Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 6: Applications of Definite Integrals - Section 6.3 - Arc Length - Exercises 6.3 - Page 335: 2


$\dfrac{8}{27}(10\sqrt {10}-1)$

Work Step by Step

We know that the formula to calculate the arc length is defined as: $L=\int_m^n \sqrt {1+[f'(x)]^2} dx$ This implies that $L=\int_0^4 \sqrt {1+[(\dfrac{3}{2})x^{(1/2)})]^2}dx=\int_0^4 \sqrt {1+(\dfrac{9}{4})x} dx$ $\implies \int_0^4 (1+(\dfrac{9}{4})x)^{1/2} dx=(\dfrac{2}{3})(\dfrac{4}{9}) [1+(\dfrac{9}{4}) (x)^{(3/2)}]_0^4=\dfrac{8}{27}(10\sqrt {10}-1)$
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.