Chapter 6 - Applications of the Integral - 6.1 Area Between Two Curves - Exercises - Page 286: 17

$A=\frac{160}{3}$

$A=\int_{-2}^2((x^3-2x^2+10)-(3x^2+4x-10))dx$ $A=\int_{-2}^2(x^3-5x^2-4x+20)dx$ $A=[\frac{1}{4}x^4-\frac{5}{3}x^3-2x^2+20x]_{-2}^2$ $A=(4-\frac{40}{3}-8+40)-(4+\frac{40}{3}-8-40)$ $A=\frac{108}{3}-(-\frac{92}{3})$ $A=\frac{160}{3}$