Question

I need help with this integral:

integral [e^-sqrt(x)]/[sqrt(x)] from 1 to infinity.

Answer 1

u =

Answer 2

see what you get when you make your -sqrt(x) your u

Answer 3

\[\int\limits_{}^{} \frac{e^{-\sqrt{x}}}{\sqrt{x}} dx \]
\[\text{Let } u=\sqrt{x} => du=\frac{1}{2 \sqrt{x}} dx => 2 du=\frac{1}{\sqrt{x}} dx\]

Answer 4

I did do the u-sub, I just didn't know where to go from there. I really suck at improper integrals.

Answer 5

\[\int\limits_{}^{}e^{-u} 2 du =2(-e^{-u})+C=-2e^{-u}+C
=-2e^{-\sqrt{x}}+C\]
so we have
\[\int\limits_{1}^{\infty}\frac{e^{-\sqrt{x}}}{\sqrt{x}} dx=\lim_{b \rightarrow \infty}[-2e^{- \sqrt{b}}-(-2e^{- \sqrt{1}})]\]

Answer 6

recall
\[\lim_{b \rightarrow \infty}e^{-b}=0\]

Answer 7

you don't have to recall simply move your e^-sq(x) to the bottom

Answer 8

Okay thank you so much! That's very helpful :)

Answer 9

well i would recall it is easy

Answer 10

\[-2e^{-x}+c=\frac{-2}{e^\sqrt{x}}\]

Answer 11

\[-2(0)+2\frac{1}{e}=\frac{2}{e}\]

Answer 12

Oh wait, I just got confused. What happens to the sqrt(x) on the bottom after you do the u-sub?

Answer 13

do you remember what was du?

Answer 14

myin has it correct but if you didn't know that e^-b goes to 0, you can just simply do what i did and solve to see that to infinity goes to 0

Answer 15

Oh right! Yes okay I see now. Sorry I forgot to move that over w/ the 2.

Answer 16

how long is it going to say "website is restarting lol"

Answer 17

its annoying

Answer 18

Oh okay, my calc one teacher drilled it into my head, so I remember that negative infinity goes to zero when it's an exponent :) Thank tho!

Answer 19

lol i know haha how do you know that e^-b goes to 0 ... do you read your calc book for fun lmfao =]

Answer 20

oh wait i exited out of it and it didn't pop back up oh wait it just popped back up :(

Answer 21

how long has that sign been up? I'm guessing longer than an hour?

Answer 22

e^{x} goes to 0 as x goes to negative infinity
so
e^{-x} goes to 0 as x goes to infinity

Answer 23

yes because
\[x^{-\infty}=\frac{1}{x^\infty}\]