username9303
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#1
Report Thread starter 13 years ago
#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

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username9816
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#2
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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.
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Jonny W
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#3
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#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
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Christophicus
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#4
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Ooohhh, two different answers! My money's on Johnny!
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username9816
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#5
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#5
(Original post by Widowmaker)
Ooohhh, two different answers! My money's on Johnny!
How cheeky of you, just because he's cleverer!
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yazan_l
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#6
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#6
(Original post by Widowmaker)
My money's on Johnny!
:ditto:
i did it without seeing the answers... and i got the same answer as Jonny_W's
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Jonny W
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#7
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#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 .
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yazan_l
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#8
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#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 .
:ditto: again!
i've just found it out too! but i refreshed: and Jonny got it first! hehehe!
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username9816
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#9
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#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?
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Jonny W
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#10
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(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.
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username9816
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#11
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#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.
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CalculusMan
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#12
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#12
(Original post by Nima)
cleverer!
No such word.
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username9816
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#13
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#13
(Original post by Matt_2K)
No such word.
I think you missed the subtle joke at myself by me saying he's cleverer...
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dvs
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#14
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(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
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CalculusMan
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#15
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#15
(Original post by Nima)
I think you missed the subtle joke at myself by me saying he's cleverer...
I feel thick tonight.
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posh_git
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#16
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#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?
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Caribou
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#17
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#17
*blank face*

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