Answer
$2\sqrt{3}$
Work Step by Step
Applying th.7, let $u=g(t)=t^{5}+2t$ (a continuous function).
Then, $\quad du=g'(t)dt=(5t^{4}+2)dt$
and, the borders change to $g(0)=0+0=0$ and $\quad g(1)=1+2=3$
$\displaystyle \int_{0}^{2\pi}\sqrt{t^{5}+2t}(5t^{4}+2)dt=\int_{0}^{3}u^{-1/2}du $
$=\displaystyle \left[ \frac{u^{-1/2+1}}{-1/2+1} \right]_{0}^{3}$
$=\displaystyle \left[ \frac{u^{1/2}}{1/2} \right]_{0}^{3}$
$=2(\sqrt{3}-0)$
= $2\sqrt{3}$
