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    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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    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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    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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    Ooohhh, two different answers! My money's on Johnny!
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    (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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    (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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    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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    (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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    (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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    (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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    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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    (Original post by Nima)
    cleverer!
    No such word.
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    (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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    (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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    (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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    (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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    *blank face*

    maths.. pah.. who needs that!
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    oh lol cos you're differentiating and not integrating.....please tell me im right!
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    (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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    lol thank you i don't feel as thick!
 
 
 
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Updated: April 14, 2006
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