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Thread: Help with antiderivatives?

  1. #1
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    Help with antiderivatives?

    If f'(x) = (-4x^2-2x)/(x^4) and f(1)=3, find f(x). what i did:
    antiderivative[(-4 x x^-2)-(2 x x^-3)]dx
    = (-4 x x^-1)-(2 x -1/2x^-2)]
    = (4/x)+(1/x^2)
    = (4/(1))+(1/(1)^2)
    =5

    i get 5 but shouldn't the answer be 3? what did i do wrong!
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  2. #2
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    Quote Originally Posted by yess View Post
    If f'(x) = (-4x^2-2x)/(x^4) and f(1)=3, find f(x). what i did:
    antiderivative[(-4 x x^-2)-(2 x x^-3)]dx
    = (-4 x x^-1)-(2 x -1/2x^-2)]
    = (4/x)+(1/x^2)
    = (4/(1))+(1/(1)^2)
    =5

    i get 5 but shouldn't the answer be 3? what did i do wrong!
    $\displaystyle f'(x) = \frac{-4x^2-2x}{x^4}$

    you forgot the constant of integration ...

    $\displaystyle f(x) = \frac{4}{x} + \frac{1}{x^2} + C $

    $\displaystyle 3 = 4 + 1 + C$

    $\displaystyle C= -2$
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  3. #3
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    Quote Originally Posted by skeeter View Post
    $\displaystyle f'(x) = \frac{-4x^2-2x}{x^4}$

    you forgot the constant of integration ...

    $\displaystyle f(x) = \frac{4}{x} + \frac{1}{x^2} + C $

    $\displaystyle 3 = 4 + 1 + C$

    $\displaystyle C= -2$
    ahh ok thank you!!
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  4. #4
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    I get

    $\displaystyle \displaystyle\int \frac{-4}{x^2}-\frac{2}{x^3}~dx = 4x^{-1}+2x^{-2}+C$

    Then

    $\displaystyle \displaystyle 3 = 4(1)^{-1}+2(1)^{-2}+C$

    Can you finish it?
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  5. #5
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    You should simplify, then use Partial Fractions...

    $\displaystyle \displaystyle \frac{-4x^2-2x}{x^4} = \frac{-4x-2}{x^3}$.


    You will need to use the decomposition of the form

    $\displaystyle \displaystyle \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3}$.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    You should simplify, then use Partial Fractions...

    $\displaystyle \displaystyle \frac{-4x^2-2x}{x^4} = \frac{-4x-2}{x^3}$.


    You will need to use the decomposition of the form

    $\displaystyle \displaystyle \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3}$.
    partial fractions?

    $\displaystyle \displaystyle \frac{-4x^2-2x}{x^4} = -\frac{4x^2}{x^4} - \frac{2x}{x^4} = -\frac{4}{x^2} - \frac{2}{x^3}$.
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  7. #7
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    Quote Originally Posted by skeeter View Post
    partial fractions?

    $\displaystyle \displaystyle \frac{-4x^2-2x}{x^4} = -\frac{4x^2}{x^4} - \frac{2x}{x^4} = -\frac{4}{x^2} - \frac{2}{x^3}$.
    That way works too
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