Answer
$a.\quad \displaystyle \frac{\sqrt{10}-3}{2}$
$ b.\quad \displaystyle \frac{3-\sqrt{10}}{2}$
Work Step by Step
$a.$
Applying th.7, let $u=g(x)=x^{4}+9$ (a continuous function).
Then, $\displaystyle \quad du=g'(x)dx=4x^{3}dx,\qquad x^{3}dx=\frac{1}{4}\cdot du$
and, the borders change to $g(0)=9$ and $g(1)=10$.
$\displaystyle \int_{0}^{1}\frac{x^{3}}{\sqrt{x^{4}+9}}dx=\int_{9}^{10 }\frac{\frac{1}{4}\cdot du}{u^{1/2}}=\frac{1}{4}\int_{9}^{10 }u^{-1/2}du$
$=\displaystyle \frac{1}{4}\left[ \frac{u^{-1/2+1}}{-1/2+1} \right]_{9}^{10}$
$=\displaystyle \frac{1}{4}\left[ \frac{u^{1/2}}{1/2} \right]_{9}^{10}$
$=\displaystyle \frac{1}{4}\left[ 2u^{1/2} \right]_{9}^{10}$
$=\displaystyle \frac{1}{2}(\sqrt{10}-3)$
$=\displaystyle \frac{\sqrt{10}-3}{2}$
$b.$
Using the same substitution, $u=x^{4}+9,\ du=4x^{3}dx$,
the borders here change to
$g(-1)=10$ and $g(0)=9$.
$\displaystyle \int_{-1}^{0}\frac{x^{3}}{\sqrt{x^{4}+9}}dx=\int_{10}^{9 }\frac{\frac{1}{4}\cdot du}{u^{1/2}}=\frac{1}{4}\int_{10}^{9 }u^{-1/2}du$
and, if we exchange the borders, we can apply the Order of Integration rule:
$=-\displaystyle \left(\frac{1}{4}\int_{9}^{10 }u^{-1/2}du \right)$
We evaluated this same integral in part (a)
$=-\displaystyle \frac{\sqrt{10}-3}{2}$
$=\displaystyle \frac{3-\sqrt{10}}{2}$
