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

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.