I'd love your help with proving that:

For continuous function $f$, $$\int_{0}^{x} \left[\int_{0}^{t}f(u) \; du \right] \; dt = \int_{0}^{x} f(u)(x-u)du$$

I'm not quite sure what I should do with this.

From Newton-Leibniz theorem I get that the left side of the equation is $(F(t)-F(0))x$, while on the right side I wanted to do an integration by parts, but I'm not sure how or if it is the direction.

Thanks a lot for the help.

  • 2
    $\begingroup$ Do you know Fubini's theorem? $\endgroup$ – Srivatsan Dec 5 '11 at 20:46
  • $\begingroup$ Yeah I do, I should try using it. Thanks! $\endgroup$ – Jozef Dec 5 '11 at 20:52

Let $F(x)=\int_{0}^{x}f(u)(x-u)du = x \int_{0}^x f(u)du -\int_{0}^x uf(u)du$.

Now, we know how to take the derivative of all these terms, $x$ and the two integrals, so we can compute:

$F'(x) = \int_0^x f(u) du + xf(x) - xf(x) = \int_0^x f(u) du$.

But that means that $F(x) = C + \int_0^x F'(u) du = C + \int_0^x (\int_0^u f(t)dt)du$ for some constant $C$. But clearly, using $x=0$, $C=0$, so you are done.

  • $\begingroup$ Thank you for the answer. Can you please show me how do you take derivative of $\int_{0}^x uf(u)du$? Thanks again $\endgroup$ – Jozef Dec 6 '11 at 9:13
  • $\begingroup$ The derivative of $\int_0^x g(u) du$ is $g(x)$. Let $g(x)=xf(x)$. $\endgroup$ – Thomas Andrews Dec 6 '11 at 13:14

You should integrate by parts the LHside with differential factor $\text{d} t$.

In fact, since $f$ is continuous then $F(t):=\int_0^t f(u)\ \text{d} u$ is of class $C^1$ and its derivative is $f(t)$; hence: $$\begin{split} \int_0^x \left[\int_0^t f(u)\ \text{d} u\right]\ \text{d} t &= t\ \int_0^t f(u)\ \text{d} u\Bigg|_0^x - \int_0^x t\ f(t)\ \text{d} t\\ &= x\int_0^x f(t)\ \text{d} t -\int_0^x t\ f(t)\ \text{d} t\\ &= \int_0^x (x-t)\ f(t)\ \text{d} t\; ,\end{split}$$ which is the claim.

  • $\begingroup$ Nice solution Pacciu.... $\endgroup$ – juantheron Nov 6 '13 at 2:14


There are couple of ways.

The first one is to change the order of integration using Fubini's theorem and this immediately gives you the answer.

The second way is a bit roundabout. Consider $$F(x) = \int_0^x \int_0^t f(u) du dt - \int_0^x f(u) (x-u) du. $$ Prove that $F'(x) = 0$ using Leibniz integration rule and hence $F(x) = F(0) = 0$ giving us the desired result.


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