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Math Help - antiderivative

  1. #1
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    antiderivative

    Consider the function . Let be the antiderivative of with .
    Then equals ?

    I am having trouble with this one, this is the answer I came up with.
    x(3sec^2(t)-10t^3)

    Thankx
    Keith
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by keith View Post
    Consider the function . Let be the antiderivative of with .
    Then equals ?

    I am having trouble with this one, this is the answer I came up with.
    x(3sec^2(t)-10t^3)

    Thankx
    Keith
    okay, we're not integrating with respect to x, or they would have asked for F(x). there are no special rules for this one. the derivative of tan(x) is sec^2(x), so just go backwards when integrating:

    f(t) = 3sec^2(t) - 10t^3
    => F(t) = int{3sec^2(t) - 10t^3}dt
    ...........= 3tan(t) - (5/2)t^4 + C

    Now F(0) = 0
    => 0 = 3tan(0) - (5/2)(0)^4 + C
    => C = 0

    so F(t) = 3tan(t) - (5/2)t^4

    so i guess you were integrating with respect to x and treating the function of t as a constant....not correct i'm afraid. if they wanted you to integrate with respect to x, they would let you know explicitly
    Last edited by Jhevon; April 27th 2007 at 12:54 PM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by keith View Post
    Consider the function . Let be the antiderivative of with .
    Then equals ?

    I am having trouble with this one, this is the answer I came up with.
    x(3sec^2(t)-10t^3)

    Thankx
    Keith
    Where did the x come from?

    F(t) = Int[3*sec^2(t) - 10t^3,t] = 3*tan(t) - (10/4)*t^4 + C

    For F(0) = 0 we have:
    0 = 3*tan(0) - (5/2)*0^4 + C = C

    Thus C = 0

    Thus
    F(t) = 3*tan(t) - (5/2)*t^4

    -Dan
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