Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 9 - Further Applications of the Integral and Taylor Polynomials - Chapter Review Exercises - Page 496: 5

Answer

$$\sqrt{a(a+1)}+\ln|\sqrt{a}+\sqrt{a+1}|$$

Work Step by Step

Given $$y= 2\sqrt{x} ,\ \ \ \ \ \ [0,a]$$ Since \begin{align*} y'&= \frac{1}{\sqrt{x}}\\ 1+y'^2&= 1+ \frac{1}{x} \end{align*} Then the arc length is given by \begin{aligned} \text{Arc Length }&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x \\ &=\int_{0}^{a} \sqrt{1+ \frac{1}{x}} d x \\ &= \int_{0}^{a} \sqrt{ \frac{x+1}{x}} d x \\ \end{aligned} Use integration by parts \begin{align*} u&= \sqrt{ \frac{x+1}{x}} \ \ \ \ \ \ \ \ \ dv=dx\\ du &= -\frac{1}{2x^{\frac{3}{2}}\sqrt{x+1}}\ \ \ \ \ \ v= x \end{align*} Then \begin{aligned} \text{Arc Length } &= \int_{0}^{a} \sqrt{ \frac{x+1}{x}} d x \\ &= x \sqrt{ \frac{x+1}{x}} +\int \frac{dx}{2x^{\frac{1}{2}}\sqrt{x+1}} \end{aligned} Let $u=\sqrt{x},\ \ \ du=\frac{1}{2\sqrt{x}}$, then \begin{align*} \int \frac{dx}{2x^{\frac{1}{2}}\sqrt{x+1}}&= \frac{1}{2}\cdot \int \frac{2}{\sqrt{u^2+1}}du\\ &=\sinh^{-1}u\\ &=\ln \left|\sqrt{u^2+1}+u\right|\\ &=\ln \left|\sqrt{x+1}+\sqrt{x}\right| \end{align*} Hence \begin{aligned} \text{Arc Length } &= \int_{0}^{a} \sqrt{ \frac{x+1}{x}} d x \\ &= \sqrt{ x(x+1)} + \ln \left|\sqrt{x+1}+\sqrt{x}\right|\bigg|_{0}^{a}\\ &=\sqrt{a(a+1)}+\ln|\sqrt{a}+\sqrt{a+1}| \end{aligned}
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