Can someone help me understand how did we get from the second line to the third one?
$$\begin{align} u(t-1)*u(t) &= \int_{-\infty}^{\infty}u(\tau-1)u(t-\tau)d\tau \\ &=\int_{1}^{t}u(t-\tau)d\tau \\ &=(t-1)u(t-1) \end{align}$$
Can someone help me understand how did we get from the second line to the third one?
$$\begin{align} u(t-1)*u(t) &= \int_{-\infty}^{\infty}u(\tau-1)u(t-\tau)d\tau \\ &=\int_{1}^{t}u(t-\tau)d\tau \\ &=(t-1)u(t-1) \end{align}$$
The integration variable is $\tau$ and the integration limits are determined by the range of $\tau$ over which both step functions under the integral are non-zero.
For the first step functions that's $\tau > 1$ and for the second one it's $\tau < t$ . This determines the integration interval. Within this interval the function to integrate is simply 1, i.e. $u(t- \tau) = 1$ for $ \tau < t$ . Integral from a to b over one is simply a-b.