[Calculus] It didn't take me long to forget how to Integrate..
 

It's been about a year since I had Calc. II and I simply cannot integrate this:

((2xe)^-x)^2

Parenthesis added for ease of reading, that's 2xe being raised to the x which is raised to the 2 power.
I was thinking substitution but its still just been too long, and I never got the 'by parts' thing down ;)

Thanks :)

you may find the problem easier if you rearrange a little:

(2xe)^-2x dx (I assume there's a dx somewhere in there...)

Yeah sorry.

That looks much easier but I dont see how you can do that... You're saying that the power -x^2 is the same as the power -2x?

No it's not, but the notation: ((2xe)^-x)^2 simplyfies to (2xe)^(-2x) not (2xe)^(-x^2)
i.e. (x^n)^m = x^(nm) != x^(n^m)

I'm not sure but someone once told me that you cannot integrate x^x. Are you sure that the question is possible?
(The derivative of x^x I can do though :)).

you can't X^x can only be done numerically, anyway, are you sure you did the () right. because if that there was no X^x term then it might be possible any way i'm looking at it right now and i don't see a clear way to do it. Search for "mathematica and integrating online" there should be a website which intergrates for you, this way you can at least know if there is a solution.

Christ,

I got straight A's in all my math through diffQ and I don't have a friggin clue.

Course, that was a few years ago...

hmm, this site(http://integrals.wolfram.com/) says:
x^2/8xe^2
I haven't got the cares to check it right now, and I can't find a formula that explains it in my differentials notes, but check it out

It's been three years since I did Calculus, and even longer since I did Algebra -- but I think Integration by Parts DOES solve it. (It's different than Wolfram's Solution, however..)

http://img445.imageshack.us/img445/3263/right0il.jpg

That would work if it's 2xe^(-2x), but it looks like it's (2xe)^(-2x).

ps. dunno how blizzak got his answer--i tried that site and came up with no answer. (looks like a neat site though!) I also wonder if this is possible without numerical methods.

There's no analytic integral for (2xe)^(-2x). If this is a class, I think the original poster used parentheses incorrectly. 2x e^(-2x) is much more reasonable.

You can't seperate an exponential from its exponent with parentheses..

What exactly are you E'ing?

hmmm jinnkai we must follow different schools of notation. looks OK to me. ever hear the expression 'please excuse my dear aunt sally?'

I think we all just ought to wait until the question is restated in a different way by the original poster. Clear it up for us...

Ack, REALLY sorry! It was my fault the parenthesis are used incorrectly. I should have just left them off completely.

2xe^-x^2


Again, my apologies :(

Using substitution I have the integral of:
e^-u du

If I remember correctly, is that just e^-u = e^-x^2?

Letting u=x^2 gives du=2x dx, and hence the integrand is now e^(-u) which integrates to -e^(-u) + C. Note the minus sign in front of exp.

gotcha! thanks :)