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 - 9.1 Arc Length and Surface Area - Exercises - Page 468: 9


$$e^2-\frac{1}{4}+\frac{1}{2}\ln \left(2e\right)$$

Work Step by Step

Given $$ y= \frac{1}{4}x^2-\frac{1}{2}\ln x$$ Then arc length is given by \begin{aligned} s=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x &=\int_{1}^{2 e} \sqrt{1+\left(\frac{1}{2} x-\frac{1}{2 x}\right)^{2}} d x \\ &=\int_{1}^{2 e} \sqrt{1+\frac{1}{4} x^{2}-\frac{1}{2}+\frac{1}{4 x^{2}}} d x \\ &=\int_{1}^{2 e} \sqrt{\frac{1}{2}+\frac{1}{4} x^{2}+\frac{1}{4 x^{2}}} d x \\ &=\frac{1}{2} \int_{1}^{2 e} \frac{x^{2}+1}{x} d x \\ &=\frac{1}{2} \int_{1}^{2 e}\left(x+\frac{1}{x}\right) d x \\ &=\frac{1}{2}\left[\frac{x^{2}}{2}+\ln |x|\right]_{1}^{2 e}\\ &=e^2-\frac{1}{4}+\frac{1}{2}\ln \left(2e\right) \end{aligned}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.