# Showing an iterated integral is equal to an integral

## Homework Statement

Let f be continuous. Show

$^{1}_{0}∫$ $^{y}_{0}∫$ $^{z}_{0}∫$ f(x) dx dz dy = $\frac{1}{2}$ $^{1}_{0}∫$ (1-x)$^{2}$ f(x) dx

Hint: Use Fubini to rewrite the left expression into an iterated integral which ends in dx.

## Homework Equations

Fubini's Theorem:
∫∫∫$_{B}$ f(x,y,z) dV = $^{s}_{r}∫$$^{d}_{c}∫$ $^{b}_{ra}∫$ f(x,y,z) dxdydz

## The Attempt at a Solution

So I've been playing around with this problem for a little bit and I'm stumped.
Fubini's theorem is the theorem that states that if f is continuous then you can treat each integral as it's own separate entity right?

So with constants, yea this would be pretty simple, but two of the ends we're measuring at are variables (namely the z and y).

Well, I tried following the hint and using Fubini's theorem and I got

$^{z}_{0}∫$$^{1}_{0}∫$$^{y}_{0}∫$f(x) dz dy dx
which if you work it out, becomes
$\frac{1}{2}$$^{z}_{0}∫$f(x)dx, which I don't know how to have equal to
$\frac{1}{2}$ $^{1}_{0}∫$ (1-x)$^{2}$ f(x) dx

Now, I think/know this is wrong. It's probably because I didn't use Fubini's theorem right or adjust the limits correctly (although involving problems with constant number values, that's what I did).

## The Attempt at a Solution

Well, this is weird.
It's not letting me edit my problem. But here's the latest update to what I have done.

Well I "cheated."
But I substituted s for f(x).

So instead of $^{1}_{0}∫$ $^{y}_{0}∫$ $^{z}_{0}∫$ f(x) dx dz dy
it's
$^{1}_{0}∫$ $^{y}_{0}∫$ $^{z}_{0}∫$ s dx dz dy

that way I followed through on the integral, and it became $\frac{1}{6}$ s or $\frac{1}{6}$ f(x) if you decide to substitute s back for f(x).
And that's the exact thing you get with the right hand side as well.

I'm not sure if that's formal enough for showing.

Dick
Homework Helper
Well, this is weird.
It's not letting me edit my problem. But here's the latest update to what I have done.

Well I "cheated."
But I substituted s for f(x).

So instead of $^{1}_{0}∫$ $^{y}_{0}∫$ $^{z}_{0}∫$ f(x) dx dz dy
it's
$^{1}_{0}∫$ $^{y}_{0}∫$ $^{z}_{0}∫$ s dx dz dy

that way I followed through on the integral, and it became $\frac{1}{6}$ s or $\frac{1}{6}$ f(x) if you decide to substitute s back for f(x).
And that's the exact thing you get with the right hand side as well.

I'm not sure if that's formal enough for showing.

That's only going to work if f(x) is a constant (like s). It's not going to work for a general function of x. You want visualize the domain of integration. It's a tetrahedron, right? Now rewrite the triple integral so the dx integration is last. See if you can get the correct limits for the dy and dz integrations.