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Homework Help: General Antiderivative

  1. Mar 17, 2008 #1
    1. The problem statement, all variables and given/known data

    Hi everyone. My Calc 1 final exam is tomorrow and due to some weather related issues we were not able to cover all material for this quarter. With that said, my professor gave us a take home quiz on material that was unable to be covered. I have done by best, but I am getting hung up on an antiderivative problem.

    Find the general antiderivative of:


    f'(x) = 1-2x-4/[tex]\sqrt{x}[/tex]+5/x-8/(1+x^2)+9/x^4

    2. Relevant equations

    Now due to having virtually no time to learn about antiderivatives (we lost a whole week due to a blizzard and instructor illness) I am really unsure where to go. Do I need to rewrite the problem on one line and then find the opposite of the derivative?

    Thanks!
     
  2. jcsd
  3. Mar 17, 2008 #2

    malawi_glenn

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    well yes, what function, if you differentiate it, will give that expression as its derivative?
     
  4. Mar 17, 2008 #3
    So basically the definition of an antiderivative is the oppositive of the derivative?

    Wait, I did not just type that last sentence. =)
     
  5. Mar 17, 2008 #4
    Do you mean:

    [tex]f'(x) = 1-2x-4\sqrt{x}+\frac{5}{x}-\frac{8}{1+x^2}+\frac{9}{x^4}[/tex] ?

    Basically when you're asked to find the anti-derivative you're trying to find the function f(x), which has this derivative f'(x), which is given.
     
    Last edited: Mar 17, 2008
  6. Mar 17, 2008 #5
    1- 2x -4[tex]\sqrt{x}[/tex]+[tex]\frac{5}{x}[/tex]-[tex]\frac{8}{1+x^{2}}[/tex]+[tex]\frac{9}{x^{4}}[/tex]
     
  7. Mar 17, 2008 #6

    malawi_glenn

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    yes!

    But you often write it given f(x), find its primitive function F(x): F' = f
     
  8. Mar 17, 2008 #7
    That is sort of how I figured that it would work. My derivative skills are giving me the most trouble at this point.
     
  9. Mar 17, 2008 #8
    Do you have a textbook? First, read the section on anti-derivatives.

    Hint: separate each quantity
    f '(x) = 1, then f(x) = x + C.
    f '(x) = 2x, then f(x) = ?
    f '(x) = 9x^-4, then f(x) = ?
    .
    .
    .
    and so on.
     
  10. Mar 17, 2008 #9
    f '(x) = 2x, then f(x) = x^2
    f '(x) = 9x^-4, then f(x) = 9x^-3/-3
    correct?
     
  11. Mar 17, 2008 #10
    The general antiderivative will have a + C at the end.
     
  12. Mar 17, 2008 #11
    So far, so good

    But Snazzy is right you need to add a +C (constant of integration) term to the anti derivative, because if f(x) had a constant value some where in it, like f(x) = 2x+5, f'(x) = 2 -- so when we integrate f'(x) we need to account for the 5. We don't necessairly know it's a 5 so that's why we add the +C
     
    Last edited: Mar 17, 2008
  13. Mar 18, 2008 #12
    It's so easy to forget the C at the end.
     
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