Question

I'm looking for help finding the limit of an improper integral.

Answer 1

Go for it.

Answer 2

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

Answer 3

Well, this is an integration by parts problem.
u=x dv=e^(-x)
du=dx v=-e^-x
\[\lim_{b \rightarrow \infty} [-e^{-x}x|_0^b+\int\limits_{0}^{b}e^{-x}dx]=\lim_{b \rightarrow \infty}[-xe^{-x}-e^{-x}]_0^b\]
Evaluating gives:
\[[-be^{-b}-e^{-b}]-[-(0)e^{-0}-e^{-0}]=-be^{-b}-e^{-b}+1\]
So you have:
\[\lim_{b \rightarrow \infty}\frac{-b}{e^b}-\lim_{b \rightarrow \infty} e^{-b}+\lim_{b \rightarrow \infty}1\]
The first one, do l'hospital's rule.
\[\lim_{b \rightarrow \infty}\frac{-1}{e^b}=0\]
The second limit is 0 as well. Then the third is 1. So the solution is 1.

Answer 4

\[\int\limits_{}^{}xe^{-x}dx=-xe^{-x}-\int\limits_{}^{}-e^{-x}dx=-xe^{-x}-e^{-x}+C\]
lets check if this is right antiderivative
\[(-xe^{-x}-e^{-x})'=-1e^{-x}+xe^{-x}+e^{-x}=xe^{-x}\]
----------ok------------------
so we have
\[\lim_{b \rightarrow \infty} \int\limits_{0}^{b}xe^{-x}dx=\lim_{b \rightarrow \infty}([-be^{-b}-e^{-b}]-[-(0)e^{-0}-e^{-0}]\] :(

Answer 5

myininaya <33 haha

Answer 6

i wasn't paying attention :(

Answer 7

Its good xP You were getting to what I had xD

Answer 8

very good now i want you to tell me how to do that symbol that says im fixing to evaluate the this function at b then minus evaluate this function at a
you know that vertical line with the superscript and the subscript

Answer 9

malevolence thats funny how we both enter in all those zeros

Answer 10

Haha, all you have to do is type "|_"a"^"b"" without the quotes, just like an integral sign, you don't need to include the {}'s unless you have something that is more than one character.
\[\huge|_{this}^{like}\]
Same with brackets ]
\[\huge[]_{easy}^{too}\]

Answer 11

I normally wouldn't, I was just doing it for explanatory purposes :P

Answer 12

\[|_a^b\]

Answer 13

Yup :)

Answer 14

me too!

Answer 15

Does that make sense to you aromfreerider?

Answer 16

Thanks again, guys.

Answer 17

No problem :D

Answer 18

thanks male for showing me that ;)

Answer 19

No problem :P

Answer 20

\[|_0^{\infty}\]

Answer 21

oh cool

Answer 22

\[|_{myininaya}^{satellite}\]

Answer 23

\[|_{\text{this}}^{\text{like}}\]

Answer 24

wow i still havent slept

Answer 25

\[\color{green}{\text{get some sleep}}\]

Answer 26

satellite did you like my formula for the vertex of a parabola
(b/2a,c-b^2/4a)

Answer 27

no. i like my formula

Answer 28

oops -b/2a

Answer 29

ok i like yours too

Answer 30

or would if it was right

Answer 31

mine is
\[(-\frac{b}{2a},f(-\frac{b}{2a}))\]

Answer 32

P=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}

Answer 33

\[a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}\]

Answer 34

too much work. is it right?

Answer 35

\[(-\frac{b}{2a},c-\frac{b^2}{4a})\]

Answer 36

yes i derived it for that one guy

Answer 37

yes i guess it is right. but if i prefer direct substitution

Answer 38

but i thought you like formulas

Answer 39

there are no interesting questions here. maybe it is time to go outside. in the elements. in any case it is time for you to sleep, otherwise you will turn into a zombie

Answer 40

its too late