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

Answer:C + ⅖|x| - (³/₂)x² Step-by-step explanation:ƒ(x) = ⅖ - 3x ; x > 0The 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.