1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How do you get from one to the other

  1. Jan 13, 2012 #1
    Hello.

    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]

    to

    [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.

    Thanks.

    Seán
     
  2. jcsd
  3. Jan 13, 2012 #2

    Mark44

    Staff: Mentor

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

    I like Serena

    User Avatar
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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
    Hello.

    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.

    Seán
     
  7. Jan 13, 2012 #6

    I like Serena

    User Avatar
    Homework Helper

    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?

    Thanks.

    Seán
     
  9. Jan 13, 2012 #8

    I like Serena

    User Avatar
    Homework Helper

    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.

    Seán
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook