Question

${\int }_a^b\left|\sin x\right|dx=8$ and ${\int }_0^{a+b}\left|\cos x\right|dx=9,$ then value of $\frac{1}{\sqrt{2}\pi }\left|{\int }_a^{b}x\sin xdx\right|$ is

Answer 1

${\int }_b^a\left|\sin x\right|dx=8\Rightarrow b-a=4\pi$

${\int }_0^{a+b}\left|\cos x\right|dx=9\Rightarrow a+b=\frac{9\pi }{2}\Rightarrow a=\frac{\pi }{4},b=\frac{17\pi }{4}$

$\frac{1}{\sqrt{2}\pi }\left|{\int }_a^{b}x\sin xdx\right|=\frac{1}{\sqrt{2}\pi }\left|{\int }_{\frac{\pi }{4}}^{\frac{17\pi }{4}}x\sin xdx\right|=2$