Why Does This Integral in a Cube Differ from Wolfram Alpha's Solution?

[aija]

This is part of an example solution to a problem about integrating a function in a [0,1]x[0,1]x[0,1] cube. I just don't understand how the midst function is integrated like in the attached picture. This is the same integral in wolfram alpha and it gives a different solution:
http://www.wolframalpha.com/input/?i=intg((yz^2)*e^(-xyz))

 

[STEMucator]

This is part of an example solution to a problem about integrating a function in a [0,1]x[0,1]x[0,1] cube. I just don't understand how the midst function is integrated like in the attached picture. This is the same integral in wolfram alpha and it gives a different solution:
http://www.wolframalpha.com/input/?i=intg((yz^2)*e^(-xyz))

If you're choosing to integrate over a nice cube, the order of integration does not matter I believe.

Your integration for dx is wrong. You may want to re-integrate it.

 

[aija]

If you're choosing to integrate over a nice cube, the order of integration does not matter I believe.

Your integration for dx is wrong. You may want to re-integrate it.

You mean the integral in the attachment is wrong? It's part of an example solution not done by me so I thought it would be right but it would help a lot to know that there is an error in the example solution.

 

[STEMucator]

You mean the integral in the attachment is wrong? It's part of an example solution not done by me so I thought it would be right but it would help a lot to know that there is an error in the example solution.

Indeed, your first integration is really :

$\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} yz^2e^{-xyz}dxdydz$
$= \int_{0}^{1} \int_{0}^{1} yz^2\int_{0}^{1} e^{-xyz}dxdydz$
$= \int_{0}^{1} \int_{0}^{1} z - ze^{-yz}dydz$

The rest shouldn't be too hard :)

 

[aija]

Ok thanks, it's clear now