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Math Help - Silly intergration question

  1. #1
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    Silly intergration question

    Hi, my intergration is ok, but I dont know why I suddenly cant do this

    \int { (t - 3)^{2}}dt

    I could expand it to form a quadratic, but isn't their a quicker method:

    add one to the bracket power, divide the whole thing by this bracket power then divide by the diffential of the bracket... it's like the chain rule but for integration

    this gives me

    \frac{(t-3)^{3}}{3}

    But that is not correct is it?

    Thanks
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  2. #2
    MHF Contributor alexmahone's Avatar
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    You only forgot to add the integration constant C.

    The answer is \frac{(t-3)^{3}}{3}+C.
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  3. #3
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    I know, but take it that im using limits Is the integration right then .. but
    http://integrals.wolfram.com/index.jsp?expr=(x-3)^2&random=false

    and in the question im doing I get the wrong answer when the limits are subbed in, yet if I expand first, then integrate I get the right answer...

    the question is

    <br /> <br />
k \int _{0}^{2}{ (t - 3)^{2}}dt = 1<br /> <br />

    prove  k = \frac{1}{9}

    if I expand the brackets and integrate I get the right answer, but If I try integrating with out expansion I get -(1/9), so I've missed something out?

    Thanks
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  4. #4
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    HANG ON. Like the title say silly intergration question. I've been such an idiot. When I was writing out the limits I forgot it's

    (k/3)([0]- [-27]) = 1

    which does mean k = 1/9

    thanks though
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  5. #5
    MHF Contributor

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    It's better to change limits when you substitute.

    If u= t- 3, then when t= 0, u= 0 -3= -3 and when t= 2, u= 2- 3= -1 so the integral becomes \int_0^2 (t-3)^2 dt= \int_{-3}^{-1} u^2 du
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