How do I go about choosing the correct bounds for this probability density function? Given the function $f(x,y) = 24xy \quad$  for $\quad 0 \le x \le 1 \quad$, $\quad 0 \le y \le 1 \quad$, and $\quad 0 \le x+y \le 1 \quad$.
I want to show that the double integral of the function within the given bounds yields 1.
My bounds are set as the following:
$\int_{0}^{1}\int_{0}^{y}24xy \quad dxdy$. 
However, this doesn't yeild 1; the bounds are set incorrectly. How do I go about choosing the correct bounds?

Checking if this integrates to 1 under the correct bounds ($\int_{0}^{1}\int_{0}^{(1-y)}$):
$\int_{0}^{1}\int_{0}^{1-y}24xy \quad dxdy$ 
=$\int_{0}^{1}(12x^2y\Big|_0^{(1-y)}) \quad dy$ 
=$\int_{0}^{1}(12(1-y)^2y) \quad dy$
=$\int_{0}^{1}(12y^3-24y^2+12y) \quad dy$ 
=$(3y^4-8y^3+6y^2)\Big|_0^1$
=$3-8+6$
=$1$

Here is an image of the original question from the text for reference, as some people think the bounds may be erroneous:

 A: You need both $x \leq y$ and $x \leq 1-y$. For $y \leq \frac  1 2 $ this gives $0 
\leq x \leq y$ and for $y>\frac   1 2 $ we get $0 \leq x \leq 1-y$. Can you now compute the integral?
A: I have attached the graph here. The intersecting points of the three regions are as shown.

So, the integral becomes,
$I = \int_{y=0}^{1/2}\int_{x=0}^{y}\ \ 24\ xy\ dxdy + \int_{y=1/2}^{1}\int_{x=0}^{1-y}\ \ 24\ xy\ dxdy$
$I = 24\big[\int_{y=0}^{1/2}y.\frac{y^2}{2}dy + \int_{y=1/2}^{1}y.\frac{(1-y)^2}{2}dy\big] $
Substitute $z = 1-y \ in \ the\ second \ integral$.
$dz = -dy\  $
At $y = 1/2, z = 1 - 1/2 = 1/2$ 
and at $y=1, z=0$
So, 
$I = 24\big[\int_{y=0}^{1/2}y.\frac{y^2}{2}dy + \int_{z=1/2}^{0}(1-z).\frac{(z)^2}{2}(-dz)\big]$
$I = 24\bigg[[\frac{y^4}{8}]^{1/2}_0 - \frac{1}{2}\int_{z=1/2}^{0} (z^2 - z^3)dz \bigg] $
$I = 24\bigg[\frac{1}{64} - \frac{1}{2}[\frac{z^3}{3} - \frac{z^4}{4}]^{0}_{1/2}\bigg]$
$I = 24\bigg[\frac{1}{64} - \frac{1}{2}[0-(\frac{1}{24} - \frac{1}{32})]\bigg]$
$I = \frac{24}{48} = \frac{1}{2}$
