Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 07/Solution 1

Since the question is asking about the area of a region in the plane, we must be careful not to compute a singed area. The easiest way to do this is to use an absolute value. $$A=\underset{0}{\overset{\pi }{\int }}|\sin (x)-\cos (x)|\mathrm{d}x.$$ By drawing a picture, we see that $\sin (0)-\cos (0)=-1$ and $\sin (\pi )-\cos (\pi )=1$, which shows us that there is some $x\in (0,\pi )$ for which $\sin (x)-\cos (x)=0$. This is precisely $x=\frac{\pi }{4}$. Therefore, we have $$\begin{array}{rl}A& =\underset{0}{\overset{\frac{\pi }{4}}{\int }}|\sin (x)-\cos (x)|\mathrm{d}x+\underset{\frac{\pi }{4}}{\overset{\pi }{\int }}|\sin (x)-\cos (x)|\mathrm{d}x\\ & =\underset{0}{\overset{\frac{\pi }{4}}{\int }}(\cos (x)-\sin (x))\mathrm{d}x+\underset{\frac{\pi }{4}}{\overset{\pi }{\int }}(\sin (x)-\cos (x))\mathrm{d}x\\ & =[\sin (x)+\cos (x){]}_{x=0}^{x=\frac{\pi }{4}}+[-\cos (x)-\sin (x){]}_{x=\frac{\pi }{4}}^{x=\pi }\\ & =\sqrt{2}-1+1+\sqrt{2}\end{array}$$

We find then that the area enclosed between the two curves is $2\sqrt{2}$.