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.)

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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.