1. Not finding help here? Sign up for a free 30min 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!

Anti Derivative (making sure its right)

  1. Dec 2, 2007 #1
    [SOLVED] Anti Derivative (making sure its right)

    1. The problem statement, all variables and given/known data

    Find anti derivative of:

    2 cos(x) - SQRT(e^2x)

    2. Relevant equations


    3. The attempt at a solution

    I got my answer as:

    - 2 sin (x) - (2/6)e^2x^(3/2)

    Is this correct? I just want to make sure!

    Thanks for your time!
  2. jcsd
  3. Dec 2, 2007 #2
    The first part of your answer is correct, but you've integrated [tex] \sqrt{e^{2x}} [/tex] incorrectly. Keep in mind that [tex] \sqrt{e^{2x}} [/tex] is just another way of saying [tex] e^{2x}^{\frac{1}{2}} [/tex]. What does this simplify to?
  4. Dec 2, 2007 #3
    So [tex] \sqrt{e^{2x}} [/tex] isnt e^2x^(1/2) ?? (since 2x is inside the sqrt)
  5. Dec 2, 2007 #4
    It is [tex](e^{2x})^{\frac{1}{2}}[/tex]

    Now power to power rule that.

  6. Dec 2, 2007 #5
    o ok so than [tex](e^{2x})^{\frac{1}{2}}[/tex] means [tex](e^{x})[/tex] since 2 and 1/2 cancel.

    So is the anti derivative just e^x since derivative of e^x = e^x * 1.

    So final answer is:
    -2 sin(x) - e^x ?
  7. Dec 2, 2007 #6
    Yes but watch your sign on the first term and don't forget your constant of integration +C. And just so you know, when you post a question, it helps us to help you if you post it exactly as stated in the text. i.e., don't forget dx. You don't need to use latex, but don't forget the dx, dy, dz and so on, as different problem statements require different solutions.

    Something like integral of [2 cos(x) - SQRT(e^2x)]dx would work perfectly!

    Welcome aboard!
  8. Dec 2, 2007 #7
    O ok. Thank you very much! The first term should be positive!

    Also sorry about that, it does say dx!

    Thanks a lot!
  9. Dec 2, 2007 #8
    Really? That is strange. Glad you got it!

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

Have something to add?