Questions about two different exponential RVs If $X\sim \operatorname{Exp}(\lambda_1)$ and $Y\sim \operatorname{Exp}(\lambda_2)$ are independent RVs. Find the following: 
a) $ P(X< Y)  $
I tried finding this with $\int_{-\infty}^{\infty} F_X(y)f_y(y)\,dy$, but the integral calculator couldn't do it.
b) Distribution of the $\operatorname{Min}(X,Y)$ (hint : think of the CDF)
I have no idea what this means. The distribution of the minimum between the 2?
c) Distribution of $X + Y$
I'm assuming this question means the closest approximate distribution to their sum?
 A: We compute $\mathbb P(X<Y)$ from the joint density:
\begin{align}
\mathbb P(X<Y) &= \int_{\mathbb R^2}f_{X,Y}\mathsf 1_{\{X<Y\}}\\
&= \int_0^\infty \int_0^y f_{X,Y}(x,y)\ \mathsf dx\ \mathsf dy\\
&= \int_0^\infty \lambda_2 e^{-\lambda_2y} \int_0^y \lambda_1e^{-\lambda_1 x}\ \mathsf dx \ \mathsf dy\\
&= \int_0^\infty \lambda_2e^{-\lambda_2y}(1 -e^{-\lambda_1 y} )\ \mathsf dy\\
&= \int_0^\infty \lambda_2 e^{-\lambda_2y}\ \mathsf dy - \lambda_2\int_0^\infty e^{-(\lambda_1+\lambda_2)y}\ \mathsf dy\\
&= 1 - \frac{\lambda_2}{\lambda_1+\lambda_2}\\
&= \frac{\lambda_1}{\lambda_1+\lambda_2}.
\end{align}
Let $Z=X\wedge Y$. Then for any $t>0$, 
$$
\mathbb P(Z>t) = \mathbb P(X>t, Y>t) = \mathbb P(X>t)\mathbb P(Y>t) = e^{-\lambda_1t}e^{-\lambda_2t}=e^{-(\lambda_1+\lambda_2)t},$$
so that $Z$ has exponential distribution with parameter $\lambda_1+\lambda_2$.
To compute the density of $W=X+Y$, we use convolution. For any $t>0$ we have
\begin{align}
f_W(t) &= (f_X\star f_Y)(t)\\
&= \int_{\mathbb R}f_X(\tau)f_Y(t-\tau)\ \mathsf d\tau\\
&= \int_0^t \lambda_1 e^{-\lambda_1\tau}\lambda_2 e^{-\lambda_2(t-\tau)}\ \mathsf d\tau\\
&= \lambda_1\lambda_2 e^{-\lambda_2 t}\int_0^t e^{-(\lambda_1-\lambda_2)\tau}\ \mathsf d\tau\\
&= \frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2}e^{-\lambda_2 t}(1 - e^{-(\lambda_1-\lambda_2) t}).\\
&= \frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2}(e^{-\lambda_2 t}-e^{-\lambda_1 t}),
\end{align}
assuming $\lambda_1\ne\lambda_2$. If $\lambda_1=\lambda_2=\lambda$ then we have
\begin{align}
(f_X\star f_Y)(t) &= \int_0^t \lambda^2 e^{-\lambda \tau}e^{-\lambda(t-\tau)}\ \mathsf d\tau\\
&= \lambda^2 e^{-\lambda t} \int_0^t \mathsf d\tau\\
&= \lambda^2 t e^{-\lambda t}.
\end{align}
