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Indefinite Integral Question; e^x

  1. Mar 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the indefinite integral (preferably using u-substitution):
    ∫(ex-e-x)2 dx


    2. Relevant equations
    N/A


    3. The attempt at a solution
    To be honest, I'm slightly confused as to which path I'm supposed to take with this, especially since I'm not sure what I should be using u to substitute here. This is one of my attempts at solving the question:

    ∫[(ex-e-x)2 dx]
    = ∫[e2x dx] - ∫[e-2x dx]
    = (e2x / 2) + (1 / 2e2x) + C

    Could someone please tell me if I have the right answer / I'm on the right track / I'm completely wrong? If the latter, I'd appreciate it if you could help get me on track!

    Thanks in advance!
     
  2. jcsd
  3. Mar 3, 2012 #2

    BruceW

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    You almost got it. But you've forgotten the cross-terms when you did the square.

    There is a good substitution you could use instead. But your method is just as good.

    Edit: p.s. welcome to physicsforums, your name is good. reminds me of the town in pokemon
     
  4. Mar 3, 2012 #3

    rock.freak667

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    If you need the u-substitution, try just putting u=ex.

    (Remember that 1/a is the same as a-1)


    EDIT: Also in your attempt you should also have


    ∫e2x dx + ∫e-2x dx +∫2 dx
     
  5. Mar 3, 2012 #4
    Before I continue asking questions, thanks for all the help! I really do appreciate it and I'm sorry if I'm a bit slow, haha.

    Now then...
    @BruceW: Thanks for the help, the welcome, and the name compliment! I'm actually a fan of the colour itself and the city in Pokemon, heh.
    @rock.freak667: Ugh, how could I forget 1/a = a-1? Thanks for the reminder. I have a question about the +∫2 dx, though. How exactly did it get there?

    I think I'm actually a bit more confused than I initially thought I was. Let's say I went with this method:

    u=ex
    du=exdx ; dx=du/ex

    ...and then what? This is what I'm leaning towards:
    =∫(u2 dx) - ∫(1/u2 dx)

    Do I substitute dx for du/ex or am I over-complicating things? Oh man, my head's gonna burst. @@
     
  6. Mar 3, 2012 #5

    SammyS

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    (ex-e-x)2 = (ex-e-x)(ex-e-x)

    Multiply it out. (Use FOIL, if that's what you're used to.)
     
  7. Mar 3, 2012 #6

    Dick

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    (e^x-e^(-x))^2=(e^x-e^(-x))*(e^x-e^(-x)). Multiply that out and you'll see where the '2' came from. rock.freak667 had the sign wrong on it.
     
  8. Mar 3, 2012 #7
    Okay, do I tried expanding it and then went through this:

    (ex-e-x)2 = (ex-e-x)(ex-e-x) = e2x + e-2x

    ∫(e2x + e-2x) dx
    u = e2x
    du = (e2x / 2) dx

    Final solution:
    (u2/e2x) + (2 / e2x)ln|e2x|
     
    Last edited: Mar 3, 2012
  9. Mar 3, 2012 #8

    Mark44

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    No, this is incorrect, as Sammy and Dick pointed out. (a - b)2 ≠ a2 - b2, which is essentially what you are doing.
    When you expand (ex - e-x)2 you should get three terms, not two.

    Also, there is an error below - you shouldn't get any log terms.
     
  10. Mar 3, 2012 #9
    Oh I see it now.
    (ex - e-x)(ex - e-x) = e2x - 2 + e-2x

    I somehow wrote "+ e0 - e0" at some point, oops. Thanks!
    Time to try again. (:
     
  11. Mar 3, 2012 #10

    Mark44

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    For you last error, I think this is what you did:

    ∫dx/e2x = ln|e2x|

    That is incorrect. Don't write the integrand as a fraction - use the negative exponents.
     
  12. Mar 3, 2012 #11
    Okay, another attempt. Hope I fixed more than I broke, this time. :/
    Thanks again everyone, I'm sorry for bothering you all. D:

    ∫(ex-e-x)2dx = ∫(e2x-2+e-2x)dx
    = ∫e2xdx + ∫e-2xdx - ∫2dx

    Let u = ex
    du = exdx

    ∫e2xdx + ∫e-2xdx - ∫2dx = (1/ex)(u3/3)+(1/ex)(u-1/-1)-2x+C
    = (1/ex)[(u3/3)+(u-1/-1)]-2x+C
    = (1/ex)[(e3x/3) - (1/ex)]-2x+C
    = (1/ex)[(e4x-3)/(3ex)]-2x+C
    = [(e4x-3)/(3e2x)]-2x+C
     
    Last edited: Mar 3, 2012
  13. Mar 3, 2012 #12

    SammyS

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    What is [itex]\displaystyle\int du \,?[/itex]
     
  14. Mar 3, 2012 #13
    ex+C?
     
  15. Mar 3, 2012 #14

    SammyS

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    Ignore my last post.

    If u = ex and du = exdx,

    then [itex]\displaystyle\int e^{2x}dx[/itex] becomes what integral in terms of u ?
     
  16. Mar 3, 2012 #15
    ∫u2dx? :x

    [EDIT] Another attempt:
    2e2x-2x-2e-2x+C

    x_x
     
    Last edited: Mar 3, 2012
  17. Mar 3, 2012 #16

    SammyS

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    Yes. You appear to be confused.

    You can't take 1/ex out of the integral. In fact you should leave it inside the integral as 1/u .

    What is u2/u ?
     
  18. Mar 3, 2012 #17
    u1 or just u? xD

    (also, I edited my previous post with another answer in case you missed it.)
     
  19. Mar 3, 2012 #18

    SammyS

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    OK, so what is [itex]\displaystyle\int u \,du \,?[/itex]
     
  20. Mar 3, 2012 #19
    (u2)/2? :x
     
  21. Mar 3, 2012 #20

    SammyS

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    What does the ":x" mean ?
     
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