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)?


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  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

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