We will evaluate $\displaystyle\int_0^1
x^2\sqrt{x^3+1}\,dx$ using the first method.

**DO:** Try this, first finding the
antiderivative using substitution, write it in terms of $x$,
then use **FTC II** to
evaluate. Don't look at
the solution below until you've tried!

$$\int x^2\sqrt{x^3+1}\,dx\overset{\fbox{$
\,\,\,\,\,\,u\,=\,x^3+1\\ \,\,\,\,du\,=\,3x^2\,dx\\
\frac{1}{3}\,du\,=\,x^2\,dx$}\\}{=}\int\frac{1}{3}\sqrt
u\,du=\frac{1}{3}\int
u^{\frac{1}{2}}\,du=\frac{1}{3}\frac{2}{3}u^{\frac{3}{2}}+c=\frac{2}{9}(x^3+1)^{\frac{3}{2}}+c.$$
(Always check:
$\frac{d}{dx}\left(\frac{2}{9}(x^3+1)^{\frac{3}{2}}+c\right)=\frac{2}{9}\frac{3}{2}(x^3+1)^{\frac{1}{2}}(3x^2)=\frac{1}{3}(x^3+1)^{\frac{1}{2}}(3x^2),$
which is our original integrand.)

Now that we have the antiderivative, we go back to
the definite integral and use **FTC II. **We have
$$\displaystyle\int_0^1
x^2\sqrt{x^3+1}\,dx=\frac{2}{9}(x^3+1)^{\frac{3}{2}}\left|\begin{array}{c}
^1 \\ _0 \end{array}\right
.=\frac{2}{9}\left((1^3+1)^{\frac{3}{2}}-(0^3+1)^{\frac{3}{2}}\right)=\frac{2}{9}\left(2^{\frac{3}{2}}-1\right),$$
where we have used the antiderivative where $c=0$ (we can use
any antiderivative in **FTC II**.)

**-------------------------------------------------------------------------------------**

Now we will evaluate the same integral, $\displaystyle\int_0^1
x^2\sqrt{x^3+1}\,dx$, using the second method.

**DO:** Try this, first using
substitution to get the integrand in terms of $u$, then change
your limits to be in terms of $u$, then use **FTC II** to evaluate. Don't
look at the solution below until you've tried!

We use the same substitution

, but we must also change
our limits of integration. When $x=0$, $u=x^3+1=0^3+1=1$,
and when $x=1$, $u=x^3+1=1^3+1=2$. We get $$\int
x^2\sqrt{x^3+1}\,dx\overset{\fbox{$ \,\,\,\,\,\,u\,=\,x^3+1\\
\,\,\,\,du\,=\,3x^2\,dx\\
\frac{1}{3}\,du\,=\,x^2\,dx$}\\}{=}\frac{1}{3}\int_{x=0}^{x=1}
u^{\frac{1}{2}}\,du\overset{\fbox{$ u(0)\,=\,1\\
u(1)\,=\,2$}\\}{=}\frac{1}{3}\int_1^2
u^{\frac{1}{2}}\,du=\frac{1}{3}\frac{2}{3}u^{\frac{3}{2}}\left|\begin{array}{c}
^2 \\ _1 \end{array}\right
.=\frac{2}{9}\left(2^{\frac{3}{2}}-1^\frac{3}{2}\right)=\frac{2}{9}\left(2^{\frac{3}{2}}-1\right)$$

If you work through both methods, you will see that the exact same
work must be done, but in a slightly different order. The
methods are equivalent. There are problems with both: in the
first,

**you must remember** to
rewrite your antiderivative in terms of $x$ before
evaluating. In the second,

**you
must remember **to change your limits of
integration.