integral of heaviside function, with variable inside heaviside part

Question

I don't understand how the integral of a heaviside function works if there is a variable inside the heaviside part.

$$\int_{0}^{t}\tau^2H(t-\tau-1)d\tau$$

Answer 1

The Heaviside function is telling you where to integrate. $$H(t-\tau-1) = \chi_{[0,+\infty]}(t-\tau-1) = \chi_{[-\infty,t-1]}(\tau)$$ Since $t-\tau-1\geq 0 \iff \tau \leq t-1$. Then the integral in question is: $$\int_0^t\tau^2H(t-\tau-1)d\tau = \int_0^{t-1}\tau^2d\tau$$ which can be solved easily.