Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculators TI-89 Titanium INTEGRATION Flaw?

  1. Feb 22, 2007 #1
    TI-89 Titanium INTEGRATION Flaw???

    Ok so this is my first post, and I am in dire need of assistance. I have a big exam tomorrow and if I can't figure out why my TI-89 is screwing up this simple integration I am in some trouble...

    I am looking for the integral of : 1/[G-(Cv)/M] with respect to "v"

    (erroneus, G=gravitational constant, C=drag coefficient, M=mass, and V=velocity)

    I know the correct answer is : -(M/C)[ln(G-(Cv)/M)]

    but my TI-89 calculator gives back the answer as:


    If anyone could please explain to me why this is so or what I am doing wrong it would be of great help...and yes I am inputting the integral in the correct fashion

    Thanks to anyone who can help!

  2. jcsd
  3. Feb 22, 2007 #2
    If you get rid of the fraction in the denominator in the original function you will arrive at the answe your calculator does, however you should notice that these only differ by a constant and that when you integrate you always have to add on a constant of integration.
  4. Feb 22, 2007 #3
    Why would I get rid of the fraction in the denominator? And if I get rid of it what do I replace it with?
  5. Feb 22, 2007 #4


    User Avatar
    Science Advisor

    You asked a question, you got an answer. Read the answer again. By the way, neither your answer nor the calculator's answer is correct!

    Both are missing the added constant. d_leet's point is that your and the calculator's answer differ only by a constant.
  6. Feb 22, 2007 #5


    User Avatar

    If we leave the constant part, the calculator is right in general, as it gives the absolute value of quantity inside logarithm.
    Last edited: Feb 22, 2007
  7. Feb 22, 2007 #6
    Maybe if it's hard for you to integrate 1/(1-ax) with respect to x then you do have a problem?

    Taken from 1/a onwards, the integral of 1/1-ax is (-1/a) ln |1-ax|. Knowing how that solve your problem without needing the calculator.
  8. Feb 23, 2007 #7
    Take the "correct" answer, and use the logarithmic rule for division: ln(a/b) = ln(a) - ln(b). Normally I'd get in trouble for telling you that, but you could have just as easily looked it up in the textbook, so I thought I'd save you a couple minutes of tedious page turning. Anyway, considering what the others here have told you, just by expanding your answer in this way, you should be convinced that your answer and the 89's answer are the same.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook