Fubini's theorem applicaction

Question

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

Answer 1

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.

Answer 2

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 $$\int_S f(z) \>{\rm d}(x,y,z)={1\over2}\int_0^a f(z)\,(a-z)^2\>dz\ .$$