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!

Annoying integral ~ Help please!

  1. Nov 17, 2012 #1
    I have to find the integral of [tex](4-x)x^{-3}[/tex]. My TI-89 says it should be [tex]\frac{x-2}{x^{2}}+C[/tex] but I can't seem to get it myself.


    I rearranged it to get [tex](4x^{-1}-1)x^{-2}[/tex] and then I used U-Substitution. And set [tex]U = 4x^{-1}-1[/tex] so that [tex]dU = -4x^{-2}dx[/tex]Then I rewrote the integral as [tex]-\frac{1}{4}\int Udu[/tex] and evaluated it to get [tex]-\frac{1}{8}U^{2}+C[/tex]or [tex]-\frac{1}{8}(4x^{-1}-1)^{2}+C[/tex] which works out to [tex]-\frac{(x-4)^{2}}{8x^{2}}+C[/tex]which is not what my calculator says it should be.

    What am I doing wrong?

    ...and if I have to make another Latex equation...
     
    Last edited: Nov 17, 2012
  2. jcsd
  3. Nov 17, 2012 #2
    1) If you hadn't dropped the [itex]\frac{1}{8}[/itex], your answer is correct.

    2) Your calculator also has a correct answer.

    3) They're actually the same answer. How can that be?

    4) You're making this integral WAY more difficult than it actually is. You don't need a substitution. Why did you just distribute [itex]x^{-1}[/itex] through instead of [itex]x^{-3}[/itex]?
     
  4. Nov 17, 2012 #3
    Right... I meant to put the 1/8 in there. But when I graph the two expressions I get two completely different tables of values. Are you sure they are the same?

    As for the unnecessary U-Substitution... that's just me being a moron I guess
     
  5. Nov 17, 2012 #4
    What do you see when you look at the graphs? How are the graphs comparable?

    What is an indefinite integral? Is it a function, or perhaps something ... broader? Whats that +C for?
     
  6. Nov 17, 2012 #5
    The table of values are completely different... on my calculator's answer, x = 1 yields -1, while on mine x = 1 yields -1.125.

    Is it not a function?

    Because an infinite number of different values of C could lead to a function with a derivative of the original function
     
  7. Nov 17, 2012 #6
    Doing the non-moronic way I get the same answer as my calculator. I'll stick with that for now but any comments will still be appreciated
     
  8. Nov 18, 2012 #7

    Mark44

    Staff: Mentor

    In other words, carry out the multiplication to arrive at this integral:
    $$ \int 4x^{-3} - x^{-2}~dx$$

    It should be pretty simple after that.
     
  9. Nov 18, 2012 #8
    Yes, I realized that. Any guess as to why my circuitous method didn't work?
     
  10. Nov 18, 2012 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    [itex]\displaystyle -\frac{(x-4)^{2}}{8x^{2}}-\frac{x-2}{x^2}=-\frac{1}{8}[/itex]

    The answers differ only by a constant, so both have the same derivative.
     
  11. Nov 18, 2012 #10

    Curious3141

    User Avatar
    Homework Helper

    To add to what SammyS said, remember that that so-called "arbitrary constant" of indefinite integration is really that - completely arbitrary. Any constant added to your answer will still give a completely valid answer. So when two expressions differ only be a constant value, they are both equally valid solutions.

    Even though you wrote the "expected" solution and yours with the constant having the same symbol [itex]C[/itex], remember that they're actually different in value (they differ by [itex]\frac{1}{8}[/itex] in this case). It would be more mathematically correct to write the constant in one case as [itex]C_1[/itex] and the other as [itex]C_2[/itex] so you don't get confused.
     
  12. Nov 18, 2012 #11
    Oh, I see it. That's interesting. So when using different methods to find an integral, the value of C (or, whats included in C) would change?
     
  13. Nov 18, 2012 #12

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    C can be different.

    Two correct answers can differ by a constant.
     
  14. Nov 18, 2012 #13

    Mark44

    Staff: Mentor

    For a different example that shows how two different techniques can produce what seem to be different antiderivatives, consider
    $$ \int sin(x)cos(x)dx$$

    If you let u = sin(x), du = cos(x)dx, you get an antiderivative of (1/2)u2 + C = sin2(x) + C1

    OTOH, if you let u = cos(x), du = -sin(x)dx, and you get an antiderivative of -(1/2)u2(x) + C2 = -(1/2)cos2(x) + C2.

    Even though the two answers appear to be significantly different, they differ by a constant (1/2). The key is the identity sin2(x) + cos2(x) = 1.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Annoying integral ~ Help please!
Loading...