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## ExamplesWe will evaluate $\displaystyle\int_0^1
x^2\sqrt{x^3+1}\,dx$ using the first method.
Now that we have the antiderivative, we go back to
the definite integral and use
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. |