How can I integrate a function like this?
$$f(t) = \int_{-\infty}^{t}\prod(\tau-1.5)-\delta(\tau-3)d\tau$$
Where $\prod(\tau-1/2)$ is a shifted rectangular pulse (i.e., it equals 1 on (1/2, 3/2)), and $\delta(\tau-3)$ is a shifted Dirac Delta Distribution. My attempt looks like this:
$$\int_{1}^{t}u(\tau)d\tau - \int_{t}^{2}u(\tau)d\tau - \int_{-\infty}^{t}\delta(\tau-3)d\tau$$
I reason that, since the rectangular pulse is centered around 1.5, and exists only on (1, 2) as a result, that this is equivalent to taking a unit step function from 1 to t (with t>1) and subtracting another unit step function from t to 2 (since t must be less than 2 to keep within the bounds of the rectangular pulse). Then, carrying out the integration:
$$\tau|^{t}_{1} - \tau|^{2}_{t} - u(\tau)|^{t}_{3}$$
Which I think should be:
$$f(t) = \begin{equation}2t-1, 1<t<2 \\ -1, 2<t \\ 0, \text{else} \end{equation}$$
But I'm really not sure if I've approached this correctly.
Just break up the integrations into intervals at the behavior changes:
$$f(t) = \begin{cases} \int_{-\infty}^t \Pi(\tau-1.5)d\tau - \int_{-\infty}^t \delta(\tau-3)d\tau &= 0 - 0 & t < 1\\ \\ \int_{1}^t d\tau - \int_{-\infty}^t \delta(\tau-3)d\tau &= t-1 - 0 & 1 \le t \le 2 \\ \\ \int_{-\infty}^t \Pi(\tau-1.5)d\tau - \int_{-\infty}^t \delta(\tau-3)d\tau &= 1 - 0 & 2 < t < 3 \\ \\ \int_{-\infty}^t \Pi(\tau-1.5)d\tau - \int_{-\infty}^t \delta(\tau-3)d\tau &= 1-1 & t > 3 \\ \end{cases}$$
More succinctly $$f(t) = \begin{cases} t-1 & 1 \le t \le 2 \\ \\ 1 & 2 < t < 3 \\ \\ 0 & \mathrm{otherwise} \\ \end{cases}$$