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How would I do this integral. I have tried the following.

$\int_0^{2} x e^{\ln(2)x}$

$u=x$ $dv=e^{\ln2(x)}$ $du=1$ $v=\frac{1}{\ln(2)}e^{\ln2(x)}$

$x\frac{1}{\ln(2)}e^{\ln2(x)}-\frac{1}{\ln(2)}\int e^{\ln2(x)}$


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I see well what I meant is 2^x= e^ln(2)x – Fernando Martinez Sep 6 '13 at 21:46
Dont just change $1$ title!!! – Jean-Sébastien Sep 6 '13 at 21:52
Please, please, please don't forget to add the $\operatorname{d}\!x$ at the end of integrals when the variable is $x$. Leaving it off is a little like saying the area of a 2cm by 3cm rectangle is 2cm! The $\operatorname{d}\!x$ refers to the "width" of the areas. – Fly by Night Sep 6 '13 at 22:00
up vote 3 down vote accepted

Looks good to me. Don't forget to evaluate the definite integral. I'd suggest "cleaning up" the final expression. For example, in terms of the corresponding indefinite integral,:

$$F(x) = \dfrac x{\ln 2}e^{\ln(2) x} - \dfrac{e^{\ln(2) x}}{(\ln 2)^2} + C = -\ln 2\cdot e^{\ln(2) x}\left(x + \ln 2\right) + C$$ for example.

Now simply evaluate $F(2) - F(0)$

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I see thanks for looking it over it helps because I can not be sure I am doing the problem correctly – Fernando Martinez Sep 6 '13 at 21:53
Yes, you seem to have the "gist" of integration by parts. Your posts today have been progressively improving! In this post, my answer is just a suggestion for presentation of the result. – amWhy Sep 6 '13 at 21:54
@FernandoMartinez: You can always differentiate an answer you get to see whether you are right. – André Nicolas Sep 6 '13 at 21:57
@amWhy: You are always supportive! +1 – Amzoti Sep 7 '13 at 1:56

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