Showing $P(X_1+X_2<1)=\frac12$ where $X_1,X_2$ are i.i.d $U(0,1)$ variables I am given that $X_1$ and $X_2$ are iid $U(0,1)$ and want to show that 
$$Pr[X_1+X_2<1]=0.5$$
My approach is to evaluate 
$$\int_0^1\int_0^{1-x_1}1 \quad dx_2dx_1$$
but there seems to be a geometric approach to this that significantly simplifies the answer.
May I have some assistance?
 A: The geometric argument (using the uniform distribution on the unit square) looks like

A: Let me call the variables X and Y.
Suppose that $X$ and $Y$ are independent, continuous random variables having probability density functions $f_X$ and $f_Y$. The cumulative distribution function of $X + Y$ is obtained as follows:
$$
P(X + Y < a) = F_{X+Y}(a)\\
= \int\int_{x+y \leq a} f_X(x)f_Y(y)dxdy = \\
= \int_{-\infty}^{\infty}\int_{\infty}^{a-y} f_X(x)f_Y(y)dx dy = \\
= \int_{-\infty}^{\infty}\int_{\infty}^{a-y} f_X(x)dx f_Y(y)dy = \\
= \int_{-\infty}^{\infty} F_X(a-y)f_Y(y)dy \\
$$
Since you have $X$ and $Y$ and they are iid $U(0,1)$ The cdf of $X$ looks like:
$$
F_X(x) =
\begin{cases}
0,&\text{if $0 \geq x$}\\
x,&\text{if $0 \leq x < 1$}\\
1,&\text{if $1 \leq x$}\\
\end{cases}
$$
and the pdf of $Y$ like:
$$
f_y(y) =
\begin{cases}
1,&\text{if $0 < x < 1$}\\
0,&\text{else}\\
\end{cases}
$$
Now you can substitute the the unknown parameter $a$ for your desired value of $1$ into the cumulative distribution function of $X+Y$.
$$
P(X + Y < 1) = F_{X+Y}(1)\\
= \int_{-\infty}^{\infty} F_X(1-y)f_Y(y)dy \\
= \int_{0}^{1} (1-y)1dy \\
= \left[ y-\frac{y^2}{2}\right]_0^1 = 0.5
$$
A: Note also that if  $X_1$ and $X_2$ are iid $U(0,1)$, then also  $1-X_1$ and $1-X_2$ are iid $U(0,1)$. So 
$$
\mathbb P(X_1+X_2<1)=\mathbb P((1-X_1)+(1-X_2)<1)=\mathbb P(X_1+X_2>1).
$$
Note that $\mathbb P(X_1+X_2=1)=0$ because $X_1+X_2$ have continuous distribution. Red probabilities here are equal:
$$
1=\color{red}{\mathbb P(X_1+X_2<1)} + \underbrace{\mathbb P(X_1+X_2=1)}_{0}+\color{red}{\mathbb P(X_1+X_2>1)} 
$$
then $\mathbb P(X_1+X_2<1)=0.5$.
