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Integrating (4x-3)^2

  1. Nov 1, 2006 #1
    The task is to find one primitive function to:

    [tex](4x-3)^2[/tex]

    This was quite straightforward. Or so I though.

    It can easily be turned into

    [tex]16x^2 - 24x + 9[/tex]

    and then integrated to

    [tex] \frac {16x^3}{3} - 12x^2 + 9x[/tex]

    choosing C = 0

    Then I started to think. Couldn't this be integrated using the chain rule backwards (with lack of better wording)?

    [tex](4x-3)^2[/tex]

    then becomes according to my integration

    [tex] \frac {(4x-3)^3}{12}[/tex]

    If I differentiate the above, I get the initial function. What I'm having a hard time understanding is that the different approaches yields different results. I plotted both of them in my graph calculator and they indeed give different results.

    Any hints on what I did wrong is greatly appreciated. Thank you for your time. Have a nice day.
     
  2. jcsd
  3. Nov 1, 2006 #2

    Office_Shredder

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    When you plot both of them, what exactly is the difference? Try expanding the second function, and see if, maybe, choosing c=0 for both cases was perhaps premature
     
  4. Nov 1, 2006 #3
    As it happens, by adding 2.25 to the latter, the both graphs coincide with eachother. How come? Why is the first method easier to get to one primitive function directly?
     
  5. Nov 1, 2006 #4

    Doc Al

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    Are you using the same C in each case? (Set x = 0 and see.)
     
  6. Nov 1, 2006 #5

    Doc Al

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    Or just expand your second answer and see the extra constant.
     
  7. Nov 1, 2006 #6
    there's no mistake. when you integrate a function, u get a constant C. this C can be any value, as any constant when differentiated gives you 0. so u can't merely take C = 0 and plot the graph that way. that's why they coincide when you add 2.25.
     
  8. Nov 1, 2006 #7

    Office_Shredder

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    The first method is easier because integrating polynomials is easy. The second method is much more useful when you have stuff like x*cos(x2), and you can't just expand it and integrate.
     
  9. Nov 1, 2006 #8
    I see.

    By expanding

    [tex] \frac {(4x-3)^3}{12}[/tex]

    I get

    [tex]5 \frac {1}{3}x^3 -4x^2 + 3x + 2.25[/tex]

    And the constant is indeed 2.25.

    So in other words, both are equally correct with respect to the initial problem (find one primitive function to [tex](4x-3)^2[/tex])?

    Both are a primitive function, only with different C values (0 and 2.25 resp.)?
     
  10. Nov 1, 2006 #9

    Office_Shredder

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    No..... you can't just throw the constant value away. Your two answers SHOULD look like

    [tex] \frac {(4x-3)^3}{12} + C[/tex]

    and

    [tex] \frac {16x^3}{3} - 12x^2 + 9x + C[/tex]

    What normally happens is you're given more information, e.g. what f(0) is, to determine what C actually is (it's not the same for the two different equations). But you can't just toss the constant out, since you're ignoring the complete set of solutions
     
  11. Nov 1, 2006 #10
    I see.

    Does that means that a satisfactory answer to the problem

    "Find one primitive function to [tex](4x-3)^2[/tex]"

    is

    [tex] \frac {16x^3}{3} - 12x^2 + 9x[/tex]

    or

    [tex] \frac {(4x-3)^3}{12}[/tex]

    or any of the two with a specified C value (ie. 1,2,3,4....)?

    where as if it had been "Find all primitive functions to [tex](4x-3)^2[/tex]", the answer would have been

    [tex] \frac {(4x-3)^3}{12} + C[/tex]

    and

    [tex] \frac {16x^3}{3} - 12x^2 + 9x + C[/tex]

    ?
     
    Last edited: Nov 1, 2006
  12. Nov 1, 2006 #11
    Yes, that's correct. [tex]\frac{(4x-3)^3}{12}[/tex] is one primitive function, where C=0. Likewise for your polynomial. [tex]\frac{(4x-3)^3}{12}+C[/tex] defines the entire family of primitive functions, where C can take any value.
     
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