Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 434: 8


$$ A=36 $$

Work Step by Step

The Area enclosed between two curves $y = f(x)$ and $y = g(x)$ over $[a,b]$ such that $f(x) \geq g(x)$ is $$\displaystyle\int_{a}^{b} f(x) - g(x) dx$$ From the graph, we can see $2x \geq x^{2} -4x$ on the interval $[0,6]$ Therefore, the area between the two curves is $$\displaystyle\int_{0}^{6} 6x-x^{2}dx$$ $$\left[ 3x^{2} - \frac{x^{3}}{3} \right]^{6}_0$$ $$ = 3\times 6^{2} - \frac{6^{3}}{3} = 36$$
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