MATH SOLVE

3 months ago

Q:
# 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

Accepted Solution

A:

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.