Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

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

Answer

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

Work Step by Step

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