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Question Number 130448 by physicstutes last updated on 25/Jan/21

$$\mathrm{Given}\mathrm{that}f\left(x\right)=f(\pi -x),\mathrm{prove}\mathrm{that}{\int }_0^{\pi }xf\left(x\right)dx=\frac{\pi }{2}{\int }_0^{\pi }f\left(x\right)dx.$$$$\mathrm{please}\mathrm{what}\mathrm{are}\mathrm{different}\mathrm{methods}\mathrm{to}\mathrm{approach}\mathrm{this}\mathrm{question}?$$

Answered by TheSupreme last updated on 25/Jan/21

$${\int }_0^{\pi }xf\left(x\right)dx$$$$u=\pi -x$$$$du=-dx$$$${\int }_{\pi }^0(\pi -u)f(\pi -u)(-du)={\int }_0^{\pi }xf\left(x\right)$$$${\int }_0^{\pi }\pi f(\pi -u)du-{\int }_0^{\pi }uf(\pi -u)du={\int }_0^{\pi }xf\left(x\right)dx$$$$than,f\left(u\right)=f(\pi -u);setI={\int }_0^{\pi }xf\left(x\right)dx$$$${\int }_0^{\pi }\pi f\left(u\right)du=2{\int }_0^{\pi }uf\left(u\right)du$$$${\int }_0^{\pi }\frac{\pi }{2}f\left(u\right)du={\int }_0^{\pi }uf\left(u\right)du$$$$$$

Commented by physicstutes last updated on 25/Jan/21

$$\mathrm{thanks}\mathrm{sir}$$

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