Question

If $\underset{a}{\overset{b}{\int }}|\sin x|dx=8$ and $\underset{0}{\overset{a+b}{\int }}|\cos x|dx=9$, then the value of $\underset{a}{\overset{b}{\int }}x\sin xdx$ is

• A. $2\sqrt{2}\pi$
• B. $-2\sqrt{2}\pi$
• C. $4\sqrt{2}\pi$
• D. $-4\sqrt{2}\pi$

Answer 1

The correct option is B $-2\sqrt{2}\pi$

We know, $\underset{a}{\overset{b}{\int }}|\sin x|dx$ represents the area under the curve from $x=a$ to $x=b$.
We also know, area from $x=a$ to $x=a+\pi$ is $2.$
$\therefore \underset{a}{\overset{b}{\int }}|\sin x|dx=8$
$\Rightarrow b-a=\frac{8\pi }{2}\cdots \left(i\right)$
Similarly, $\underset{0}{\overset{a+b}{\int }}|\cos x|dx=9$
$\Rightarrow a+b-0=\frac{9\pi }{2}\cdots \left(ii\right)$
From equations $\left(i\right)$ and $\left(ii\right),$
$a=\frac{\pi }{4},b=\frac{17\pi }{4}$
Hence, $\underset{a}{\overset{b}{\int }}x\sin xdx$ $=\underset{\pi /4}{\overset{17\pi /4}{\int }}x\sin xdx$
$={[-x\cos x]}_{\pi /4}^{17\pi /4}+\underset{\pi /4}{\overset{17\pi /4}{\int }}\cos xdx$
$=-\frac{17\pi }{4}\cos \frac{17\pi }{4}+\frac{\pi }{4}\cos \frac{\pi }{4}+[\sin x{]}_{\pi /4}^{17\pi /4}$
$=-\frac{4\pi }{\sqrt{2}}=-2\sqrt{2}\pi$