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Question Number 50413 by Abdo msup. last updated on 16/Dec/18

$$letf\in C^0(R,R)/\mathrm{\forall }x\in Rf(a+b-x)=f\left(x\right)$$$$1)find{\int }_a^{b}xf\left(x\right)dxintermsof{\int }_a^{b}f\left(x\right)dx$$$$2)calculate{\int }_0^{\pi }\frac{xdx}{1+sinx}$$

Commented by Abdo msup. last updated on 24/Dec/18

$$1){\int }_a^{b}xf\left(x\right)dx={\int }_a^{b}xf(a+b-x)dx$$$${=}_{a+b-x=t}-{\int }_b^a(a+b-t)f\left(t\right)dt$$$$={\int }_a^b(a+b)f\left(t\right)-{\int }_a^{b}tf\left(t\right)dt\Rightarrow$$$$2{\int }_a^{b}xf\left(x\right)dx=(a+b){\int }_a^{b}f\left(t\right)dt\Rightarrow$$$${\int }_a^{b}xf\left(x\right)dx=\frac{a+b}{2}{\int }_a^{b}f\left(x\right)dx$$$$2)wehave{\int }_0^{\pi }\frac{xdx}{1+sinx}={\int }_0^{\pi }xf\left(x\right)dxwith$$$$f\left(x\right)=\frac{1}{1+sinx}wehavef(0+\pi -x)=\frac{1}{1+sin(\pi -x)}$$$$=f\left(x\right)\Rightarrow {\int }_0^{\pi }xf\left(x\right)dx=\frac{\pi }{2}{\int }_0^{\pi }\frac{dx}{1+sinx}$$$${=}_{tan\left(\frac{x}{2}\right)=t}\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}\frac{1}{1+\frac{2t}{1+t^2}}\frac{2dt}{1+t^2}$$$$=\pi {\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2+2t}=\pi {\int }_0^{\mathrm{\infty }}\frac{dt}{{(t+1)}^2}$$$$=\pi {[-\frac{1}{t+1}]}_0^{+\mathrm{\infty }}=\pi .$$

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