# Integral question Watch

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Completely stuck on this one. I tried to solve the integral using the substitution t = e

Any help would be much appreciated

^{-x}but I just got lost in my own workings.Any help would be much appreciated

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#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.

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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#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

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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#5

(Original post by

Ooohhh, two different answers! My money's on Johnny!

**Widowmaker**)Ooohhh, two different answers! My money's on Johnny!

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#6

(Original post by

My money's on Johnny!

**Widowmaker**)My money's on Johnny!

i did it without seeing the answers... and i got the same answer as Jonny_W's

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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 .

I made several similar errors in my solution, but thanks to Mathematica corrected them before I posted .

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#8

(Original post by

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 .

**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 .

i've just found it out too! but i refreshed: and Jonny got it first! hehehe!

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#9

(Original post by

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

**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

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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#10

(Original post by

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?

**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?

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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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#13

(Original post by

No such word.

**Matt_2K**)No such word.

**cleverer**...

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#14

(Original post by

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.

**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.

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#15

(Original post by

I think you missed the subtle joke at myself by me saying he's

**Nima**)I think you missed the subtle joke at myself by me saying he's

**cleverer**...
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#16

**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 .

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#18

oh lol cos you're differentiating and not integrating.....please tell me im right!

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#19

(Original post by

oh lol cos you're differentiating and not integrating.....please tell me im right!

**posh_git**)oh lol cos you're differentiating and not integrating.....please tell me im right!

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