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I need to show this,

Suppose that $f,g: [a,b] \rightarrow \mathbb{R}$ are continous. Show that exist $\eta \in (a,b)$ such that $$g(\eta)\int_a^\eta f(x)dx=f(\eta)\int_\eta^b g(x)dx $$

I defined $F(x):= \int_a^x f(x)dx$, $G(x):= \int_a^x g(x)dx$ and $ h: [a,b] \rightarrow \mathbb{R}$, $h(x)=F(x)(G(x)-G(b))$, $h'$ exist, so

$$h'(x)=F'(x)(G(x)-G(b))+ G'(x)F(x)$$ $$=f(x) \left( \int_a^x g(x)dx -\int_a^b g(x)dx \right)+g(x) \int_a^x f(x)dx$$ $$=f(x)\left(-\int_x^b g(x)dx \right)+g(x) \int_a^x f(x)dx$$ $$= g(x) \int_a^x f(x)dx-f(x)\int_x^b g(x)dx $$

As $h$ is continuous and differentiable en $(a,b)$ and $h(a)=h(b)=0$, exists $c \in (a,b)$ such that $h'(c)=h(c)$.

$$h'(c)=g(c) \int_a^c f(x)dx-f(c)\int_c^b g(x)dx$$ and $$h(c)=\left(\int_a^c f(x)dx \right)\left(-\int_c^b g(x)dx \right)$$

In this step i'm stuck, can someone help me?

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  • $\begingroup$ By Rolle’s theorem there exists $c$ such that $h’(c) = 0$. With this you are done. $\endgroup$ – RRL Mar 12 at 5:03

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