## anonymous 5 years ago what is the integral of xe^-x with limits 0 to infinity?

1. myininaya

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

2. myininaya

before we look at the limits lets check this

3. myininaya

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

4. myininaya

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

5. myininaya

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

6. anonymous

we are using limits because it is improper integral?

7. myininaya

yes! :)

8. anonymous

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

9. myininaya

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

10. myininaya

now we have

11. myininaya

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

12. myininaya

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

13. myininaya

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

14. myininaya

so we have -0-0+1=1

15. anonymous

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

16. myininaya

np