Question

evaluate the integral from negative infinity to infinity of xe^(x^(2)) dx

Answer 1

0 i believe

Answer 2

yeah, its zero but i dont know how to evaluate it

Answer 3

to get zero

Answer 4

Let u=x^2
du/dx=2x
du/2=x dx

\[\int{xe^{x^2}}dx=1/2\int{e^u}du = \frac{e^u}{2}+C=\frac{e^{x^2}}{2}+C\]
When x approaches infinity (or minus infinity, since it is squared), e^(x^2) approaches infinity as well. So the answer should be infinite.

(Do you really mean e^(x^2) or is it maybe e^(-x^2)?)

Answer 5

yeh it doesnt converge i remember learning about this forgot the reason though

Answer 6

Intuitively, the function xe^(x^2) is odd, so the answer should be automatically zero.

Answer 7

i has something to do with the limits... from -R to R it is 0 but something strange happens at infinity

Answer 8

its: e^(-x^2)

Answer 9

because e^(x^2) is obviously an even function, which is multiplied by the odd function x, and hence the function is odd. Odd functions always have that property.

Answer 10

yes for finite values your right but not for infinite values

Answer 11

thats a good way to look at it

Answer 12

Oh! In that case, e^(-x^2) approaches zero for both positive infinity and negative infinity, so the answer is zero.

Answer 13

oh ok :) thank u!

Answer 14

no matter how good you get at maths it always comes down to signs to screw everything up