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# Integral question watch

1. Completely stuck on this one. I tried to solve the integral using the substitution t = e-x but I just got lost in my own workings.

Any help would be much appreciated

2. f(x) = Sqrt[e^(-x) - e^(-2x)] has a defined limit as x -> Inf (zero) and is also defined at the lower limit x = 0. Hence, Int (0 to Inf) f(x) dx will have a defined value and thus is convergent.

I = Int (0 to Inf) Sqrt[e^(-x) - e^(-2x)] dx
= Int (0 to Inf) e^(-x/2).Sqrt[1 - e^(-x)] dx

Let u = e^(-x/2) = > du/dx = -(1/2)e^(-x/2) => dx = -2e^(x/2)
x = 0 => u = 1/e^0 = 1
x = Inf => u = 1/Inf = 0

:. I = Int (1 to 0) e^(-x/2) . Sqrt[1 - u^2] . (-2)e^(x/2) du
= 2 Int (0 to 1) Sqrt[1 - u^2] du

Let u = [email protected] => du/[email protected] = [email protected] => du = [email protected] [email protected]
u = 0 => @ = sin^-1(0) = 0
u = 1 => @ = sin^-1(1) = Pi/2

:. I = 2 Int (0 to Pi/2) cos^2(@) [email protected]
= Int (0 to Pi/2) ([email protected] + 1) [email protected]
= [([email protected])/2 + @] (0 to Pi/2)
= [sin(Pi)/2 + Pi/2]
= Pi/2

EDIT: I made a silly error, correct answer as above, previously was Pi/8.
3. t = e^(-x)
dt/dx = -t

(int from 0 to infinity) sqrt(e^(-x) - e^(-2x)) dx
= (int from 1 to 0) sqrt(t - t^2) * -(1/t) dt
= (int from 0 to 1) sqrt(1/t - 1) dt

t = cos^2(u)
dt/du = -2cos(u)sin(u)

= (int from pi/2 to 0) sqrt(sec^2(u) - 1) * -2cos(u)sin(u) du
= 2 (int from 0 to pi/2) tan(u)cos(u)sin(u) du
= 2 (int from 0 to pi/2) sin^2(u) du
= (int from 0 to pi/2) 1 - cos(2u) du . . . . . cos(2u) = 1 - 2sin^2(u)
= [ u - (1/2) sin(2u) ] (from 0 to pi/2)
= pi/2
4. Ooohhh, two different answers! My money's on Johnny!
5. (Original post by Widowmaker)
Ooohhh, two different answers! My money's on Johnny!
How cheeky of you, just because he's cleverer!
6. (Original post by Widowmaker)
My money's on Johnny!

i did it without seeing the answers... and i got the same answer as Jonny_W's
7. Nima's mistake was here: "u = e^(-x/2) = > du/dx = -2e^(-x/2)"

I made several similar errors in my solution, but thanks to Mathematica corrected them before I posted .
8. (Original post by Jonny W)
Nima's mistake was here: "u = e^(-x/2) = > du/dx = -2e^(-x/2)"

I made several similar errors in my solution, but thanks to Mathematica corrected them before I posted .
again!
i've just found it out too! but i refreshed: and Jonny got it first! hehehe!
9. (Original post by Jonny W)
t = e^(-x)
dt/dx = -t

(int from 0 to infinity) sqrt(e^(-x) - e^(-2x)) dx
= (int from 1 to 0) sqrt(t - t^2) * -(1/t) dt
= (int from 0 to 1) sqrt(1/t - 1) dt

t = cos^2(u)
dt/du = -2cos(u)sin(u)

= (int from pi/2 to 0) sqrt(sec^2(u) - 1) * -2cos(u)sin(u) du
= 2 (int from 0 to pi/2) tan(u)cos(u)sin(u) du
= 2 (int from 0 to pi/2) sin^2(u) du
= (int from 0 to pi/2) cos(2u) + 1 du
= [ (1/2) sin(2u) + u ] (from 0 to pi/2)
= pi/2
cos2x = cos^2x - sin^2x = 1 - 2sin^2x => 2sin^2x = 1 - cos2x => sin^2x = (1 - cos2x)/2
However, you replaced sin^2u with (1 + cos2x)/2.

I corrected my error, and I got Pi/2 as well - But it seems like you made an error in your working?
10. (Original post by Nima)
cos2x = cos^2x - sin^2x = 1 - 2sin^2x => 2sin^2x = 1 - cos2x => sin^2x = (1 - cos2x)/2
However, you replaced sin^2u with (1 + cos2x)/2.

I corrected my error, and I got Pi/2 as well - But it seems like you made an error in your working?
You are right of course.
11. I think we have truly discovered the power of Jonny W. Even if he makes an error in the working, a true rarity, he'll still get the right answer hehe.
12. (Original post by Nima)
cleverer!
No such word.
13. (Original post by Matt_2K)
No such word.
I think you missed the subtle joke at myself by me saying he's cleverer...
14. (Original post by Nima)
I think we have truly discovered the power of Jonny W. Even if he makes an error in the working, a true rarity, he'll still get the right answer hehe.
rofl
15. (Original post by Nima)
I think you missed the subtle joke at myself by me saying he's cleverer...
I feel thick tonight.
16. (Original post by Jonny W)
Nima's mistake was here: "u = e^(-x/2) = > du/dx = -2e^(-x/2)"

I made several similar errors in my solution, but thanks to Mathematica corrected them before I posted .
OK im feeling thicker...why is that wrong?
17. *blank face*

maths.. pah.. who needs that!
18. oh lol cos you're differentiating and not integrating.....please tell me im right!
19. (Original post by posh_git)
oh lol cos you're differentiating and not integrating.....please tell me im right!
You are.
20. lol thank you i don't feel as thick!

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