Chapter 8.2, Problem 31E

To determine

To Find:

To find the area of the given regions enclosed by the lines $y=\sin \left(\frac{\pi x}{2}\right)$ and $y=x$ .

Answer to Problem 31E

The area of the given regions enclosed by the lines is $\frac{4}{\pi }-1$ .

Explanation of Solution

Given information:

The given function is $y=\sin \left(\frac{\pi x}{2}\right)$ and $y=x$ .

The red line is the graph of $y=\sin \left(\frac{\pi x}{2}\right)$ which is below the graph of $y=x$ between $x=\pm 1$

  

Both the functions are symmetrical across the y-axis. Therefore, we can find the area between the curves in the first quadrant and multiply it by 2 to get the total area.

If a region is bounded by a right curve $x_R=g_{L\left(y\right)}$ and a left curve $x_L=g_{R\left(y\right)}$ from $y=c$ to $y=d$ , then the area of this region is

  $A={\displaystyle \underset{c}{\overset{d}{\int }}\left(x_R-x_L\right)}dy$

  $A=2{\displaystyle \underset{0}{\overset{1}{\int }}\left(\sin \left(\frac{\pi x}{2}\right)-x\right)dx}$

  $\begin{array}{l}A=2{\left[-\frac{2}{\pi }\cos \left(\frac{\pi x}{2}\right)-\frac{1}{2}{x}^2\right]}_0^1\\ A=2\left(-\frac{2}{\pi }\cos \left(\frac{\pi }{2}\right)-\frac{1}{2}-\left(-\frac{2}{\pi }\cos 0-0\right)\right)\\ A=2\left(-\frac{1}{2}+\frac{2}{\pi }\right)\\ A=\frac{4}{\pi }-1\end{array}$

Therefore, the area of the given regions enclosed by the lines is $\frac{4}{\pi }-1$ .