<s>[INST] Joint PDF P[X+Y<=0.5] I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:
Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.
I need to find the $P(x+y\leq 0.5)$.
For the double integration, I have the following bounds:


*

*Outer bound is respectively to $x$ and is from $0$ to $0.5$.

*Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.
Here is the initial set up:
$\int_0^.5 \int_0^{.5-x}(x +y) dydx$.
My steps of integration:
1: $\int_0^.5 \left(xy + \frac{y^2}{2}\right)\Big|_0^{.5-x}dx$.
2: $\int_0^.5$ $\left(\frac{x}{2} -x^2+\frac{(0.5-x)^2}{2}\right) dx $.
3: $(\frac{x^2}{4} - \frac{x^3}{3} + \frac{(0.5-x)^3}{6})\Big|_0^{0.5}$.
I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?
 [/INST] Your made the sign error in step 2 (as indicated by KaviRamaMurthy).
As an alternative, you can simplify in step 2 before integrating:
$$\int_0^{0.5}\left(\frac{x}{2} -x^2+\frac{(0.5-x)^2}{2}\right) dx=
\int_0^{0.5}\left(\require{cancel}\cancel{\frac{x}{2}} -x^2+\frac{1}{8}-\cancel{\frac x2}+\frac{x^2}{2}\right) dx=\\
\int_0^{0.5}\left(-\frac{x^2}{2}+\frac18\right) dx=
\left(-\frac{x^3}{6}+\frac x8\right)|_0^{0.5}=
-\frac{1}{48}+\frac{1}{16}=
\frac 1{24}.$$
Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $\frac1{48}$:
$$(\frac{x^2}{4} - \frac{x^3}{3} \color{red}{-} \frac{(0.5-x)^3}{6})\Big|_0^{0.5}=\left(\frac{1}{16}-\frac1{24}-0\right)-\left(0-0-\frac1{48}\right)=\frac1{24}.$$
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