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Assume $f(x),g(x)$ is continuous on $[a,b]$. show that there exists $\xi \in [a,b]$, such that $$g(\xi)\int_a^\xi f(x)\text{d}x=f(\xi)\int_\xi^b g(x)\text{d}x$$

I tried to use intermediate value theorem to $F(t) = g(t)\int_a^t f(x)\text{d}x-f(t)\int_t^b g(x)\text{d}x$. but I failed to find two opposite values.

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Let $$ F(t)=\int_a^tf(x)\:dx\int_t^bg(x)\:dx $$ Then $F$ is differentiable on $[a,b]$ for $f, g$ are continuous on $[a,b]$ and $$ F'(t)=f(t)\int_{t}^bg(x)\:dx-g(t)\int_a^{t}f(x)\:dx $$ Since $F(a)=F(b)=0$, by Roll's theorem, there is a $\: \xi\in (a,b)$ such that $$ F'(\xi)=f(\xi)\int_{\xi}^bg(x)\:dx-g(\xi)\int_a^{\xi}f(x)\:dx=0 $$ i.e. $$ f(\xi)\:dx\int_{\xi}^bg(x)\:dx=g(\xi)\int_a^{\xi}f(x)\:dx $$

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Let $F(e)=\int_a^ef(x)dx$ and $G(e)=\int_e^bg(x)dx$. We have $F'(e)=f(e)$ and $G'(e)=-g(e)$ so $$g(e)F(e)-f(e)G(e)=-G'(e)F(e)-F'(e)G(e)=-(GF)'(e).$$ Now $F(a)=G(b)=0$ so $(FG)(a)=(FG)(b)=0$ so by Rolle's theorem (a special case of IVT)$$\exists e\in (a,b) : (FG)'(e)=0.$$

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