Convolution limits of positive function

Question

Why are the limits of the convolution of two positive functions $0$ and $t$?

$$(f \star g)(t) = \int_0^t f(x)g(t-x)\,dx \text{ for } f,g : [0,\infty) \rightarrow \mathbb{R}$$

I understand the lower limit of 0 because both are non-negative functions, but why $t$?

Thank you!

Answer 1

Consider the integrand term $f(x)g(t-x)$. If $x<0$, then $f(x)=0$. If $x>t$, $g(t-x)=0$. Thus for a fixed $t$, $f(x)g(t-x)$ is 0 outside the interval $x\in [0,t]$.