Chapter 7, Problem 14RE

To determine

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

Answer to Problem 14RE

The area of the region enclosed by the lines and the curves is $2.1043$ .

Explanation of Solution

Given information:

  $\begin{array}{l}y={\sec }^{2}x\\ y=3-\left|x\right|\end{array}$

Calculation:

Let's graph the given area in Desmos and draw some conclusions based on it.

  

From the graph, it is seen that the area is enclosed above by the line $y=3-\left|x\right|$ and below by the $y={\sec }^{2}x$ from $x=-0.8255\text{ to }x=0.8255$ .

Demonstrate to get the integration boundaries:

The point of intersection is found by equating the given equations to get,

  $\begin{array}{c}{\sec }^{2}x=3-\left|x\right|\\ {\sec }^{2}x+\left|x\right|-3=0\end{array}$

Since, there is no simple method to solve the above equation using the Wolfram Alpha.

By entering the equation to get the required values,

Numerical solutions

  $x\approx \pm 0.8255405194204$

The area of the given region is,

  $A={\displaystyle \underset{-0.8255}{\overset{0.8255}{\int }}\left[3-\left|x\right|-{\sec }^{2}x\right]}dx$

Because the area is symmetric with respect to the $y$ -axis, the absolute value will be lost in the sub integral function, making it easier to solve the derived integral. Hence,

  $A=2\cdot {\displaystyle \underset{-0.8255}{\overset{0.8255}{\int }}\left[3-x-{\sec }^{2}x\right]}dx$

Using the trigonometric formula and the power formula, let's proceed with the calculation:

  $\begin{array}{c}A=2\cdot {\displaystyle \underset{-0.8255}{\overset{0.8255}{\int }}\left[3-x-{\sec }^{2}x\right]}dx\\ =2\cdot {\left.\left[3x-\frac{x^2}{2}-\tan x\right]\right|}_0^{0.8255}\\ =2\cdot \left[2.4765-0.34072-1.0836\right]\\ =2.1043\end{array}$

Therefore, the area of the region enclosed by the lines and the curves is $2.1043$ .