Question

need help with one last integral..

Answer 1

Determine if converges or diverges, if converges find its value. \[\int\limits_{0}^{\infty} xe^(-x^2) dx\]

Answer 2

it doesnt look right but its xe^-x^2...not sure how to show it

Answer 3

\[\huge \int\limits_{0}^{\infty}xe^{-x^2}dx\]

Answer 4

I obviously know I need to use the limit. Should i use by parts..?not sure

Answer 5

you are correct sam

Answer 6

u sub

Answer 7

yep, u=-x^2

Answer 8

u=-x^2
du=-2xdx
du/-2 = xdx

Answer 9

damn it.so u sub not by parts? spring break really killed my brain with these..

Answer 10

\[\int\limits_{0}^{\infty}\frac{-du}{2}e^u\]
\[-\frac{1}{2}\int\limits\limits_{0}^{\infty}e^udu\]

Answer 11

change your limits

Answer 12

so i would take the limit correct? so would the -1/2 be on the outside of limit?or does that matter?

Answer 13

if \(u=-x^2\) then as \(x\to \infty\) we have \(u\to -\infty\)

Answer 14

so doesnt that mean it diverges, and cant find value?

Answer 15

Anyone know for sure about the diverges thing?

Answer 16

it converges...the answer is 1/2