Chapter 6 - Applications of the Integral - 6.1 Area Between Two Curves - Exercises - Page 286: 1

$A=102$

$y_1=3x^2+12$ $and$ $y_2=4x+4 $ $A= \int_{-3}^3 (y_1 - y_2)dx $ $A = \int_{-3}^{3} (3x^2+12-4x-4) dx $ $A= \int_{-3}^{3} (3x^2-4x+8)dx $ $A = [x^3-2x^2+8x]_{-3}^3 $ $A = [27-18+24]-[-27-18-24] $ $A = [33]-[-69] $ $A=33+69$ $A=102$