Answer
$a.\quad \displaystyle \frac{45}{8} \\
b.\quad -\displaystyle \frac{45}{8}$
Work Step by Step
$a.$
Applying th.7, let $u=g(t)=t^{2}+1$ (a continuous function).
Then, $\displaystyle \quad du=g'(t)dt=2tdt,\qquad dt=\frac{du}{2}$
and, the borders change to $g(0)=1$ and $g(\sqrt{7})=8$.
$\displaystyle \int_{0}^{\sqrt{7}}t(t^{2}+1)^{1/3}dt=\int_{1}^{8}u^{1/3}\cdot\frac{u}{2}du$
$=\displaystyle \frac{1}{2}\left[ \frac{u^{1+1/3}}{1+1/3} \right]_{1}^{8}$
$=\displaystyle \frac{1}{2}\left[ \frac{u^{4/3}}{4/3} \right]_{1}^{8}$
$=\displaystyle \frac{1}{2}\cdot\frac{3}{4}(8^{4/3}-1^{4/3})$
$=\displaystyle \frac{3}{8}((2^{3})^{4/3}-1)$
$=\displaystyle \frac{3}{8}(2^{4}-1)$
$=\displaystyle \frac{3}{8}\cdot 15 =\frac{45}{8}$
$b.$
Applying th.7, let $u=g(t)=t^{2}+1$ (a continuous function).
Then, $\displaystyle \quad du=g'(t)dt=2tdt,\qquad dt=\frac{du}{2}$
and, the borders change to $g(-\sqrt{7})=8$ and $g(0)=1$.
$\displaystyle \int_{-\sqrt{7}}^{0}t(t^{2}+1)^{1/3}dt=\int_{8}^{1}u^{1/3}\cdot\frac{u}{2}du$
$=\displaystyle \frac{1}{2}\left[ \frac{u^{1+1/3}}{1+1/3} \right]_{8}^{1}$
$=\displaystyle \frac{1}{2}\left[ \frac{u^{4/3}}{4/3} \right]_{8}^{1}$
$=\displaystyle \frac{1}{2}\cdot\frac{3}{4}(1-8^{4/3})$
$=\displaystyle \frac{3}{8}(1-(2^{3})^{4/3})$
$=\displaystyle \frac{3}{8}(1-2^{4})$
$=\displaystyle \frac{3}{8}\cdot(-15) $
$=-\displaystyle \frac{45}{8}$
