Answer
$a.\quad \displaystyle \frac{10}{3}$
$ b.\quad \displaystyle \frac{70}{27}$
Work Step by Step
$a.$
Applying th.7, let $u=g(v)=1+v^{3/2}$ (a continuous function).
Then, $\displaystyle \quad du=g'(v)dv=\frac{3}{2}v^{1/2}dv,\qquad \sqrt{v}dv=\frac{2du}{3}$
and, the borders change to $g(0)=1$ and $g(1)=2$.
$\displaystyle \int_{0}^{1}\frac{10\sqrt{v}}{(1+v^{3/2})^{2}}dr=\int_{1}^{2}\frac{10}{u^{2}}\cdot\frac{2du}{3}=\frac{20}{3}\int_{1}^{2}u^{-2}du$
$=\displaystyle \frac{20}{3}\left[ \frac{u^{-2+1}}{-2+1} \right]_{1}^{2}$
$=\displaystyle \frac{20}{3}\left[-u^{-1} \right]_{1}^{2}$
$=\displaystyle \frac{20}{3}(-\frac{1}{2}+1)$
$=\displaystyle \frac{20}{3}\cdot\frac{1}{2}$
$=\displaystyle \frac{10}{3}$
$b.$
Using the same substitution,
$u=g(v)=1+v^{3/2}, \displaystyle \ \ =\frac{3}{2}v^{1/2}dv,\qquad \sqrt{v}dv=\frac{2du}{3}$
here, the borders change to $g(1)=2$ and
$g(4)=1+4^{3/2}=1+(2^{2})^{3/2}=1+2^{3}=9$.
$\displaystyle \int_{0}^{1}\frac{10\sqrt{v}}{(1+v^{3/2})^{2}}dr=\int_{2}^{9}\frac{10}{u^{2}}\cdot\frac{2du}{3}=\frac{20}{3}\int_{2}^{9}u^{-2}du$
$=\displaystyle \frac{20}{3}\left[ \frac{u^{-2+1}}{-2+1} \right]_{2}^{9}$
$=\displaystyle \frac{20}{3}\left[-u^{-1} \right]_{2}^{9}$
$=\displaystyle \frac{20}{3}(-\frac{1}{9}+\frac{1}{2})$
$=\displaystyle \frac{20}{3}\cdot\frac{7}{18}$
$=\displaystyle \frac{70}{27}$
