Computing $\int_0^1 xe^{2x}\, dx$

Question

My approach to compute $\int_0^1 xe^{2x}\, dx$ is via integration by parts by setting $$ u'(x)=x^2,\qquad v(x)=e^{2x} $$ which gives me $$ \int_0^1xe^{2x}\, dx=\frac{e^2}{2}-\int_0^1x^2e^{2x}\, dx. $$ Then, doinf again integration by parts by setting $$ u'(x)=x^2,\qquad v(x)=e^{2x}, $$ I get $$ \int_0^1 x^2e^{2x}\, dx=\frac{e^3}{3}-\frac{2}{3}\int_0^1x^3e^{2x}\, dx $$ which gives me $$ \int_0^1 xe^{2x}\, dx=\frac{e^2}{6}+\frac{2}{3}\int_0^1 x^3e^{2x}\, dx. $$

I guess now I have to do another integration by parts and so on but this won't come to an end. So what am I doing wrong?

Answer 1

There is a rule of thumb, known as the ILATE or the LIATE rule, where

I - Inverse trigonometric, L - Logarithmic, A - Algerbraic, T - Trigonometric, E - Exponential,

It suggests that the type of function that appears first in the acronym be taken as $v(x)$ and the other as $u'(x)$.

In this case, $x$ is algebraic and $e^{2x}$ is exponential. This suggests a reverse order from what you chose.

$$I=\int_0^1xe^{2x}dx=\frac{e^2}{2}-\frac12\int_0^1e^{2x}dx$$

which can be easily simplified.

Also see this question and this question.

Answer 2

You can easily compute the integral by parts: recall that $\int fg'=fg-\int f'g$. Then you choose $f(x)=x$, so that $f'(x)=1$, and $g'(x)=\mathrm{e}^{2x}$, getting thus $g(x)={1\over 2}\mathrm{e}^{2x}$. Compute $$ \int x\mathrm{e}^{2x}\,dx={1\over 2}x\mathrm{e}^{2x}-{1\over 2}\int \mathrm{e}^{2x}\,dx={1\over 2}x\mathrm{e}^{2x}-{1\over 4}\mathrm{e}^{2x}+C. $$ Thus you have: $$ \int_0^1 x\mathrm{e}^{2x}\,dx=\left[{1\over 2}x\mathrm{e}^{2x}-{1\over 4}\mathrm{e}^{2x}\right]_0^1={1\over 4}\mathrm{e}^2+{1\over 4}. $$