Correct bounds when finding PDF of $X + Y$, where $X,Y$ i.i.d. 
Let $X,Y \sim \mathcal{U}([0,a])$ be independent identically continuously uniform distributed real random variables.
  Find the PDF of $Z := X + Y$.

I know that this can be accomplished via convolution:
For $x \in \mathbb{R}$ we have
\begin{align*}
\int_{\mathbb{R}} f_{X}(x - y) f_{Y}(y) \ \text{dy}
& = \frac{1}{(a - 0)^2} \int_{\mathbb{R}} 1_{[0,a]}(x - y) \cdot 1_{[0,a]}(y) \ \text{dy} \\
& = \frac{1}{a^2} \int_{0}^{a} 1 \ \text{dy} \cdot 1_{[0,2a]}(x)
= \frac{1_{[0,2a]}(x)}{a}.
\end{align*}
I obtained the bounds like this.
From the second integral in the first line we have
$$
0 \le x - y \le a
\qquad \text{and} \qquad
0 \le y \le a.
$$
Adding both inequalities yields $y \le x + y \le 2a + y$ or
$$
0 \le x \le 2 a.
$$
Therefore $x < a$ and the bounded on the dy-integral are [0, a] and furthermore we know that $x \in [0,2a]$.
Is this (result and procedure) correct and / or is there an easier way to do this?

In this task I faced a similar challenge:
  Let $X,Y \sim \text{Exp}(\lambda)$ for some $\lambda > 0$ be independent and identically distributed.
  Find the density of $X + Y$.

\begin{align*}
(f_{X} \ast f_{Y})(x)
& = \int_{\mathbb{R}} f_{X}(x - y) f_{Y}(y) \ \text{dy} \\
& = \int_{\mathbb{R}} \lambda e^{-\lambda (x - y)} 1_{[0, \infty)}(x - y) \cdot \lambda e^{-\lambda y} 1_{[0, \infty)}(y) \ \text{dy} \\
& = \lambda^2 \int_{0}^{x} e^{-\lambda (x - y)} \cdot e^{-\lambda y} \ \text{dy} \cdot 1_{[0, \infty)}(x)
= \lambda^2 \int_{0}^{x} e^{-\lambda x} \ \text{dy} \cdot 1_{[0, \infty)}(x) \\
& = \lambda^2 x \cdot e^{-\lambda x} \cdot 1_{[0, \infty)}(x).
\end{align*}
Here the same question as above applies.
 A: First part (uniform). If you want to use the CDF method to find $P(Z \le z)$ by integration,
the following plots (made from a simulation), may help you find the
limits on integrals. I used $a = 2.$ 
The left-hand plot shows the region of integration
for $P(Z \le 1.5).$ For values of $z$ above $2$ it seems easiest
to break the integral into two parts. [However, because the joint distribution is uniform, this can be a high school geometry/algebra problem, if you like.]
The middle plot shows the empirical CDF (jump of $1/10^5$ at each
of $10^5$ simulated values of $Z),$ which suggests the form of
the CDF you'll get.
The right-hand plot shows a histogram of $10^5$ simulated values of
$Z.$ [A discrete analogue of this is the 'triangular' PDF from the sum upon rolling two fair dice.]

Second part (exponential). If you know about moment generating
functions, that's probably the easiest way to show that the sum of
two independent exponential random variables, each with rate $\lambda$
is $\mathsf{Gamma}(\text{shape}=2,\text{rate}=\lambda).$ However, the PDF and convolution methods also work. [Perhaps see Wikipedia
on gamma distributions to check your answer.]
Another simulation to indicate the result:
set.seed(708);  m = 10^6
x = rexp(m, .1); y = rexp(m, .1);  z = x+y
hist(z, prob=T, br=50, col="skyblue2")
 curve(dgamma(x, 2, .1), add=T, lwd=2, col="red")


Notes: (a) This exponential relationship is much used in applications, including queueing theory. [Q. A person is next in line for a bank teller whose service times are exponential with rate one every five minutes, what is the probability that person finishes being served within 8 minutes? A. In R, code pgamma(8, 2, 1/5) returns 0.4750691.] 
(b) If the two exponential random variables have different
rates, the problem is very much messier. (If interested, perhaps google
'sum of exponentials different rates'.)
(c) In case you want it, the R code for making the figure is provided
below:
par(mfrow=c(1,3))
set.seed(2019)
 m = 30000
 x = runif(m, 0, 2);  y = runif(m, 0, 2)
 plot(x, y, pch=".")
  event = (x + y < 1.5)
   points(x[event], y[event], pch=".", col="green2")
m = 10^5
 X = runif(m, 0, 2);  Y = runif(m, 0, 2)
 Z = X+Y
 plot(ecdf(Z))
 hist(X+Y, prob=T, br=50, col="skyblue2")
par(mfrow=c(1,1))

