Science:Math Exam Resources/Courses/MATH103/April 2010/Question 01 (b)/Solution 1

We want to integrate

$\int x^{2}e^{2x}dx$

with integration by parts. We have three possible products:

a) $1\cdot x^{2}e^{2x}$

b) $x\cdot xe^{2x}$

c) $x^{2}\cdot e^{2x}$

The first product doesn't make sense to choose since the second term is neither easy to derivate nor to integrate.

The second product also doesn't make sense because the second term is not easy to integrate. Also choosing $xe^{2x}$ as $u$ and $x$ as $dv$ is not good since integrating $x$ AND derivating $xe^{2x}$ makes the equation more complicated.

So it must be the third product. It doesn't matter, if we choose $e^{2x}$ as $u$ or $dv$ because the integral is the same (excapt for constants). If we choose $x^{2}$ as $dv$, we get $x^{3}/3$ as $v$, what makes the equation more complicated. So we take $x^{2}$ as $u$ and get $du=2x$ and $e^{2x}$ $dv$ and get $v=e^{2x}/2$. That simplified the equation. So we take answer ii.

Integration would work now like this:

${\begin{aligned}\int x^{2}e^{2x}dx=&{\frac {1}{2}}x^{2}e^{2x}-\int 2x{\frac {1}{2}}e^{2x}dx\\=&{\frac {1}{2}}x^{2}e^{2x}-\underbrace {\int xe^{2x}dx} _{\text{integrate by parts again}}\\=&{\frac {1}{2}}x^{2}e^{2x}-{\frac {1}{2}}xe^{2x}+\int {\frac {1}{2}}e^{2x}dx\\=&{\frac {1}{2}}x^{2}e^{2x}-{\frac {1}{2}}xe^{2x}+{\frac {1}{4}}e^{2x}+c\end{aligned}}$