Chapter 7.2, Problem 39E

To determine

To find: The area of the triangular region in the first quadrant bounded on the left by the $y$ -axis and the right by the given curves.

Answer to Problem 39E

The area is: $A=\sqrt{2}-1$

Explanation of Solution

Given information:

The curves are:

  $\begin{array}{c}y=\sin x\\ y=\cos x\end{array}$

Calculation:

In the image below, the region that need to examine is coloured in green. As seen in the image, the graph of $\cos (x)$ is the bottom border and the graph of $\sin (x)$ is the top border.

Additionally, utilise the lowest positive solution of the equation $\sin (x)=\cos (x)$ to locate the intersection of those graphs.

The intersection point has a coordinate system because $x=\frac{\pi }{4}$ is the smallest positive solution to this equation.

  $\left(\frac{\pi }{4},\frac{\sqrt{2}}{2}\right)$

So, conclude that the given area represented by the integral:

  

The graph is shown below:

  

Therefore, the required area of the triangular region in the first quadrant bounded on the left by the $y$ -axis and the right by the given curves is $\sqrt{2}-1$ .