## gabyem 2 years ago f(t)=t^2(1+t^3)^4 interval [0,2] what is the average value of the function?

$f(t)=t^{2}(1+t^{3})^{4}$ $\frac{1}{2-0}\int\limits_0^2 t^2(1+t^3)^4dt=\frac{1}{2}\int\limits_0^2 t^2(1+t^3)^4dt$ $u=t^3$ $du=3t^2dt$ $\frac{1}{2}\int\ (1+t^3)^4t^2dt$ $t^2dt=\frac{du}{3}$ $\frac{1}{2}\int\limits \frac{(1+u)^4du}{3}=\frac{1}{6}\int\limits (1+u)^4du$ $\frac{1}{6}\int\limits (1+u)^4du=\frac{1}{6}[\frac{(1+u)^5}{5}]$ $\frac{1}{30}[(1+t^3)]_0^2=\frac{1}{30}[(1+8)-(1)]=\frac{4}{15}$