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 - Page 287: 40


$\dfrac{128 \sqrt 2}{15} \approx 12.068$

Work Step by Step

The area under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by: $Area, A= \int_{2}^{4} 2x \sqrt {x-2} \ dx \\ = 2 \int_{2}^{4} x \sqrt {x-2} \ dx \\= 2 \int_2^4 (x-2+2) \sqrt {x-2} \ dx$ Let us apply the substitution method such that: $a=x-2 \implies dx=da$ and $x=a+2$ Now, $A= 2 \int_0^2 (a+2) \times a^{1/2} da \\=\dfrac{4 a^{5/2}}{5}+\dfrac{8 a^{3/2}}{3}]_0^2 \\=\dfrac{128 \sqrt 2}{15} \approx 12.068$
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