Use Fubini's Theorem to find a function $g(z)$ such that $\int_0^a\int_0^x\int_0^y f(z)\,dz\,dy\,dx=\int_0^a g(z)f(z)\,dz$ where $a$ is a real number.
I tried to use Fubini's Theorem,
$$\int_0^a\int_0^x\int_0^y f(z)\,dz\,dy\,dx =\int_0^a\int_z^a\int_z^x f(z)\,dy\,dx\,dz =\int_0^a (z^{2}/2+a^{2}/2-2a)f(z)\,dz$$
Hence, $g(z)=z^{2}/2+a^{2}/2-2a$.
Is it correct? Or I need to write more steps? Can any give me an answer?
\begin{align} & \int_0^a \left( \int_0^x \left( \int_0^y f(z)\,dz \right) \,dy \right) \,dx \\[10pt] = {} & \iiint\limits_{(x,y,z)\,:\, 0\,\le\,z\,\le\,y\,\le\,x\,\le\,a} f(z)\, d(z,y,x) \\[10pt] = {} & \int_0^a \left( \int_z^a \left( \int_z^x f(z) \, dy \right) \,dx \right) \, dz \end{align} As $y$ goes from $z$ to $x$ and $x$ goes from $z$ to $a,$ the value of $f(z)$ does not change, so it can be pulled out of the two inner integrals: $$ \int_0^a f(z) \left( \int_z^a \left( \int_z^x 1 \, dy \right) \, dx \right) \, dz $$ So \begin{align} g(z) = {} & \int_z^a \left( \int_z^x 1 \, dy \right) \, dx \\[8pt] = {} & \int_z^a (x-z) \, dx = \left[ \frac {x^2} 2 - zx \right]_{x:=z}^{x:=a} \\[8pt] = {} & \left( \frac {a^2} 2 - az \right) - \left( \frac{z^2} 2 - z^2 \right) = \frac{a^2-z^2} 2 - (az-z^2) \\[8pt] = {} & \frac{(a-z)^2} 2. \end{align}