Blogs, Calculus, Study Help

Integration by substitution

As part of my Calculus 2 class, I have created a practice exam to help my students study.  In doing so, I felt that providing the questions and the solutions in a blog series would be helpful not just to my students, but also to anyone that may be taking, or teaching, calculus 2.  I hope you find the following solutions helpful in your studies.

You can also view me presenting the solution on YouTube at

 

Find \[\begin{align*} \int_{0}^{1}x^{3}(x^{4}+2)^{3}dx. \end{align*}\]

First we find the anti-derivatives
\[\begin{align*}
\int x^{3}(x^{4}+2)^{3}dx.
\end{align*}\]

Here, we can simplify and integrate, or we can use substitution. I will use substitution with \(i=x^{4}+2\), then \(di=4x^{3}dx\) so \(dx=\frac{di}{4x^{3}}\). We then substitute in \(i\) and \(dx\) and get
\[\begin{align*}
\int x^{3}(x^{4}+2)^{3}dx&=\int x^{3}i^{3}\frac{di}{4x^{3}} \\
&=\frac{1}{4}\int i^{3}di \\
&=\frac{1}{4}\frac{i^{4}}{4}+c \\
&=\frac{(x^{4}+2)^{4}}{16}+c.
\end{align*}\]

Using this anti-derivative, we can then find that
\[\begin{align*}
\int_{0}^{1}x^{3}(x^{4}+2)^{3}dx&=\frac{(x^{4}+2)^{4}}{16}|_{0}^{1} \\
&=\frac{3^{4}}{16}-\frac{2^{4}}{16}=\frac{65}{16}.
\end{align*}\]

Other Resources

I have it linked above, but you can find the practice exam here.  You can find the other solutions to the practice exam in the other posts available here.

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