Question

Let $f:\left(0,1\right]\to R$ be a continuous function such that ${\int }_0^{\pi }f\left(\sin x\right)\mathrm{dx}=2018,\mathrm{then}{\int }_0^{\pi }xf\left(\sin x\right)\mathrm{dx}$ is equal to

• A. $1009\pi$
• B. $1008\pi$
• C. $2017\pi$
• D. $2016\pi$

Answer 1

The correct option is A $1009\pi$

Let $I={\int }_0^{\pi }xf\left(\sin x\right)\mathrm{dx}\Rightarrow I={\int }_0^{\pi }\left(\pi -x\right)f\left(\sin \right(\pi -x\left)\right)\mathrm{dx}\Rightarrow I={\int }_0^{\pi }\left(\pi -x\right)f\left(\sin x\right)\mathrm{dx}\Rightarrow I=\pi {\int }_0^{\pi }f\left(\sin x\right)\mathrm{dx}-{\int }_0^{\pi }xf\left(\sin x\right)\mathrm{dx}\Rightarrow 2I=\pi {\int }_0^{\pi }f\left(\sin x\right)\mathrm{dx}=2018\pi \Rightarrow I=1009\pi$