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.

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92