## Thomas' Calculus 13th Edition

Published by Pearson

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

#### Answer

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

#### 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 $L=\int_0^4 \sqrt {1+[(\dfrac{3}{2})x^{(1/2)})]^2}dx=\int_0^4 \sqrt {1+(\dfrac{9}{4})x} dx$ $\implies \int_0^4 (1+(\dfrac{9}{4})x)^{1/2} dx=(\dfrac{2}{3})(\dfrac{4}{9}) [1+(\dfrac{9}{4}) (x)^{(3/2)}]_0^4=\dfrac{8}{27}(10\sqrt {10}-1)$

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