Question

what is the integral of xe^-x with limits 0 to infinity?

Answer 1

\[\int\limits_{}^{}xe^{-x}dx=-xe^{-x}-\int\limits_{}^{}-e^{-x}dx=-xe^{-x}-e^{-x}\]

Answer 2

before we look at the limits lets check this

Answer 3

(-1)e^{-x}--x(-e^{-x})+e^{-x}
=xe^{-x}
YAY!
now for the limists...

Answer 4

\[\lim_{b \rightarrow \inf}(-be^{-b}-e^{-b}+e^0)\]

Answer 5

e^0=1
-e^{-b}->0 as b->inf
now let's look at -be^{-b}

Answer 6

we are using limits because it is improper integral?

Answer 7

yes! :)

Answer 8

ahh i see. i was wondering where the e^0 came from?

Answer 9

i plugged in ther limits
the bottom limit is 0
-[0e^0-e^0]=+e^0=+1

Answer 10

now we have

Answer 11

\[\lim_{b \rightarrow \inf}be^{-b}=\lim_{b \rightarrow \inf} \frac{b}{e^b}=\lim_{b \rightarrow \inf} \frac{1/e^b}\]

Answer 12

\[\lim_{b \rightarrow \inf}be^{-b}=\lim_{b \rightarrow \inf} \frac{b}{e^b}\]

Answer 13

use l'hospital's rule
so we have
\[\lim_{b \rightarrow \inf} \frac{1}{e^b}=\lim_{b \rightarrow \inf}e^{-b}=0\]

Answer 14

so we have -0-0+1=1

Answer 15

ah i see! thank you so much for your help :)

Answer 16

np