# Riemman-Stieltjes Integration-IVT

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?

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