$$f(x)=\begin{cases}\frac13e^{-x/3},&x>0\\0,&\text{otherwise}\end{cases}$$
(a) If two of these fuses have independent useful lives $X_1$ and $X_2$. One fuse in part (a) is in a primary system and the other is in the backup system entering use only if primary system fails. Find the density function of the total effective life of the two fuses.
I have this:
$$\int_{0}^{1}\int_{0}^{1-x_2}\frac13e^{-x_1/3}\cdot\frac13e^{-x_2/3}dx_1dx_2$$
But I'm not sure...
In general, the sum of i.i.d. exponential random variables has an Erlang distribution - see here for a proof: Sum of exponential random variables?
In this case $\lambda=\frac13$, so the density of $X_1+X_2$ is given by $$ f_{X_1+X_2}(x) = \frac19 xe^{-\frac13 x}\cdot\mathsf 1_{(0,\infty)}(x). $$
You are almost right. Start by finding the distribution function $F_{X_1+X_2}(t)=P(X_1+X_2\le t)$.
For $t\le0, F_{X_1+X_2}(t)=0$.
For $t>0,P(X_1+X_2\le t)$ is found by integrating $f_{X_1X_2}(x_1,x_2)$ over the region $x_1+x_2\le t\in\mathbb R^+$.
$$F_{X_1+X_2}(t)=\int_0^t\int_0^{t-x_1}\left(\frac13e^{-x_1/3}\right)\left(\frac13e^{-x_2/3}\right)dx_2~dx_1\\=\frac13\int_0^t[e^{-x_1/3}-e^{-t/3}]dx=1-e^{-t/3}\left(1+\frac t3\right)$$
The density function can be obtained by differentiating $F_{X_1+X_2}(t)$.
