Calculate an unconditional probability given only the conditional probability Let X and Y be jointly absolutely continuous random variables. Suppose X~Exponential(2) and P(Y > 5 | X = x) = $$e^{-3x}$$. Compute P(Y > 5).

I've computed P(Y > 5) as follows:

P(Y > 5) = P(Y > 5 | X = $x_1$) + ... + P(Y > 5 | X = $x_n$) = $\int_0^\infty e^{-3x} dx = \frac13$

However, I'm not sure the answer is correct, since I did not have to use the fact that X~Exponential(2). Can someone help?
If $P\{Y &gt; 5\mid X = x\} = \exp(-3x)$, then 
$$P\{Y &gt; 5\} = \int_{-\infty}^\infty P\{Y &gt; 5\mid X = x\} f_X(x)\,\mathrm dx
= \int_0^\infty \exp(-3x) \cdot 2\exp(-2x)\,\mathrm dx$$
and so the answer does depend on the distribution of $X$. The above 
is  an application of
the law of total probability for continuous random variable as @Patrick says in
his answer, but your answer of $\frac{1}{3}$ is incorrect. You are
adding up conditional probabilities conditioned on different events,
and that is not what the law of total probability does.
Your integrated needs to be: $e^{-3x}f_X(x)$ and so the result follows from the law of total probability. Also, the sum you wrote as being equal to the integral is not correct.  I am not the best person to explain the theory behind the integral being equal to a limit of Riemann sums but just know that your sum does not make sense because the sample space for $X$ is not discrete.