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23 November, 01:33

Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative. Remember to use absolute values where appropriate.) f (x) = 2 5 - 3 x, x > 0

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  1. 23 November, 02:41
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    C + ⅖|x| - (³/₂) x²

    Step-by-step explanation:

    ƒ (x) = ⅖ - 3x; x > 0

    The antiderivative of ƒ (x) is a function that you can differentiate to get ƒ (x). It is the indefinite integral of ƒ (x).

    Thus, if ƒ (x) = ⅖ - 3x, we must find ∫ (⅖ - 3x) dx.

    ∫ (⅖ - 3x) dx = ∫⅖dx - ∫3xdx = ⅖∫dx - 3∫xdx

    (a) Integrate the first term

    ⅖∫dx = ⅖x

    (b) Integrate the second term

    ∫xdx = ½x², so

    3∫xdx = (³/₂) x²

    (c) Add the constant of integration, C

    Remember that dC/dx = 0 when C is a constant.

    We don't know if there's a constant term in the antiderivative (there may or may not be one), so we include the constant term C to make it general.

    (d) Combine the three terms

    ∫ (⅖ - 3x) dx = C + ⅖x - (³/₂) x²

    We must have x > 0, so

    ∫ƒ (x) dx = C + ⅖|x| - (³/₂) x²

    Check:

    (d/dx) (C + ⅖|x| - (³/₂) x²) = 0 + ⅖ - 3x = ⅖ - 3x

    Differentiating the integral returns the original function.
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