The function g(x)=12x^2-sinx is the first derivative of f(x). If f(0)=-2, what is the value of f(2pi)?

Question

a)30pi+2
b)8pi^2-2
c)4pi^3+8
d)32pi^3-2
e)2pi^2+5

Answer 1

so f'(x) = 12x^2 - sinx
f(x) = 4x^3 + cosx + c
f(0) = -2
-2 = 0 + cos0 + c
-2 = 1 + c
c = -3
f(x) = 4x^3 + cosx - 3
f(2π) = 32π^3 + 1 - 3
= 32π^3 - 2