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Homework Help: Part 2 of the Fundamental Theorem of Calculus

  1. Sep 6, 2006 #1

    I have a problem that I am getting stuck simplifying further.

    The problem asks me to find the integral if it exists using Part 2 of the FTC.

    I know that the second part of the FTC says:[itex]\int_{a}^{b} f(x)dx = F(b) - F(a)[/itex] Where F is the anti-derivative of f.

    Here is the problem:
    [itex]\int_{9}^{4} [/itex] x ^-1 dx

    While [itex]f(x) = [/itex] x^ -1 is not continuous throughout it is continuous on the interval [4,9]. So therefore it does exist.

    So here is my solution:
    [itex]f(x) =[/itex] x^ -1
    [itex]F(x) = ln|x| + C[/itex]
    [itex]F(4) = ln4 + C[/itex]
    [itex]F(9) = ln9 + C[/itex]

    Here is where I am having trouble simplifying. I would then use [itex]F(b) - F(a)[/itex]

    so when I do that I have [itex]ln4 + C - ln9 + C[/itex]

    Could I just write that as [itex]ln4 - ln9 + C[/itex] ? Do I still need to simplify further?

    Thanks in advance. :smile:
    Last edited: Sep 6, 2006
  2. jcsd
  3. Sep 6, 2006 #2
    I'm thinking that I would probably need to simplify further and write it as:

    [itex]ln 4/9 + C[/itex]

    Would this be correct?
  4. Sep 6, 2006 #3


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    The C's cancel out. Think about it, [itex]\int_9 ^4x^{-1}dx[/itex] is a single number, it doesn't depend on a constant. There is no such thing as THE antiderivative of f, but there is a family of antiderivatives. x |-> ln|x| is one of the antiderivatives. Any function of the form x |-> ln|x| + C is in the family, for any real C. Moreover, only functions of that form are in the family. The theorem should say that if F is ANY antiderivative of f, then

    [tex]\int _a ^b f(x)dx = F(b) - F(a)[/tex]

    By choosing F(x) = ln|x| + C, you choose an arbitrary antiderivative, but even in this case, when you write F(9) = ln(9) + C and F(4) = ln(4) + C, it's the same C, so they cancel out when you do F(4) - F(9). Alternatively, you could have chosen a particular antiderivative, e.g. the one where C = 0.
    Last edited: Sep 6, 2006
  5. Sep 6, 2006 #4
    No, you forgot to flip your limits of integration. Here:

    integral(x^-1 dx, x = 9 to x = 4) = -integral(x^-1 dx, x = 4 to x = 9)
    = F(9) - F(4) = ln(9) + C -(ln(4) + C) = ln(9) - ln(4) = ln(9/4) = ln((3/2)^2) = 2ln(3/2)

    Notice the constant of integration canceled out. This is always the case, so you can just omit it in your calculation.
  6. Sep 6, 2006 #5
    Oh right, how did I miss the cancellation of the constant C? And forget to flip the limits? I kept thinking to myself it looked odd, but never thought to flip it. lol I will be sure not to forget next time. Thanks for all your help!
  7. Sep 6, 2006 #6
    Actually I have a similiar problem that I could use a bit of help on as well.

    I have to find the integral using the same method.

    Here's the problem:
    [itex]\int_{-\pi}^{\pi} sinx dx[/itex]

    So basically it boils down to:
    = [itex]-cos\pi + C - (cos\pi + C)[/itex]
    = [itex]-cos\pi - cos\pi[/itex]
    = 2
    Last edited: Sep 6, 2006
  8. Sep 6, 2006 #7
    That part is wrong, you are missing a - sign, do you see where?
  9. Sep 6, 2006 #8
    Uhmm not really: Let me try to break it down further.

    [itex]F(\pi) = - cos\pi + C[/itex]

    [itex]F(-\pi) = cos\pi + C[/itex]

    I'm assuming for [itex]F(-\pi)[/itex] that since cos is already negative and that's paired with a negative [itex]\pi[/itex] then that combo would be positive. I'm guessing that is not correct.
  10. Sep 6, 2006 #9
    cosx is even. This means cos(x) = cos(-x) for every x. So for you, cos(-pi) = cos(pi) = -1. I think you made the mistake and said cos(-pi) = -cos(pi), which is incorrect, since cos(-pi) = -1 and -cos(pi) = -(-1) = 1.
  11. Sep 6, 2006 #10
    And you should get 0 as the answer btw when you do it so you can check yourself.
  12. Sep 6, 2006 #11
    I think I'm getting confused. I mean I believe I understand what you are saying. I do recall learning that, but as I've proven before nothing really sticks with me. lol

    So if I keep all negative signs in the equations it would look something like:

    [itex]F(\pi) = -cos\pi + C[/itex]

    [itex]F(-\pi) = -cos(-\pi) + C[/itex]

    Which in this case would give me an answer of 0.

    Oh you beat me to it. lol Thanks for your help. Next time I better just leave all my negative signs where they belong. :)
  13. Sep 6, 2006 #12
    Umm... Why do you think he needs to flip the limits? The integral as written above is from 9 to 4, so the anwer would be ln(4/9)=2*ln(2/3).
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