Chapter 6 - Applications of the Integral - 6.1 Area Between Two Curves - Exercises - Page 286: 7

$A=\frac{343}{3}$

$f(x)=-x^2+x+20$ $and$ $g(x)=x^2-5x$ $When$ $f(x)=g(x),$ $x=-2,5$ $A=\int_{-2}^5(f(x)-g(x))dx$ $A=\int_{-2}^5(-2x^2+6x+20)dx$ $A=[-\frac{2x^3}{3}+3x^2+20x]_{-2}^5$ $A=[-\frac{250}{3}+75+100]-[\frac{16}{3}+12-40]$ $A=[\frac{275}{3}]-[-\frac{68}{3}]$ $A=\frac{343}{3}$