Prove: $\int_a^b f(x) g(x) dx = f(a) \int_a^c g(x) dx + f(b) \int_c^b g(x) dx$ 
Let $f$ and $g$ be real-valued continuous functions on a closed and bounded interval $[a,b]$. Given that $f$ is
  increasing and $g$ is positive on $[a,b]$. Prove the following:  $$
 \int_a^b f(x) g(x) dx = f(a) \int_a^c g(x) dx + f(b) \int_c^b g(x) dx
 $$ where $c$ $ϵ$ $[a,b]$.

I would have attempted this by working from the RHS, which gives the following: 
$$f(a) \int_a^c g(x) dx + f(b) \int_c^b g(x) dx = f(a) [g(c)-g(a)] + f(b) [g(b)-g(c)]$$
by Fundamental Theorem of Calculus (Part II). 
Subsequently, i would expand it as: 
$$ f(a)g(c)- f(a)g(a) + f(b)g(b) - f(b)g(c) $$
Working from LHS, 
$$ \int_a^b f(x) g(x) dx = f(b)g(b) - f(a)g(a) $$
As such, 
$ f(a)g(c) - f(b)g(c) \implies g(c) [f(a) - f(b)] $ must be $0$. 
How do i derive the above to complete this proof? 

Write $H(x)=f(a)\int_a^xg(x)dx+f(b)\int_x^bg(x)dx-\int_a^bf(x)g(x)dx$. $H(a)\geq 0$ and $H(b)\leq 0$.
$H(a)=\int_a^b(f(b)-f(x))g(x)\geq 0$ since $f(b)\geq f(x)$ and $g(x)\geq 0$ $H(b)=\int_a^b(f(a)-f(x))g(x)\leq 0$, so there exists $c$ such that  $H(c)=0$ i.e
$f(a)\int_a^cg(x)d(x)+f(b)\int_c^bg(x)dx=\int_a^bf(x)g(x)dx$.
