Find $P(X+Y > \frac{3}{2})$ given that X is uniformly distributed on [0,1] and the conditional distribution of Y given X=x is uniform on [1-x,1]. Suppose $X$ has the uniform distribution on the interval [0,1] and that the distribution of random variable $Y$ given that $X=x$ is uniform on the interval $[1-x,1].$
Find the probability that $P(X+Y > \frac{3}{2})$ given to 4 decimal places.
So far this is what I have:
$P(X+Y > \frac{3}{2}) = 1 - P(X+Y < \frac{3}{2})= 1 - P(Y < \frac{3}{2} - X)$
$P(Y < \frac{3}{2} - X) = \int_0^1 \int_{1-x}^{\frac{3}{2}-x} f_{XY}(x,y) \,dy \,dx = \int_0^1 \int_{1-x}^{\frac{3}{2}-x} \frac{1}{x} \,dy \,dx $
$= \int_0^1 [\frac{y}{x}]_{1-x}^{\frac{3}{2}-x} \,dx = \int_0^1 \frac{1}{2x} \, dx = [\frac{1}{2}ln(2x)]_0^1$
As $ln(2x)$ can't be evaluated at $0$ I can't go any further.
Have I done anything wrong and what is a better method to answer this question so that this problem doesn't occur?
Note: I calculated $f_{XY}(x,y)$ from the distributions of $f_X(x)$ and $f_{Y|X}(y|x)$ as $f_{XY}(x,y)=f_{Y|X}(y|x)f_X(x).$

I get
\begin{eqnarray}
P[X+Y > {3 \over 2}] &=& \int_{x={1 \over 2}}^1 \int_{y={3 \over 2}-x}^1 {1 \over x} dy dx \\
&=& \int_{x={1 \over 2}}^1   (1-{1 \over 2x}) dx \\
&=& (x-{ 1\over 2} \ln x) \mid_{1 \over 2}^1 \\
&=& {1 \over 2}(1+\ln {1 \over 2}) \\
&\approx& 0.1534
\end{eqnarray}
Where did the bounds come from?
First, note that the distribution is supported in $[0,1]^2$.
So we are interested in the set
$D=\{(x,y)| x+y > {3 \over 2} \} \cap [0,1]^2$.
Note that if $x \le {1 \over 2}$, then $x+y \le {3 \over 2}$ so
we can write
$D = \{(x,y)| x+y > {3 \over 2} , x \in ({1 \over 2}, 1], y \in [0,1] \} = \{(x,y)| x+y > {3 \over 2} , x \in ({1 \over 2}, 1], y \in ({3 \over 2} -x,1] \} $.
Just as a check, if $x \in ({1 \over 2}, 1]$ and $y \in ({3 \over 2} -x,1]$ then $y \in (1-x,1]$, so for $(x,y) \in D$, the partial pdf for $Y$ is $ {1 \over 1-(1-x)} = { 1\over x}$ and the partial for $X$ is $1$.
