You can compute the integral
$$
f(t) = \int_{t-1/2}^{t}\chi_{[0,1]}(u)du
$$
in cases, for different ranges of $t$.
I think it's easier (personally) if we manipulate the rectangular pulse function instead of doing the substitution for $u$. I'm assuming that
$$
\chi_{[a,b]}(x) = \begin{cases}
\hfill 1 \hfill & \text{ if } x\in [a,b] \\
\hfill 0 \hfill & \text{ if } x\notin [a,b] \\
\end{cases}.
$$
With this, we can see that
\begin{align}
\chi_{[0,1]}(t-s) &= \begin{cases}
\hfill 1 \hfill & \text{ if } t-s\in [0,1] \\
\hfill 0 \hfill & \text{ if } t-s\notin [0,1] \\
\end{cases}
\\ &= \begin{cases}
\hfill 1 \hfill & \text{ if } s\in [t-1,t] \\
\hfill 0 \hfill & \text{ if } s\notin [t-1,t] \\
\end{cases}
\\ &= \chi_{[t-1,t]}(s).
\end{align}
So we have,
$$
f(t) = \int_{0}^{1/2}\chi_{[0,1]}(t-s)ds = \int_{0}^{1/2}\chi_{[t-1,t]}(s)ds.
$$
For $t\leq 0$, $\chi_{[t-1,t]} = 0$ in the range of integration, so $f(t) = 0$.
For $t\in (0,\frac{1}{2}]$, $f(t) = \int_{0}^{t}1 ds = t$.
For $t \in (\frac{1}{2},1)$, $f(t) = \int_{0}^{1/2}1 ds = \frac{1}{2}$.
For $t \in [1,\frac{3}{2})$, $f(t) = \int_{t-1}^{1/2}1 ds = \frac{1}{2} - (t-1) = \frac{3}{2} - t$.
For $t \geq \frac{3}{2}$, $f(t) = 0$.
Visually, the function $f(t)$ looks like a trapezoid. It's the amount of overlap between the two rectangular pulses as you slide them over one another (with the variable $t$). The overlap takes on a constant value of $\frac{1}{2}$ when the small rectangle is completely inside the larger rectangle.
