Answer:
64
Explanation:
The convolution of two functions is given by:
[tex](f*g)(t)=\int\limits^t_0 f(\tau)g(t-\tau) \, d\tau[/tex] (1)
In this case:
[tex]f(t)=6t\\g(t)=t[/tex]
So, let's replace the functions in the equation (1):
[tex](f*g)(t)=\int\limits^t_0 6\tau*(t-\tau) \, d\tau[/tex]
[tex](f*g)(t)=\int\limits^t_0 6\tau t \, d\tau - \int\limits^t_0 6\tau^{2} \, d\tau[/tex]
Integrating with respect to τ
[tex]6t(\frac{\tau^{2} }{2} )\left \{ {{t=t} \atop {t=0}} \right. - 6(\frac{\tau^{3} }{3} ) \left \{ {{t=t} \atop {t=0}} \right.[/tex]
Evaluating the integrals:
[tex](f*g)(t)=3t^{3} -2t^{3} =t^{3}[/tex]
Finally evaluating f*g at t=4:
[tex](f*g)(4)=4^{3} =64[/tex]
