(Core 3 level) Integrate the function f(x) = 2 -cos(3x) between the bounds 0, pi/3.

f(x) = 2 - cos(3x)integrate function x term2 -> 2x (raise power of x then divide by new power for polynomial functions of x) -cos(3x) -> -(1/3)(sin(3x)) (using a substitution of 3x = u, then cos(u) integrates to sin(u)) Answer:2x - (1/3)(sin(3x)) + csubstituting bounds,(2*(pi/2) - (1/3(sin(3pi/3))) - (20 - (1/3)(sin(30/3)) = 2pi/3 - 0 = 2pi/3.