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Math Help - Define integral

  1. #1
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    Define integral

    The value of is
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xcelxp View Post
    The value of is
    are you sure you have the limits right? usually they put the smaller number at the bottom. should the -2 be 2?

    int{-2:0}[(x + 2)^2]dx

    we proceed by substitution

    let u = x + 2
    => du = dx

    so our integral becomes

    int{u^2}du
    = (1/3)u^3 + C
    = (1/3)(x + 2)^3 + C

    evaluating between -2 and 0 we get
    int{-2:0}[(x + 2)^2]dx = (1/3)(0 + 2)^3 - (1/3)(-2 + 2)^3
    ................................= 8/3
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    I am 100% sure that is the correct question. That doesnt seem to be the right answer.
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    The answer is -8/3.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xcelxp View Post
    I am 100% sure that is the correct question. That doesnt seem to be the right answer.
    i'm sure the integral is correct. switching around the limits to the order you used them just changes the answer to a negative value

    int{0:-2}[(x + 2)^2]dx

    we proceed by substitution

    let u = x + 2
    => du = dx

    so our integral becomes

    int{u^2}du
    = (1/3)u^3 + C
    = (1/3)(x + 2)^3 + C

    evaluating between 0 and -2 we get
    int{0:-2}[(x + 2)^2]dx = (1/3)(-2 + 2)^3 - (1/3)(0 + 2)^3
    ................................= -8/3
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    Forum Admin topsquark's Avatar
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    In general:

    Int[f(x) dx, b, a] = - Int[f(x) dx, a, b]

    There is no logical restriction that a > b or b > a. It just depends on the kind of problem that you are doing. In Physics I have seen many instances where a > b.

    -Dan
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