Question

See anything wrong in evaluating this integral ?

Answer 1

\[\begin{align}

\int xe^x\,dx &= \dfrac{x^2}{2}e^x - \int \dfrac{x^2}{2}e^x \,dx ~\\~\\

&= \dfrac{x^2}{2}e^x - \dfrac{x^3}{3!}e^x + \int \dfrac{x^3}{3!}e^x \,dx ~\\~\\

&= \dfrac{x^2}{2}e^x - \dfrac{x^3}{3!}e^x + \dfrac{x^4}{4!}e^x -\cdots + C ~\\~\\

&= \color{red}{-e^x+xe^x} + e^x\left(\color{red}{1-x}+\dfrac{x^2}{2}e^x - \dfrac{x^3}{3!}e^x + \dfrac{x^4}{4!}e^x -\cdots\right) + C ~\\~\\

& = \color{red}{-e^x+xe^x} + e^x\left(e^{-x}\right) + C\\~\\
&=\color{red}{-e^x+xe^x} + C\\~\\
\end{align}\]

Answer 2

Technically, you need to prove that:
\[ \lim_{n \rightarrow \infty} \int \frac{x^n}{n!} e^x dx = 0\]

Answer 3

Other than your typo in line 4, it seems fine

Answer 4

\[\begin{align}
\int xe^x\,\mathrm dx
&&&\color{gray}{\textstyle \int u\,\mathrm dv=uv-\int v\,\mathrm du}\\
&&\color{gray}{u}&\color{gray}{\, =x\qquad\mathrm dv= e^x\,\mathrm dx}\\
&&\color{gray}{\mathrm du} \, &\color{gray}{=\mathrm dx\qquad v=e^x}\\
&= xe^x-\int e^x\,\mathrm dx\\
&=xe^x-e^x+C
\end{align}\]

Answer 5

I see the e^x 's in the parentheses are the afore mentioned typo.
But, how did you get the +C to be outside of these parentheses?

Answer 6

Ahh right, those e^x 's inside the parenthesis are a typo.

The rest of the expression without C is one antiderivative of xe^x,
so I have simply added a C since all the antiderivatives differ by a constant ?

Answer 7

@Bobo-i-bo Yeah, I think I have used the convergence of e^x series without explicitly stating...

Answer 8

No, wait. I see what you mean now

Answer 9

This result needs to be proven at step3 !
\[ \lim_{n \rightarrow \infty} \int \frac{x^n}{n!} e^x dx = 0\]

Answer 10

Yup! :)