Question

Hi I need help evaluating an integral, it is:

0 to 1 squareroot(1+e^-x)/e^x dx,

I'm not sure if I substitute twice or just once?

Answer 1

I find staring at it for a while helps...
\[\Large \int\limits_{0}^1\frac{\sqrt{1+e^{-x}}}{e^x}dx\]

Answer 2

Well, can we make any u-substitutions here? Looks like 1+e^-x might be a fruitful look if we see that 1/(e^x) is really e^-x.

Answer 3

Here, I'll make it easier to see :P
\[\Large \int\limits_{0}^1\frac{\sqrt{1+e^{-x}}}{e^x}dx = \int\limits_0^1e^{-x}\sqrt{1+e^{-x}}dx \]

Answer 4

\[\Large \int\limits_{0}^1\frac{\sqrt{1+e^{-x}}}{e^x}dx =\Large \int\limits_{0}^1(\sqrt{1+e^{-x}})e^{-x}dx \]

Answer 5

Ok let's not completely spoon feed him now! lol

Answer 6

Thanks for making it look viewable! So I'm pretty sure that u = e^-x so du = e^x dx

Answer 7

one sec

Answer 8

sorry.. wrong derivative...think chain rule :D

Answer 9

And besides, that u-substitution won't help you... don't forget that you still have that constant term 1 in the radical that you have to worry about...

Answer 10

Consider \(\sqrt{1+e^{-x}}=\sqrt{u(x)}\)

Answer 11

-e^-x

Answer 12

There you go... keep in mind that \(\large e^{-x}\) isn't the only function with \(\large -e^{-x}\) for its derivative....

Answer 13

quick question, wouldn't i get a weird interval with this sub?

Answer 14

like -1/e to 0 lol

Answer 15

Where did you get that interval?

Answer 16

u=1=-e^-1

Answer 17

+1

Answer 18

\[\Large \int\limits_a^b(f\circ g)(x)g'(x)dx \]
Let u = g(x)
du = g'(x)dx

\[\Large \int\limits_{g(a)}^{g(b)}f(u)du \]

Answer 19

Yes I understand that but the interval is still what I said right?

Answer 20

Check it... if x = 0, what is \(\large 1+e^{-x}\)?

Answer 21

No, @iambatman you've chosen the wrong substitution. Pick:

u=1+e^(-x)

And hopefully you'll see why that works out well for you. It'll just take some practice to get the hang of "seeing" it.

Answer 22

2

Answer 23

ahh thanks kai lol

Answer 24

and what about when x = 1?
Work from there :D

Answer 25

1+1/e?

Answer 26

Yup, that's the one. I guess it is rather weird, but hey, that's the way things are :P

Answer 27

I got that before and thought I was wrong since I've never seen anything like it before lol

Answer 28

so I have a negative integral from 1+1e to 2 (square root (u)) du at the moment, so shall i go and get the anti deriv. of it now? :P

Answer 29

negative integral from 1+1/e to 2?
How?

Answer 30

du = -e^-x dx, du/ -e^-x = dx and the intervals were changed to 2 and 1+1/e because u = 0 and u = 1

Answer 31

Remember that THIS was your integral...
\[\Large \int\limits_{0}^1(\sqrt{1+e^{-x}})e^{-x}dx\]

Answer 32

Now, you let \(\large u = 1+e^{-x }\)

Answer 33

yeah I did that du/dx = -e^-x

Answer 34

Yes... -e^-x... something that you don't actually HAVE in your integrand...

Answer 35

Thank god for algebra, you can multiply both sides by -1. Disaster averted.

Answer 36

To rectify that, we multiply it by (-1) two times...
\[\Large \color{red}-\int\limits_{0}^1(\sqrt{1+e^{-x}})(\color{red} -e^{-x})dx\]

Answer 37

Now, you may proceed normally.

Answer 38

yeah thats what i have

Answer 39

Then... continue it.

Answer 40

Show us what you're made of bat man. Help us help you by putting in the effort.

Answer 41

one sec

Answer 42

ok i'm getting some weird answer
4sqaureroot(2)/3 -2/3(1+1/e)^3/2

Answer 43

Perfect!

http://www.wolframalpha.com/input/?i=integral+0+to+1+of+e%5E%28-x%29sqrt%281%2Be%5E%28-x%29%29dx

Answer 44

Sweet, thanks for the help guys!