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Math Help - last question i promise- another antiderivative

  1. #1
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    last question i promise- another antiderivative

    Find f. f '(x) = √x(9 + 5x)
    f(1) = 11
    f =

    thats what I got, and obviously its wrong.. what am I doing wrong here?
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  2. #2
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    Hello, rsanford!

    f'(x) \:= \:\sqrt{x}\,(9 + 5x),\quad f(1) \,=\,11

    Find f(x)

    We have: . f'(x) \;=\;9x^{\frac{1}{2}} + 5x^{\frac{3}{2}}

    Integrate: . f(x)\;=\;\int\left(9x^{\frac{1}{2}} + 5x^{\frac{3}{2}}\right)\,dx \;=\;6x^{\frac{3}{2}} + 2x^{\frac{5}{2}} + C \;=\;2x^{\frac{3}{2}}(3 + x) + C

    Since f(1) = 11, we have: . 2\!\cdot\!1^{\frac{3}{2}}(3 + 1) + C \:=\:11 \quad\Rightarrow\quad C \,=\,3


    Therefore: . f(x)\;=\;2x^{\frac{3}{2}}(3 + x) + 3

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