Answer
$ \int^3_0 (-2x^2 +6x) dx$
Work Step by Step
On the interval [0,3], $y_2 \geq y_1$
set up the definite integral
$ \int^3_0 (y_2-y_1)dx$
$ \int^3_0 ((-x^2 +2x +3) - ( x^2 -4x +3))dx$
$ \int^3_0 (-2x^2 +6x) dx$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 7 - Applications of Integration - 7.1 Exercises - Page 442: 3
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