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: 18

Answer

$$A = \frac{1}{6}$$
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Work Step by Step

From the graph, we can see that $\sqrt {x-1} \geq x-1$ on the interval $[1,2]$; therefore, the area between the two curves is $$A = \displaystyle\int_{1}^{2} \sqrt {x-1} - (x-1)dx$$ $$A=\displaystyle\int_{1}^{2}(x-1)^{\frac{1}{2}} - x+1dx$$ $$A=\left[\frac{(x-1)^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^{2}}{2}+x\right]^{2}_1$$ $$A=\Bigg[\frac{2(2-1)^{\frac{3}{2}}}{3}-\frac{2^{2}}{2}+2\Bigg] - \Bigg[\frac{2(1-1)^{\frac{3}{2}}}{3} - \frac{1}{2} +1\Bigg]$$ $$A=\bigg[\frac{2}{3}-2+2\bigg] - \frac{1}{2}$$ $$A = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6}=\frac{1}{6}$$
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