Convolution of two rectangular pulses

$y(t) = x(t) * h(t) = \displaystyle\int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau$, with $x(\tau) = 1$ for $-1 < \tau < 1$, and $h(t - \tau) = 1$ for $-1 < t - \tau < 1$, i.e. $t - 1 < \tau < t + 1$.

The result is non-zero only where the two pulses overlap. Each has width 2, so the output extends from $t = -2$ to $t = 2$.

       x(tau)            h(t-tau)
     _________        _________
    |         |      |         |
 ---+----+----+--  --+----+----+--> tau
   -1    0    1    t-1    t   t+1

Case 1: $t < -2$

$t + 1 < -1$: no overlap, $y(t) = 0$.

Case 2: $-2 \le t < 0$ (partial overlap from left)

Overlap from $\tau = -1$ to $\tau = t + 1$:

$$y(t) = \int_{-1}^{t+1} 1\,d\tau = t + 2$$

Case 3: $0 \le t < 2$ (partial overlap from right)

Overlap from $\tau = t - 1$ to $\tau = 1$:

$$y(t) = \int_{t-1}^{1} 1\,d\tau = 2 - t$$

Case 4: $t \ge 2$

$t - 1 > 1$: no overlap, $y(t) = 0$.

Result

$$y(t) = \begin{cases} t + 2, & -2 \le t < 0 \\ 2 - t, & 0 \le t < 2 \\ 0, & |t| \ge 2 \end{cases} \;=\; (2 - |t|)\ \text{for } |t| < 2$$

 y(t)
   2 |      /\
     |     /  \
   1 |    /    \
     |   /      \
  ---+--/---+----\---> t
      -2    0     2

Answer: a triangular pulse of height 2 at $t = 0$, base from $-2$ to $2$. Convolving two equal rectangles always gives a triangle whose width is the sum of the widths (2 + 2 = 4) and whose peak equals the area of overlap (2).