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Question Number 130448 by physicstutes last updated on 25/Jan/21
Given that f(x) = f(π−x), prove that ∫_0 ^π xf(x)dx = (π/2)∫_0 ^π f(x)dx. please what are different methods to approach this question?
$$\mathrm{Given}\:\mathrm{that}\:{f}\left({x}\right)\:=\:{f}\left(\pi−{x}\right),\:\mathrm{prove}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\pi} {xf}\left({x}\right){dx}\:=\:\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {f}\left({x}\right){dx}. \\ $$$$\mathrm{please}\:\mathrm{what}\:\mathrm{are}\:\mathrm{different}\:\mathrm{methods}\:\mathrm{to}\:\mathrm{approach}\:\mathrm{this}\:\mathrm{question}? \\ $$
Answered by TheSupreme last updated on 25/Jan/21
∫_0 ^π xf(x)dx u=π−x du=−dx ∫_π ^0 (π−u)f(π−u)(−du)=∫_0 ^π xf(x) ∫_0 ^π πf(π−u)du−∫_0 ^π uf(π−u)du=∫_0 ^π xf(x)dx than, f(u)=f(π−u); set I=∫_0 ^π xf(x)dx ∫_0 ^π πf(u)du=2∫_0 ^π uf(u)du ∫_0 ^π (π/2)f(u)du=∫_0 ^π uf(u)du
$$\int_{\mathrm{0}} ^{\pi} {xf}\left({x}\right){dx} \\ $$$${u}=\pi−{x} \\ $$$${du}=−{dx} \\ $$$$\int_{\pi} ^{\mathrm{0}} \left(\pi−{u}\right){f}\left(\pi−{u}\right)\left(−{du}\right)=\int_{\mathrm{0}} ^{\pi} {xf}\left({x}\right) \\ $$$$\int_{\mathrm{0}} ^{\pi} \pi{f}\left(\pi−{u}\right){du}−\int_{\mathrm{0}} ^{\pi} {uf}\left(\pi−{u}\right){du}=\int_{\mathrm{0}} ^{\pi} {xf}\left({x}\right){dx} \\ $$$${than},\:{f}\left({u}\right)={f}\left(\pi−{u}\right);\:{set}\:{I}=\int_{\mathrm{0}} ^{\pi} {xf}\left({x}\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \pi{f}\left({u}\right){du}=\mathrm{2}\int_{\mathrm{0}} ^{\pi} {uf}\left({u}\right){du} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{\pi}{\mathrm{2}}{f}\left({u}\right){du}=\int_{\mathrm{0}} ^{\pi} {uf}\left({u}\right){du} \\ $$$$ \\ $$
Commented by physicstutes last updated on 25/Jan/21
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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