## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 9 - Further Applications of the Integral and Taylor Polynomials - Chapter Review Exercises - Page 496: 3

#### Answer

$$4\sqrt{17}$$

#### Work Step by Step

Given $$y=4 x-2, \quad[-2,2]$$ Since $$y' = 4$$ Then the arc length is given by \begin{aligned} s&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x \\ &=\int_{-2}^{2} \sqrt{1+16} d x \\ &=\int_{-2}^{2} \sqrt{17} d x\\ &=\sqrt{17}x\bigg|_{-2}^{2}\\ &= 4\sqrt{17} \end{aligned}

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