The area of the region (x,y):y -|x|,y |x x|,y 0 is:

Question

The area of the region ${(x,y):y\le \pi -|x|,y\le |x\sin x|,y\ge 0}$ is:

Accepted Answer

We need the area of the region

${(x,y):y\le \pi -|x|,\, y\le |x\sin x|,\, y\ge 0}.$

This means the required area is the area under the curve

$y=min(\pi -|x|,\, |x\sin x|)$

and above the $x$-axis.

So the area is

$A=\int min(\pi -|x|,\, |x\sin x|)\, dx$

over the values of $x$ where the region exists.

Since $y\ge 0$ and $y\le \pi -|x|$, we must have

$\pi -|x|\ge 0\, \implies \, |x|\le \pi .$

Hence the region lies in $[-\pi ,\pi ]$.

Also, both functions $\pi -|x|$ and $|x\sin x|$ are even functions, so the region is symmetric about the $y$-axis. Therefore,

$A=2{\int }_{\pi }^{0}min(\pi -x,\, x\sin x)\, dx.$

Now for $0\le x\le \pi $, we have $\sin x\ge 0$, so

$|x\sin x|=x\sin x.$

Thus,

$A=2{\int }_{\pi }^{0}min(\pi -x,\, x\sin x)\, dx.$

Now compare $\pi -x$ and $x\sin x$ on $[0,\pi ]$.

At $x=0$,

$\pi -x=\pi ,\, x\sin x=0.$

At $x=\pi $,

$\pi -x=0,\, x\sin x=0.$

Check where they meet:

$x\sin x=\pi -x.$

It is easy to see that $x=\frac{\pi }{2}$ satisfies this, because

$\frac{\pi }{2}\sin \frac{\pi }{2}=\frac{\pi }{2}=\pi -\frac{\pi }{2}.$

Also, on $[0,\pi /2]$, we have $x\sin x\le \pi -x$, and on $[\pi /2,\pi ]$, we have $\pi -x\le x\sin x$.

So,

$A=2[{\int }_{\pi /2}^{0}x\sin x\, dx+{\int }_{\pi }^{\pi /2}(\pi -x)\, dx].$

Now evaluate each integral.

First,

$\int x\sin x\, dx=-x\cos x+\int \cos x\, dx=-x\cos x+\sin x.$

Therefore,

${\int }_{\pi /2}^{0}x\sin x\, dx={[-x\cos x+\sin x]}_{\pi /2}^{0}=1.$

Next,

${\int }_{\pi }^{\pi /2}(\pi -x)\, dx.$

Let $u=\pi -x$. Then as $x$ goes from $\pi /2$ to $\pi $, $u$ goes from $\pi /2$ to $0$. Or directly,

${\int }_{\pi }^{\pi /2}(\pi -x)\, dx={[\pi x-\frac{{x}^{2}}{2}]}_{\pi }^{\pi /2}.$

Compute:

At $x=\pi $,

${\pi }^{2}-\frac{{\pi }^{2}}{2}=\frac{{\pi }^{2}}{2}.$

At $x=\frac{\pi }{2}$,

$\pi \cdot \frac{\pi }{2}-\frac{1}{2}{(\frac{\pi }{2})}^{2}=\frac{{\pi }^{2}}{2}-\frac{{\pi }^{2}}{8}=\frac{3{\pi }^{2}}{8}.$

So,

${\int }_{\pi }^{\pi /2}(\pi -x)\, dx=\frac{{\pi }^{2}}{2}-\frac{3{\pi }^{2}}{8}=\frac{{\pi }^{2}}{8}.$

Hence,

$A=2(1+\frac{{\pi }^{2}}{8})=2+\frac{{\pi }^{2}}{4}.$

Therefore, the correct answer is

$2+\frac{{\pi }^{2}}{4}$

So, Option B is correct.

Solutions · 1