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So, I thought I had a good grasp on how Fubini's theorem works, but then I found this exercise:

Let $\;f: [0,a] \rightarrow \mathbb{R}$ be a continuous function Use Fubini's theorem to prove

\begin{align} \int_{0}^{a}\left(\int_{0}^{x}\left(\int_{0}^{y} f(z) dz \right)dy \right)dx = \frac{1}{2}\int_{0}^{a}f(z)(a-z)^2dz \end{align}

So... I tried changing the integration order in every possible way, but the integral on the right never came out. I thought about using the fundamental theorem by defining $\;F(y) = \int_{0}^{y} f(z)dz$ and going from there, but to be honest I'm kind of clueless.

Any insight would be greatly appreciated

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    $\begingroup$ Try doing the iterated integral in the following order $$ \int_0^a f(z)\left[\int_z^a\left[\int_y^a 1\,dx\right]\,dy\right]\,dz. $$ $\endgroup$ Dec 5 '16 at 1:39
  • $\begingroup$ Oh, I think I got this now. So the whole point of this question is to represent the region in a equivalent way? For instance, \begin{align} R = \{(x,y,z) \in R^3 | y \leq x \leq a , z \leq y \leq a, 0 \leq z \leq a\} \end{align} This definition is equivalent to that implied by the integration limits in my original question, and the integral over it would be \begin{align} \int_{y}^{a} \int_{z}^{a} \int_0^{a} f(z) dz dy dz \end{align} And then we can apply Fubini's theorem $\endgroup$
    – user396072
    Dec 5 '16 at 2:31
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When integrating over a triangle, you can use an indicator function of the triangle and apply Fubini Theorem: $\newcommand{\dif}{\mathrm d}$ $$\begin{align*} \int_0^a \int_0^x \int_0^y f(z)\,\dif z\,\dif y\,\dif x &= \int_0^a \int_0^x \int_0^x [z \leq y]f(z)\,\dif z\,\dif y\,\dif x \\ &= \int_0^a \int_{[0,x]^2} [z \leq y]f(z)\,\dif (z \times y)\,\dif x \\ &= \int_0^a \int_0^x \int_0^x [z \leq y]f(z)\,\dif y\,\dif z\,\dif x \\ &= \int_0^a \int_0^x f(z) \left(\int_0^x [z \leq y] \dif y \right)\dif z\,\dif x \\ &= \int_0^a \int_0^x f(z) (x - z)\,\dif z\,\dif x \\ \end{align*}$$ Can you continue from here? $[\cdot]$ is the Iverson bracket.

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  • $\begingroup$ Huh, what strange notations. $\endgroup$ Dec 5 '16 at 1:47
  • $\begingroup$ Hm, I was not familiar with such notation. I take it that by "indicator function" you mean "characteristic function"? $\endgroup$
    – user396072
    Dec 5 '16 at 2:13
  • $\begingroup$ @user396072 Yes. Indicator function is the same as characteristic function. $[z \leq y]$ can also be interpreted as the indicator of a triangle. $\endgroup$ Dec 5 '16 at 2:14
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Your domain of integration $S$ is determined by the inequalities $$0\leq z\leq y\leq x\leq a\ ,\tag{1}$$ and is a simplex with vertices $(0,0,0)$, $(a,0,0)$, $(a,a,0)$, $(a,a,a)$. The function to be integrated depends on $z$ alone. Fubini's theorem, aka "Cavalieri's principle", then says that $$\int_S f(z) \>{\rm d}(x,y,z)=\int_0^a f(z)\,{\rm area}(S_z)\>dz\ ,\tag{2}$$ whereby $S_z$ is the shape cut out from $S$ by a horizontal plane at level $z$. Now from $(1)$ it follows that $$S_z=\{(x,y)\,|\,z\leq y\leq x\}\qquad(0\leq z\leq a)\ ,$$ hence $S_z$ is a triangle of area ${1\over2}(a-z)^2$. Plugging this into $(2)$ gives $$\int_S f(z) \>{\rm d}(x,y,z)={1\over2}\int_0^a f(z)\,(a-z)^2\>dz\ .$$

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