Area — Question 7

Question 7

Find the exact area of the region enclosed by the curves $$y = \sin x \quad\text{and}\quad y = \cos x$$ on the interval $$\frac{\pi}{4} \le x \le \frac{5\pi}{4}.$$

Question 7 - Solution

First find the points of intersection by solving $$\sin x = \cos x.$$

This gives $$\tan x = 1 \quad\Rightarrow\quad x=\frac{\pi}{4}+k\pi.$$

The smallest two consecutive intersection points occur at $$x=\frac{\pi}{4} \quad\text{and}\quad x=\frac{5\pi}{4}.$$

On the interval between these points, we have $$\sin x \ge \cos x.$$

Thus the enclosed area is $$A=\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)\,dx.$$

Compute: $$\int \sin x\,dx=-\cos x, \qquad \int \cos x\,dx=\sin x.$$

So $$A=\left[-\cos x-\sin x\right]_{\pi/4}^{5\pi/4}.$$

Evaluate: $$-\cos\!\left(\frac{5\pi}{4}\right)-\sin\!\left(\frac{5\pi}{4}\right) = \sqrt{2},$$ $$-\cos\!\left(\frac{\pi}{4}\right)-\sin\!\left(\frac{\pi}{4}\right) = -\sqrt{2}.$$

Thus, $$A=\sqrt{2}-(-\sqrt{2})=2\sqrt{2}.$$

$$\boxed{2\sqrt{2}}$$