definite integration

can someone find the end result of the definite integral of xe^(-2x), from 1 to infinity?

i did it on a recent exam, and i secured the answer 3/4e^2.  although the answer was/is 'wrong', i got the question right since it was checking for convergence or divergence(either way, it would've ended up converging).  my professor said that the true answer was 1/4e^2.  i'm checking over my exam, and i don't see where i went wrong with the signs unless i'm missing something? i used classic FTC, f(b) - f(a), and i still don't know how he was able to subtract 1 from 2.

fyi, this is calc II material.  the class is calc III, but he adds in series-questions every now and then.  can someone help me out?

http://www.wolframalpha.com/input/?i=integrate+xe^%28-2x%29+dx+from+x%3D1+to+infinity

u r winner ha ha ha

They give you that stuff in Calc 3?

Here, we got shit like that in our equivalent of Calc 1.

Is this Lebesgue integration shit?

>>4

No. It's really simple. Just find the indefinite integral by parts, by letting, say, u = x, dv/dx = e^(-2x) => du/dx = 1, v = -e^(-2x)/2
This gives you that the antiderivative of (xe^(-2x) dx) = -xe^(-2x)/2 - antiderivative of (-e^(-2x)/2 dx) = -e^(-2x)/2 * (x + 1/2) + C

Then, to find whether it converges or not:
integral of (xe^(-2x) dx) from 1 to infinity = -1/2 * (lim (e^(-2a) * (a + 1/2)) as a -> infinity) - (-e^(-2)/2 * (1 + 1/2))
= -1/2 * (lim (a + 1/2) / e^2a as a -> infinity) + (3/2) / (2e^2)
= -1/2 * (lim 1 / 2e^2a as a -> infinity) + 3/4e^2
= 3/4e^2


Conclusion: OP's professor is a dumbass.

>>4
It's a continuous function on interval. So trivially within the sphere of Riemann integration...

>>6
Technically, Riemann integration is defined only on bounded intervals, no?

Is an improper integral like this still said to be "Riemann integrable"? Or is it technically only "Lebesgue integrable", but just able to be computed by Riemann means (and taking the limit of the endpoint of the interval)?

I'm just trying to see what the point of Lebesgue integration is...

>>7

Lebesgue integration is used to deal with really weird stuff. Like the integration of the cantor set. You can also marry it with distribution theory and integrate the Dirac Delta.

The point of Lebesgue integration is really to just to deal with the weirder things. It doesn't care about discontinuities or crazy jumps. That's why it's very useful in a lot of problems.