Question

Suppose you have a function f(x) = e^(-Ax), where A is a positive real number.

Show that the integral from 1 to 2 of x*f(x) dx approaches 0 as A approaches infinity

Answer 1

Otherwise:
\[f(x) = e ^{-Ax}\], A is a positive real number
Show:
\[\int\limits_{1}^{2}xf(x)dx\] approaches 0 as A approaches infinity

Answer 2

Try to consider that
\[\int\limits_{}^{}e^{nx}dx=\frac{ e^{nx} }{n}\]
As n approaches negative infinity, you get
\[\frac{ e^{-\infty x} }{-\infty}=-\frac{1}{e^{\infty x} \infty}\]
Which for x = 1 and x = 2 approaches 0

Answer 3

That was actually the first part of the problem. Wouldn't integrating by x times the function be more different that just that answer, even though in the end it will come down to a bunch of fractions that will ultimately reach zero as a goes to infinity?

Answer 4

Oh, didn't read that x, sorry.
I'll just post the WolframAlpha approach, since I have no idea on how to integrate by parts.
http://www3.wolframalpha.com/Calculate/MSP/MSP76951a39f64hh908gi9d000029a17226icg50i1h?MSPStoreType=image/png&s=1&w= [[1]]&h=[[2]]
From that it should be easy.

Answer 5

(You have to copy-paste the full URL into your browser, not just the part in blue)

Answer 6

Ah. Thank you for the that. I actually integrated that using integration by parts and that just confirmed my answer. Thanks!

Answer 7

If that's all you needed you can try to go on WolframAlpha.com next time.