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Homework Help: 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


    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


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    [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


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    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


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    C can be different.

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


    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.
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