How do I go about choosing the correct bounds for this probability density function?

Question

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?

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Checking if this integrates to 1 under the correct bounds ($\int_{0}^{1}\int_{0}^{(1-y)}$):

$\int_{0}^{1}\int_{0}^{1-y}24xy \quad dxdy$

=$\int_{0}^{1}(12x^2y\Big|_0^{(1-y)}) \quad dy$

=$\int_{0}^{1}(12(1-y)^2y) \quad dy$

=$\int_{0}^{1}(12y^3-24y^2+12y) \quad dy$

=$(3y^4-8y^3+6y^2)\Big|_0^1$

=$3-8+6$

=$1$

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

Answer 1

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?

Answer 2

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] $

$I = 24\bigg[\frac{1}{64} - \frac{1}{2}[\frac{z^3}{3} - \frac{z^4}{4}]^{0}_{1/2}\bigg]$

$I = 24\bigg[\frac{1}{64} - \frac{1}{2}[0-(\frac{1}{24} - \frac{1}{32})]\bigg]$

$I = \frac{24}{48} = \frac{1}{2}$