By definition,
$$\phi\ast\psi(t)=\int_{-\infty}^{\infty}\phi(t-x)\psi(x)\;dx=\int_{-\infty}^{\infty}\phi(x)\psi(t-x)\;dx$$
$\phi(x)=0$ unless $0\leq x<1$, in which case $\phi(x)=1$. Therefore
$$ \int_{-\infty}^{\infty}\phi(x)\psi(t-x)\;dx=\int_0^1\psi(t-x)\;dx$$
Now consider a few cases. If $t\geq 2$ then $t-x\geq t-1\geq 1$ for all $x\in[0,1]$, so the integral is zero. Similarly, if $t<0$ then $t-x\leq t<0$ for all $x\in[0,1]$, so again the integral is zero.
So that leaves $0\leq t<2$. If $0\leq t<1$ then $t-x\in [0,1)$ when $0\leq x\leq t$, so
$$ \int_0^1\psi(t-x)\;dx=\int_0^t\psi(t-x)\;dx=\int_0^t\psi(x)\;dx$$
and this integral is equal to $t$ if $0\leq t<\frac{1}{2}$, and is equal to $\frac{1}{2}-(t-\frac{1}{2})=1-t$ if $\frac{1}{2}\leq t<1$.
Finally, if $1\leq t<2$, then $t-x\in[0,1)$ when $t-1<x\leq 1$, hence
$$\int_0^1\psi(t-x)\;dx=\int_{t-1}^1\psi(x)\;dx$$
which is equal to $\frac{1}{2}-(t-1)-\frac{1}{2}=1-t$ if $1\leq t<\frac{3}{2}$, and is equal to $-[1-(t-1)]=t-2$ if $\frac{3}{2}\leq t<2$.
we have:
$f(t) = \begin{cases} 2 & 0< t< 3 \\ 0 &\text{elsewhere} \end{cases}$
$g(t) = \begin{cases} 2 & 0< t< 1 \\ 0 & \text{elsewhere} \end{cases}$
so let's take a look over the intervals $\Bbb R,(0,3),(0,1)$:
Using those interval we can see $4$ cases: $(-\infty,0],(0,1),[1,3),[3,\infty)$, I want to point out that there are 4 cases but we will use $[1,3),[3,\infty)$ as one:
if $\tau\in(-\infty,0]$ then $g(\tau)=0$
if $\tau\in[1,3)\cup[3,\infty)=[1,\infty)$ then $g(\tau)=0$
so we can reduce the integral: $\int_{-\infty}^{+\infty} f(t - \tau) g(\tau) \ d\tau\to \int_{0}^{1} f(t - \tau) g(\tau) \ d\tau$
now in the interval $(0,1),\ g(\tau)$ is a constant, so we can take it out:$\int_{0}^{1} f(t - \tau) g(\tau) \ d\tau\to2\int_{0}^{1} f(t - \tau)d\tau$
now we have $g*f(t)=2\int_{0}^{1} f(t - \tau)d\tau$, set $\omega=t-\tau\implies d\omega=-d\tau,\omega(0)=t,\omega(1)=t-1$ so we get: $2\int_{0}^{1} f(t - \tau)d\tau\to-2\int_{t}^{t-1} f(\omega)d\omega\to2\int_{t-1}^{t} f(\omega)d\omega$
Can you solve this kind of intervals?
I learnt that playing with functions can help a lot for intuition.
for example: try: $t=-5,t=0.5,t=1,t=2,t=3.5,t=5$ and see what happens. after that try to find intervals for which the function gives the same value, see how it connects to the original $4$ cases i presented, what is the same and what is different.
Best Answer
Hint. Note that $f(x-t)=2$ when $0\leq x-t\leq 1$, i.e. $x-1\leq t\leq x$, otherwise it is zero. Hence $$\int_{-\infty}^{\infty} g(t)f(x-t)dt=\int_{x-1}^x g(t)2 dt.$$ Now if $x\in[0,1)$ then what is $g(t)$ for $t\in [x-1,0]$? And for $t\in [0,x]$?