Theorem related with mean value Integral for integrals 
Let $f,g: [a,b] \longrightarrow \mathbb{R},$ $f$ is monotone with $f(a)=0$, and $g$ is continuous.
Then there is $\theta \in [a,b] $ such that:
$$\displaystyle \int_a^b f(x)g(x)dx = f(b)\int_{\theta}^bg(x)dx.$$

My attempt:
If I consider that $\displaystyle \int f'(x)dx $ exists, then I define
$$G(x)= - \displaystyle \int_x^b g(t)dt $$
By partial integration :
$$\displaystyle \int_a^bf(x)g(x) = f(x)G(x)\bigg|_a^b - \int_a^bf'(x)G(x)dx\\[2em]
\implies \displaystyle \int_a^bf(x)g(x) =  - \int_a^bf'(x)G(x)dx$$
and by general mean value integral it's done, but I don't have the proof for the hypothesis that $\displaystyle \int f'(x)dx $ exists.
So, is there an other way to solve it?
 A: Presumably you mean that $f$ is monotone increasing for this to be true, and you are given no hypothesis regarding the existence of $f'$. 
Of course, a monotone increasing function is differentiable almost everywhere and the derivative is integrable.  However, we also would need to know that 
$$\int_a^b f'(x) \, dx = f(b) - f(a)$$
for your approach, and that may not always be the case without additional assumptions, e.g., $f$ absolutely continuous.
Fortunately, no information about $f'$ is really needed to prove this. One approach is to use Riemann sums.  This is straightforward but tedious.  
Another way is to use Riemann-Stieltjes integrals and partial integration.
Defining $G(x) = \int_a^x g(t) \, dt$, we have since G(a) = 0,
$$\int_a^b f(x) g(x) \, dx = \int_a^b f \, dG = f(b)G(b)- f(a) G(a)  - \int_a^b G \, df \\ = f(b)G(b)  - \int_a^b G \, df $$
Since $G$ is continuous, by the first mean value theorem for integrals there exists $\theta$ such that 
$$\int_a^b G \, df = G(\theta)\int_a^b df = G(\theta)(f(b) - f(a))$$
Thus, since $f(a) = 0$ we have
$$\int_a^b f(x) g(x) \, dx = f(b)G(b) - f(b)G(\theta) = f(b) \int_\theta^b g(x) \, dx$$
