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## Homework Statement

Let f be continuous. Show

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

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

## Homework Equations

Fubini's Theorem:

∫∫∫[itex]_{B}[/itex] f(x,y,z) dV = [itex]^{s}_{r}∫[/itex][itex]^{d}_{c}∫[/itex] [itex]^{b}_{ra}∫[/itex] 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

[itex]^{z}_{0}∫[/itex][itex]^{1}_{0}∫[/itex][itex]^{y}_{0}∫[/itex]f(x) dz dy dx

which if you work it out, becomes

[itex]\frac{1}{2}[/itex][itex]^{z}_{0}∫[/itex]f(x)dx, which I don't know how to have equal to

[itex]\frac{1}{2}[/itex] [itex]^{1}_{0}∫[/itex] (1-x)[itex]^{2}[/itex] 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).