Explain the convolutions of setp function

Question

$$ t* u(t-a)= \frac{1}{2}(t-a)^2 u(t-a)$$

attempt:

$$ t* u(t-a)= \int_{-\infty}^{+\infty} tu(t-a)dt$$

I don't quiet understand what i write before. What mean that t is a causal function?

$$ t* u(t-a)= \int_{0}^{t} (t-\tau)u(\tau-a)d\tau $$

$$ t* u(t-a) = \frac{1}{2} (t-a)^2...$$

Answer 1

Let \begin{align} f(t)&=u(t-a)\\ g(t)&=t\\ \end{align} Therefore, \begin{align} f(t)*g(t)&=\int_{0}^{t}f(\tau)g(t-\tau)d\tau\\ &=\int_{0}^{t} u(\tau-a)(t-\tau)d\tau\\ &=\int_{a}^{t}(t-\tau)d\tau \,\,\,\,\,\,\,\,\,\,\,\,\text{assuming}(t\ge a)\\ &=-\frac{1}{2}(t-\tau)^2\biggl|_a^t u(t-a)\\ &=\frac{1}{2}(t-a)^2 u(t-a) \end{align}