how to find PDF of $X+Y$ 
Suppose $X$ has probability density function $f$, and $Y$ is uniformly
distributed on [0,1]. Supposing X and Y are independent, find the
probability density of $X + Y$

$$ f_{X+Y}(t) = \int_{-\infty}^{\infty} f(s)f_Y(t-s)ds $$
We know $f_Y(t-s)=1$ for $0 \leq t-s \leq 1$ and 0 elsewhere.
How do I use this information to simplify the above integral? I feel like I need to change bounds so that $f_Y(t-s)$ disappears, but I'm not sure how. I believe I need to substitute $s$ for $-\infty$ and $s+1$ for $\infty$ because $0 \leq t-s \leq 1 \leftrightarrow s \leq t \leq s+1$
So $$ f_{X+Y}(t) = \int_{-\infty}^{\infty} f(s)f_Y(t-s)ds =  \int_{s}^{s+1} f(s)ds$$ is the answer.
(I'm quite shaky on this material; apologize if this is poorly put)
 A: Note that we are fixing a value of $t$; $s$ is a dummy integration variable so you cannot substitute it with anything. We have (due to independence)
\begin{align}
f_{X+Y}(t)&=(f*f_Y)(t)\\
&=\int_{-\infty}^{\infty}f(s)f_Y(t-s)\,ds\\
&=\int_{\{s\in\Bbb{R}\,:\, 0\leq t-s\leq 1\}}f(s)\cdot 1\,ds
\end{align}
The last equality is precisely for the reason you mentioned.
So, we're integrating over a certain set of $s$ values. This set depends on $t$, so you just have to rearrange the inequality "to make $s$ the subject of the inequality". For example, I'm sure you can verify that
\begin{align}
\{s\in\Bbb{R}\,:\, 0\leq 3-s\leq 1\}&= \{s\in\Bbb{R}\,:\, -1\leq s-3\leq 0\}\\
&=\{s\in\Bbb{R}\,:\, 2\leq s\leq 3\}\\
&=[2,3]
\end{align}
So, what is the result for a general value of $t$?
A: there are several way to solve  your problem. One of these ways is to use the convolution but you must pay attention on the integral bounds.
To understand how convolution works in this case, observe that from $Z=X+Y$ you get
$$y=z-x$$
that is
$$0<z-x<1$$
now, when $0<z<1$ you must have that $0<x<z$ but when $z>1$ you have that $z-1<x<1$
Putting together these information, by integration you get
$$f_Z(z)=z\cdot\mathbb{1}_{[0;1)}(z)+(2-z)\cdot\mathbb{1}_{[1;2]}(z)$$
that is a triangular density
In a more compact form,
$$f_Z(z)=[1-|1-z|]\cdot\mathbb{1}_{[0;2]}(z)$$

A different way to understand the correct integral bounds when integrating in dx is to observe that the joint domain $(X,Z)$ is the following

