This is exercise 2.2.1 from Achim Klenke: »Probability Theory — A Comprehensive Course«
Let $X$ and $Y$ be independent random variables with $X \sim \exp_\theta$ and $Y \sim \exp_\rho$ for certain $\theta,\rho > 0$. Show that $$\mathbf{P}[X < Y] = \frac{\theta}{\theta +\rho}\, .$$
Now, in practice, this exercise is easy. $\exp_\theta$-distribution is defined as $$ \mathbf{P}[X \leq x] = \int_0^x \theta e^{-\theta t} \, dt \quad \text{ for } x \geq 0\, .$$
We just have to evaluate the integral: $$\int_0^\infty \mathbf{P}[X \leq x] \cdot \rho e^{-\rho x} \, d x = \int_0^\infty \Bigl(\int_0^x \theta e^{- \theta t} \, d t \Bigr) \cdot \rho e^{-\rho x} \, d x\, ,$$
which gives $\frac{\theta}{\theta +\rho}$.
But how does one do it rigorously?
Why is the following possible: $$\mathbf{P}[X < Y] = \int_0^\infty \mathbf{P}[X \leq x]\cdot \mathbf{P}[Y = x] \, d x \\ \text{ and using } \mathbf{P}[Y = x] = \rho e^{-\rho x} \, ?$$
Convolution of real valued random variables hasn't been defined yet.