We are asked to prove the following integral;

$\int_0^x(\int_0^tf(u) du) dt=\int_0^xf(u)(x-u)du$

I know that we have to use integration by parts which when we differentiate a function will help us use the fundamental theorem of calculus. I just don't know when and were to apply this.

Thanks in advance!

  • $\begingroup$ The first step: $\int_0^x(\int_0^tf(u)\mathrm{d}u)\mathrm{d}t=[t(\int_0^tf(u)\mathrm{d}u)]_0^x-\int_0^xt\mathrm{d}(\int_0^tf(u)\mathrm{d}u)$ $\endgroup$ – Yuta Apr 30 at 5:54

This is analogous to the summation

$\begin{array}\\ \sum_{i=0}^n \sum_{j=0}^i f(j) &=\sum_{j=0}^n \sum_{i=j}^n f(j)\\ &=\sum_{j=0}^n f(j)\sum_{i=j}^n 1\\ &=\sum_{j=0}^n f(j)(n-j+1)\\ \end{array} $


$\begin{array}\\ \int_0^x(\int_0^tf(u) du) dt &=\int_0^x\int_u^xf(u) dt du\\ &=\int_0^xf(u)\int_u^x1 dt du\\ &=\int_0^xf(u)(x-u) du\\ \end{array} $

The change in the order of integration follows from the fact that $0 \le t \le x $ and $0 \le u \le t$ is the same as $0 \le u \le x$ and $u \le t \le x$.

  • $\begingroup$ I'm not quite sure what you are doing here. I haven't seen this summation before. $\endgroup$ – Artem Pulemotov Apr 30 at 6:09

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