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Thread: Antiderivative

  1. #1
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    Antiderivative

    hi,

    Q1. Write the gerneral form of the antiderivative of
    f(x) = 4x - 2x^2 + x*squareroot(x) + e^x

    b) Hence calculate

    f (4x - 2x^2 + x*squareroot(x) + e^x) dx

    *where f is an antiderivative, a = 0, b = 1.


    thanks,

    moon
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  2. #2
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    Quote Originally Posted by Moon Hoplite View Post
    hi,

    Q1. Write the gerneral form of the antiderivative of
    f(x) = 4x - 2x^2 + x*squareroot(x) + e^x

    b) Hence calculate

    f (4x - 2x^2 + x*squareroot(x) + e^x) dx

    *where f is an antiderivative, a = 0, b = 1.


    thanks,

    moon

    a) f(x) = 4x - 2x^2 + x \sqrt{x} + e^x.

    Remember that x \sqrt{x} = x \times x^{\frac{1}{2}} = x^{\frac{3}{2}}.

    So F(x) = \int{f(x)\,dx} = \int{4x - 2x^2 + x^{\frac{3}{2}} + e^x\,dx}

     = 2x^2 - \frac{2}{3}x^3 + \frac{2}{5}x^{\frac{5}{2}} + e^x + C.


    For b) evaluate F(1) - F(0).
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  3. #3
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    Quote Originally Posted by Prove It View Post
    a) f(x) = 4x - 2x^2 + x \sqrt{x} + e^x.

    Remember that x \sqrt{x} = x \times x^{\frac{1}{2}} = x^{\frac{3}{2}}.

    So F(x) = \int{f(x)\,dx} = \int{4x - 2x^2 + x^{\frac{3}{2}} + e^x\,dx}

     = 2x^2 - \frac{2}{3}x^3 + \frac{2}{5}x^{\frac{5}{2}} + e^x + C.


    For b) evaluate F(1) - F(0).

    so does b) = 3.1516??
    Last edited by Moon Hoplite; Jan 10th 2009 at 04:08 AM.
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  4. #4
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    Quote Originally Posted by Moon Hoplite View Post
    so does b) = e??
    F(1) = 2(1)^2 - \frac{2}{3}(1)^3 + \frac{2}{5}(1)^{\frac{5}{2}} + e^1 + C

     = 2 - \frac{2}{3} + \frac{2}{5} + e + C

     = \frac{26}{15} + e + C.


    F(0) = 2(0)^2 - \frac{2}{3}(0)^3 + \frac{2}{5}(0)^{\frac{5}{2}} + e^0 + C

     = 1 + C.


    So F(1) - F(0) = \frac{26}{15} + e + C - (1 + C)

     = \frac{11}{15} + e.


    So no, b) is not e. It's \frac{11}{15} + e.
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