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: 7

Answer

$$A = \frac{9}{2} $$
1517714535

Work Step by Step

$ y = (x-2)^{2}$, $y = x$ In this case, it will be easier to integrate along the x-axis. Find the intersections, which will be the integration limits. $$(x -2)^{2} = x$$ $$x^{2} - 4x + 4 = x$$ $$x^{2} - 5x + 4 = 0$$ $$(x - 1) (x - 4) = 0 $$ $$x = 1, 4$$ Integrate, subtracting the "minor" function $(y=(x-2)^{2})$ from the "major" function $( y = x)$ ${\displaystyle \int_{1}^{4}} (x - (x^{2} - 4x + 4 )) dx$ = ${\displaystyle \int_{1}^{4}} (x - x^{2} +4x -4) dx$ $${\displaystyle \int_{1}^{4}} (-x^{2} +5x-4)dx$$ $$\left[-\frac{1}{3}x^{3}+\frac{5}{2} x^{2} - 4x \right]^{4}_1$$ $$= -\frac{1}{3} (4)^{3} + \frac{5}{2} (4)^{2} - 4(4) - \bigg( -\frac{1}{3}+\frac{5}{2} - 4 \bigg)$$ $$= -\frac{64}{3}+\frac{80}{2} - 16 + \frac{1}{3} - \frac{5}{2} +4$$ $$= \frac{- 128+240-96+2-15+24}{6} $$ $$\frac{27}{6} = \frac{9}{2}$$
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