Assume that $f:[a,b]\rightarrow\mathbb{R}$ is continuous on $[a,b]$ and $g:[a,b]\rightarrow\mathbb{R}$ is integrable and $g(x)\geq0$ for all $x\in[a,b]$. Then there exists a $c\in(a,b)$ such that $$ f(c)\int_{a}^{b}g(x)dx=\int_{a}^{b}g(x)f(x)dx$$
Suppose that $f:[a,b]\rightarrow\mathbb{R}$ is continuous on $[a,b]$ and $g:[a,b]\rightarrow\mathbb{R}$ is integrable on $[a,b]$. Since $[a,b]$ is a compact set, then by the extreme value theorem, there exist $x_1,x_2\in[a,b]$ such that $f(x_1)\leq f(x)\leq f(x_2)$. As $g(x)\geq 0$ for all $x\in[a,b]$, thus $g(x)f(x_1)\leq g(x)f(x)\leq g(x)f(x_2)$. Since $g$ is integrable on $[a,b]$, so $\int_{a}^{b}g(x)dx>0$; thus $$\int_{a}^{b}g(x)f(x_1)dx\leq\int_{a}^{b}g(x)f(x)dx\leq\int_{a}^{b} g(x)f(x_2)dx\Longrightarrow f(x_1)\leq\frac{\int_{a}^{b}f(x)g(x)dx}{\int_{a}^{b}g(x)dx}\leq f(x_2)$$ Now, apply the Intermediate Value Theorem for $f$, there exists a $c\in(a,b)$ satisfy $$f(c)=\frac{\int_{a}^{b}f(x)g(x)dx}{\int_{a}^{b}g(x)dx}\Longrightarrow f(c)\int_{a}^{b}g(x)dx=\int_{a}^{b}g(x)f(x)dx$$
Can someone give me a hint to write a different proof? Thanks
The proof contains a small gap, which can be fixed, I guess. You say that $g(x) \geq 0$ yet $\int\limits_{a}^{b} g(x) dx >0$. What happens if $g(x)$ is identically zero?