Let $f:R\to R$ be a continuous function, and let $F:R\to R$ given by: $F(x)=\int_a^x xf(t)dt$, justify if it's differentiable

Question

So we know that for a function to be differentiable, the next must limit exists: $$\lim\limits_{h\to 0}f(x)=\frac{f(x+h)-f(x)}{h}$$ Then, $$\lim\limits_{h\to 0}\frac{(x+h)f(x+h)-xf(x)}{h}$$ And we can interpret the factor $(x+h)$ accordingly to the variable of wich the function depends, so $(x+\Delta t)$

Then we have: $$\lim\limits_{\Delta t\to 0}\frac{(x+\Delta t)f(x+\Delta t)-xf(x)}{\Delta t}$$ $$\lim\limits_{\Delta t\to 0}\frac{1}{\Delta t}[(x+\Delta t)f(x+\Delta t)-xf(x)]$$

Now, $$\lim\limits_{\Delta t\to 0}\frac{1}{\Delta t}(\int_a^{(x+\Delta t)}xf(t)dt - \int_a^x xf(t)dt)$$ $$\lim\limits_{\Delta t\to 0}\frac{1}{\Delta t}(\int_x^{x+\Delta t} xf(t)dt)$$

Until know, I think I need to justify why it is possible to let $dt$ transform into $\Delta t$, by the approximation of the integral, with the product $xf(t)\Delta t$, but I'm not quite sure about this. So, is anything wrong with the process until now? My last thought is correct?

Answer 1

Let denote $\phi$ the function such that:

$$\forall x\in\mathbb{R}, \phi(x) := \int_a^x f(t)dt.$$

You have then that $\phi$ is differentiable and more precisely:

$$\forall x\in\mathbb{R},\phi'(x) = f(x).$$

Your function $F$ can be written as:

$$\forall x\in\mathbb{R},F(x)=x\phi(x).$$

The function $F$ is then the product of two differentiable functions, so is $F$. This answer your question, but you can go further ! :) We have that:

$$\forall x\in\mathbb{R},F'(x)=\phi(x)+x\phi'(x).$$

By what has been said before, you have then that:

$$\forall x\in\mathbb{R},F'(x)=\phi(x)+xf(x).$$

You can rewrite it as:

$$\forall x\in\mathbb{R},F'(x)=\frac{F(x)}{x}+xf(x).$$

You have then a ODE on the function $F$ not that hard to solve... This additional comment is just here to show how one can use the derivative of integrals to find nice ways to determine explicit solution...