Chapter 7.2, Problem 32E

To determine

To find: The area of the region enclosed by the lines and curve.

Answer to Problem 32E

The area is: $A=\frac{\pi }{2}$

Explanation of Solution

Given information:

The curves are:

  $\begin{array}{c}y={\sec }^{2}x,\\ y={\tan }^{2}x,\\ x=-\pi /4,\\ x=\pi /4\end{array}$

Calculation:

To find the intersection $($ if any $)$ of the graphs of the functions

  $\begin{array}{c}y={\sec }^2(x)\\ y={\tan }^2(x)\end{array}$

On the interval

  $\left[-\frac{\pi }{4},\frac{\pi }{4}\right]$

From the equation:

  ${\sec }^2(x)={\tan }^2(x)\mathrm{.....}(1)$

Therefore, from the equation $(1)$ , obtain that $1={\sin }^2(x)$ then equality does not hold for any $x$ in the interval $\left[-\frac{\pi }{4},\frac{\pi }{4}\right]$ .

As a result, the two graphs do not overlap, but know that

  $\begin{array}{c}{\sec }^2(x)=\frac{1}{{\cos }^2(x)}\ge \frac{{\sin }^2(x)}{{\cos }^2(x)}\\ ={\tan }^2(x)\end{array}$

And the integral represents the area of the indicated region:

  

The graph is shown below:

  

Therefore, the required area of the region enclosed by the lines and curve is $\frac{\pi }{2}$ .