Answer
True
Work Step by Step
We have $(f *g)(t)=\int^t_0 f(t- \tau)g(\tau) d\tau\\
(g*f)(t)=\int^t_0 g(t-\tau)f(\tau) d\tau$
We can notice that if we substitute $u=t-\tau$ to $(g*f)(t)=\int^t_0 g(t-\tau)f(\tau) d\tau$ then the equation will become $(f *g)(t)=\int^t_0 f(t- \tau)g(\tau) d\tau$.
Hence, it is true to say that $f*g=g*f$
