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Homework Help: How do you get from one to the other

  1. Jan 13, 2012 #1

    I am going through some worked examples for a class I have, and there is one step I don't understand, and I hope someone can help me with that.

    It goes from

    [itex]\frac{1}{D_pB}\int e^{x/d}(-de^{-W_B/d} + de^{-x/d}) dx[/itex]


    [itex]\frac{1}{D_pB}\int (1 - e^{x-W_B/d}) dx[/itex]

    The limits are 0 to WB

    I have tried to work through a few things, such as multiplying out the brackets and all, but I am not getting it right, so I am doing something wrong.


  2. jcsd
  3. Jan 13, 2012 #2


    Staff: Mentor

    Show us what you have done.
  4. Jan 13, 2012 #3

    I like Serena

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    Hi SMOF! :smile:

    Multiplying the brackets out is the way to go.
    What did you get?
    Did you consider to use that ##e^x e^y = e^{x+y}##?

    Btw, there is a typo in your example.
    It appears you dropped a factor "d".
  5. Jan 13, 2012 #4


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    It looks like in going from the first expression to the second they just made a factor of [itex]d[/itex] disappear. I don't think that's right.
  6. Jan 13, 2012 #5

    Sorry, there was a typo, the second line should have been

    [itex]\frac{d}{D_pB}\int e^{x/d}(-de^{-W_B/d} + de^{-x/d}) dx[/itex]

    Working out the brackets, I get

    [itex]e^{x/d}(-de^{-W_B/d} + de^{-x/d}) dx[/itex]

    goes to (I think)

    [itex]-de^{x-W_B/d} + de^{d} [/itex]

    I am not sure about the de^d part.

  7. Jan 13, 2012 #6

    I like Serena

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    Not quite.

    You should get:
    $$-de^{(x/d)-(W_B/d)} + de^{(x/d) - (x/d)}$$
    Perhaps you can simplify that?
  8. Jan 13, 2012 #7
    Hello, thanks for the reply!

    Right, am I right in saying that the de^(x-d)-(x-d) comes to e^0 = 1?

    But the why is the other other exponential all over d, and not d - d?


  9. Jan 13, 2012 #8

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    I do not understand why you are using minus signs now instead of division slashes.
    Any particular reason?

    Anyway, you need to evaluate ##e^{x/d}e^{-W_B/d}##.

    Considering that the power formula is ##\displaystyle e^a e^b = e^{a+b}##.
    Can you substitute ##a=x/d## and ##b=-W_B/d## in this formula?
  10. Jan 13, 2012 #9
    Yes, two. One, I am a bit of an idiot, and two, I should have been asleep hours ago! Sorry about that :redface:

    Yes, I think I have it now.

    Thanks for everyones help! And I shall make sure I read over what I have written before I submit it in future!

    Again, many thanks.

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