Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.4 Length Of A Plane Curve - Exercises Set 5.4 - Page 374: 2

Answer

$\text{See explanation.}$

Work Step by Step

$\text{According to the Theorem of Pythagoras, the length of the segment is}$ \begin{align} L = \sqrt {1 + 5^2} =\sqrt {26} \end{align} $\text{(a) Formula (4)}$ \begin{align} L = \int_0^1 \sqrt {1+5^2} \ dx = \sqrt {26} \end{align} $\text{(b) Formula (5)}$ \begin{align} L = \int_0^5 \sqrt {1+\frac{1}{25}} \ dy = \sqrt {26} \end{align} $\text{Thus, we have confirmed that the results are correct.}$
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