Find the integral between 1 and -2 for (4-x^2-3x^3)
= [4x - 1/3x3 - 3/4x4]1-2 = (4(1) - 1/3(1)3 - 3/4(1)4) - (4(-2) - 1/3(-2)3 - 3/4(-2)4) =(2 11/12) - (-8 + 8/3 - 12) =(2 11/12) - (-17 1/3) = 20 1/4
AG
= [4x - 1/3x3 - 3/4x4]1-2 = (4(1) - 1/3(1)3 - 3/4(1)4) - (4(-2) - 1/3(-2)3 - 3/4(-2)4) =(2 11/12) - (-8 + 8/3 - 12) =(2 11/12) - (-17 1/3) = 20 1/4
