Chapter 8.2, Problem 39E

To determine

To Find:

To find the area of the triangular region in the first quadrant bounded by y axis and on the right by curves $y=\sin x$ and $y=\cos x$ .

Answer to Problem 39E

The area of the region in the first quadrant bounded by curves is $\sqrt{2}-1$ .

Explanation of Solution

Given information:

The given function is $y=\sin x$ , $y=\cos x$ and bounded by y-axis.

The red curve is the graph of $y=\sin x$ and the blue line is the graph of $y=\cos x$ , shading the region in the first quadrant enclosed by two curves.

  

Now finding the value of $x$ by equating $y=\sin x$ and $y=\cos x$ ,

  $\cos x=\sin x$

As a result, they intersect at $x=\frac{\pi }{4}$ .

To find the area of the shaded region we have to calculate the integral enclosed between $y=\sin x$ and $y=\cos x$ .

Since $\cos x$ is above $\sin x$ , calculate the integral of $\left(\cos x-\sin x\right)$

Now take the integral of $x$ from $0$ to $\frac{\pi }{4}$ ,

Therefore, the area of the region in the first quadrant bounded is,

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

Therefore, the area of the region in the first quadrant bounded by curves is $\sqrt{2}-1$ .