Finding joint density function of two independent random variables 
Let $X$ have density $f_X(x) = 2x$ for $x\in(0,1)$ and zero otherwise. Let $Y$ be uniform on the interval $[1,2].$ Assume that $X$ and $Y$ are independent.
a) Find the joint PDF of $(X,Y)$. Use this to compute $\mathbb P(Y-X\geq3/2).$
b) Find the PDF of $X+Y$.

$\textbf{My Thoughts:}$ For part (a), I have the following. Since both variables are independent, I compute
$$f_{XY}(x,y)=f_X(x)f_Y(y)=\begin{cases}2x&\text{if }x\in(0,1)\,\text{and}\,y\in[1,2]\\0&\text{otherwise. }\end{cases}$$
Then I have
\begin{align*}\mathbb P(Y-X\geq3/2)&=1-\mathbb P(Y-X<3/2)\\
&=1-\int_{1/2}^{2}\int^{1/2}_{0}2x\,dx\,dy\end{align*}
For part (b) I compute, for $z\in [2,3]$
\begin{align*}
f_{X+Y}(z)&=\int_{-\infty}^{\infty}f_{X}(t)f_{Y}(z-t)\,dt\\
&=\int^{1}_{0}2t\,dt\\
&=1,
\end{align*}
and I have that $f_{X+Y}=0$ otherwise from the above calculation.

Is any of the above correct, especially the bounds on the double integral in part (a)? I am having trouble visualizing the region over which I have to integrate.

Thank you for your time and for any help.
 A: First of all for $(a)$ $$\begin{align} \Bbb P \left (Y-X<\frac 3 2 \right ) & = \int_{1}^{\frac 3 2} \int_{0}^{1} 2x\ \text {dx dy} + \int_{\frac 3 2}^{2} \int_{y-\frac 3 2}^{1} 2x\ \text {dx dy}. \\ & = \frac  1 2 + \frac {11} {24}. \\ & = \frac {23} {24}. \end{align}$$ So $$\Bbb P \left (Y-X \geq \frac 3 2 \right ) = 1 - \frac {23} {24} = \frac {1} {24}.$$
Finally for $(b)$ if $Z=X+Y$ and $f_Z(z)$ denotes the PDF of $Z$ then
$$
f_Z(z) = \left\{
        \begin{array}{ll}
            z^2 - 2z + 1 & \quad 1 < z < 2 \\
            4z - z^2 - 3 & \quad 2 \leq z < 3 \\
            0 & \quad \text {elsewhere}
        \end{array}
    \right.
$$
A: For (a) see the answer of DBchatto67.
Let $\left[x+y\leq z\right]$ denote the function $\mathbb{R}^{3}\to\mathbb{R}$
that takes value $1$ if $x+y\leq z$ and takes value $0$ otherwise.
Then for the CDF of $Z:=X+Y$ we find that 
$\begin{aligned}F_{Z}\left(z\right)=P\left(X+Y\leq z\right) & =\int\int\left[x+y\leq z\right]f_{X}\left(x\right)f_{Y}\left(y\right)dydx\\
 & =\int_{0}^{1}\int_{1}^{2}\left[x+y\leq z\right]2xdydx\\
 & =2\int_{0}^{1}x\int_{1}^{2}\left[y\leq z-x\right]dydx
\end{aligned}
$
Now we discern cases.


*

*If $z\leq1$ then $P\left(X+Y\leq z\right)=0$.

*If $1<z<2$ then 
$\begin{aligned}P\left(X+Y\leq z\right) & =2\int_{0}^{z-1}x\int_{1}^{z-x}dydx\\
 & =2\int_{0}^{z-1}x\left(z-x-1\right)dx\\
 & =\frac{1}{3}\left(z-1\right)^{3}
\end{aligned}
$


*

*If $2\leq z<3$ then 


$\begin{aligned}P\left(X+Y\leq z\right) & =2\int_{0}^{1}x\int_{1}^{\min\left(2,z-x\right)}dydx\\
 & =2\int_{0}^{z-2}x\int_{1}^{2}dydx+2\int_{z-2}^{1}x\int_{1}^{z-x}dydx\\
 & =2\int_{0}^{z-2}xdx+2\int_{z-2}^{1}x\left(z-x-1\right)dx\\
 & =z-\frac{5}{3}-\frac{1}{3}\left(z-2\right)^{3}
\end{aligned}
$


*

*If $z\geq3$ then $P\left(X+Y\leq z\right)=1$
We find the PDF by differentiating the CDF
$f_{Z}\left(z\right)=\left(z-1\right)^{2}$ if $1<z<2$ and $f_{Z}\left(z\right)=1-\left(z-2\right)^{2}=\left(3-z\right)\left(z-1\right)$
if $2\leq z<3$ and $f_{Z}\left(z\right)=0$ otherwise.
