# 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?

## 1 Answer

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

$$\int_a^b f(x) g(x) \, dx = f(b)G(b) - f(b)G(\theta) = f(b) \int_\theta^b g(x) \, dx$$

• Was this of any help to you? – RRL Apr 2 at 13:58