\int _0^4\:\left(3x^2-x^3\right)dx

Question

compute the integration using the normal definit integration method (Actual)

area= \int _0^4\:\left(3x^2-x^3\right)dx

Answer 1

Area = int [0..4] (3x^2 - x^3) dx

Area = (x^3 - x^4/4)[0..4]

Area = {(4^3 - 4^4/4) - (0 - 0)}

Area = {4^3 - 4^3}

Area = 0