan }^{2}x$ .

The red line is the graph of $y={\sec }^{2}x$ which is above the graph of $y={\tan }^{2}x$ .

  

The functions $y={\sec }^{2}x$ and $y={\tan }^{2}x$ do not intersect on the interval $-\frac{\pi }{4}\le x\le \frac{\pi }{4}$ .

  $A={\displaystyle \underset{-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}\left({\sec }^{2}x-{\tan }^{2}x\right)dx}$

  $A={\displaystyle \underset{-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}1dx}$

  $\begin{array}{l}A={\left[x\right]}_{-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\\ A=\frac{\pi }{2}\end{array}$

Therefore, The area of the given regions enclosed by the lines is $\frac{\pi }{2}$ .