Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
Do you know Fubini's theorem? –  Srivatsan Dec 5 '11 at 20:46
    
Yeah I do, I should try using it. Thanks! –  Jozef Dec 5 '11 at 20:52

3 Answers 3

up vote 6 down vote accepted

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.

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

HINT

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.

share|improve this answer

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.

share|improve this answer
    
Nice solution Pacciu.... –  juantheron Nov 6 '13 at 2:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.