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?

  • 3
    $\begingroup$ Bring the $x$ out of the integral. Then product rule and FTC answer your question. $\endgroup$
    – peek-a-boo
    Sep 18 at 21:18

1 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...


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .