Need help regarding the convolution of two signals The two functions $x(t)=\cases{t \quad -1\leq t\leq 1\\ 0 \quad\text{else}}$ and $h(t)=\cases{1 \quad 0\leq t\leq 2\\ 0 \quad \text{else}}$ are given. Now, I want to compute $$g(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)\;\mathrm{d}\tau.$$
First, I need to substitute $t=\tau$ and mirror $h(\tau)$ which is $h(-\tau)$. Now I need to shift it by $t$: $h(t-\tau)$. Our function look like this now: $x(\tau)=\cases{\tau \quad -1\leq \tau\leq 1\\ 0 \quad\text{else}}$ and $h(t-\tau)=\cases{1 \quad t-2\leq \tau\leq t\\ 0 \quad \text{else}}$
We have 4 cases:
Case1: $-\infty < t \leq -1$ where $g(t) = 0$
Case2: $-1< t\leq 1$ with $$g(t)=\int_{-1}^{t}\tau\;\mathrm{d}\tau=\left[\frac{\tau^2}{2}\right]^t_{-1}=\frac{t^2}{2}-\frac12.$$
Case3: $1<t\leq 3$ with $$g(t)=\int_{t-2}^{1}\tau\;\mathrm{d}\tau=\left[\frac{\tau^2}{2}\right]^{1}_{t-2}=\frac12-\frac{t^2-4t+4}{2}.$$
Case4: $3<t<\infty$ where $g(t)=0$
The function $g(t)$ is now defined as $g(t)=\cases{\frac{t^2}{2}-\frac12 & $-1<t\leq 1$\\\frac12-\frac{t^2-4t+4}{2} & $1<t\leq 3$ \\0 &\text{else}}$
Is this correct? How can I find out on my own whether I have performed the convolution correctly or not?
 A: What you've written is correct. I can't think of any elementary method to find out on your own if you made a mistake appart from those pointed out by Chrystomath in the comments.
On a more personal note using characteristic functions has made me less prone to making mistakes in the past. Maybe it will help you too ?
Rewrite $x$ and $h$ using characteristic functions :
$$ x(t) = t\; \mathbb 1_{[-1,1]}(t) \qquad h(t) = \mathbb 1_{[0,2]} (t).$$
This means that
$$ h(t - \tau) = \mathbb{1}_{[0,2]} (t-\tau) = \mathbb{1}_{[t-2,t]}(\tau).$$
Therefore
$$ g(t) = \int_{\mathbb R} \tau\; \mathbb 1_{[-1,1]}(\tau) \mathbb{1}_{[t-2,t]}(\tau) \; d \tau = \int_\mathbb R \tau \mathbb{1}_{[t-2,t] \cap [-1,1]}(\tau) \; d \tau  = \int_{[-1,1] \cap [t-2,t]} \tau  \; d\tau$$
But $$[-1,1] \cap [t-2,t] = 
\begin{cases} \varnothing  & \text { if } t < -1 \text{ or } t-2 > 1 \\
 [-1,t]  & \text{ if } -1 \leq t \leq 1 \\
 [t-2,1] & \text { if } -1 \leq t-2 \leq 1 \\ \end{cases}
$$
so $$g(t) = \begin{cases} 0 & \text { if } t < -1 \text{ or } t-2 > 1  \\
\int_{-1}^{t} \tau \;d\tau &  \text{ if } -1 \leq t \leq 1 \\
\int_{t-2}^1 \tau \; d \tau  & \text { if } -1 \leq  t-2 \leq 1. 
 \end{cases}  $$
