Determining the probability of $X_2 \ge X_1$ given they have different probability functions Suppose that I have a random variable $X_1$ which is normally distributed, and a random variable $X_2$ having the density function shown in the figure below. How would I determine ${\rm P}(X_1 \le X_2)$? 

 A: Assuming that random variables $X_1$ and $X_2$ have joint density $f_{X_1, X_2}$ and marginal densities $f_{X_1}$ and $f_{X_2}$, we have
$$
P(X_1 \leq X_2)
  = \int_{-\infty}^{\infty}\int_{-\infty}^{x_2} f_{X_1, X_2} (x_1, x_2) dx_1 dx_2 .
$$
If the two random variables are independent, the probability is
\begin{align*}
P(X_1 \leq X_2)
  & = \int_{-\infty}^{\infty} \int_{-\infty}^{x_2} f_{X_1}(x_1) f_{X_2}(x_2) dx_1 dx_2\\
  & = \int_{-\infty}^{\infty} f_{X_2}(x_2) \left\{\int_{-\infty}^{x_2} f_{X_1}(x_1)
dx_1\right\} dx_2 \\
  & = \int_{-\infty}^{\infty} f_{X_2}(x_2) F_{X_1}(x_2) dx_2 \\
  & = E\left\{ F_{X_1}(X_2) \right\}.
\end{align*}
Similarly, we have $P(X_2 \leq X_1) = E\left\{ F_{X_2}(X_1) \right\}$, and hence
$$
P(X_1 \leq X_2) = E\left\{ F_{X_1}(X_2) \right\} = 1 - E\left\{ F_{X_2}(X_1) \right\}.
$$
Note that if $X_1$ and $X_2$ have the same distribution, then
$U = F_{X_1}(X_2) = F_{X_2}(X_2)$ follows a uniform $U(0, 1)$ distribution, and
hence $P(X_1 \leq X_2) = E(U) = 0.5$ as one would expect.
Now, based on your graphic and the information provided in one of your comments, it seems that a shifted exponential distribution would be a reasonable choice for the distribution of $X_2$.
So, if $X_1 \sim N(\mu, \sigma^2)$ and $X_2$ follows a shifted exponential distribution with rate $\lambda > 0$ and location parameter $a$, that is $F_{X_2}(x) = 1 - \exp\{-\lambda(x-a)\}$ if $x > a$ and $F_{X_2}(x) = 0 $ otherwise, then the probability is
\begin{align}
P(X_1 \leq X_2)
 &= 1 - \int_{-\infty}^{\infty} F_{X_2}(x) f_{X_1}(x) dx \\
 &= 1 - \int_{a}^{\infty} [ 1 - \exp\{-\lambda(x-a)\} ] f_{X_1}(x) dx \\
 &= 1 - P(X_1 > a) + \int_{a}^{\infty} \exp\{-\lambda(x-a)\} f_{X_1}(x) dx \\
 &= \Phi\left( \frac{a - \mu}{\sigma} \right)
      + \exp \left\{ \lambda (a - \mu) + \frac{\lambda^2\sigma^2}{2} \right\}
        \Phi\left( \frac{\mu - a - \lambda \sigma^2}{\sigma} \right),
\end{align}
where $\Phi(\cdot)$ denotes the standard normal distribution function.
In your case, numerical values of the parameters could be $a = 2$ and $\lambda = 1$ (assuming that $X_2$ corresponds to "Strain" $\times 1000$).
A: The difference between two independent variables has a distribution given by the convolution of the two individual distributions. So basically just smooth that curve above with the normal distribution. It will extend into the negative region and have a smooth peak somewhere in the positive region. That's about all there is to say about it. 
A: If your graph above represents a probability density, then you can find the density for $X_2 - X_1$ via the convolution integral. Then $P(X_2 <= X_1)$ is equivalent to the probability that $P(X_2 - X_1 <= 0)$, using the density function derived from convolution. If you have discrete distributions, then there's an equivalent process using a sum instead of an integral.
For the analytic approach, see if this helps: http://courses.washington.edu/bioen316/Assignments/316_SCP.pdf
If you can sample from both, then simulations of $X_2 - X_1$ will give a working approximation. I'm only familiar with how to do that with an inverse CDF, but there's a computational method mentioned here that may interest you: http://blog.quantitations.com/tutorial/2012/11/20/sampling-from-an-arbitrary-density/ 
