## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{13 \delta}{6}$
The formula to calculate the arc length is as follows: $L=\int_m^n \sqrt {1+[y']^2} dx$ Now, $m_x=\delta \int_0^2 (\sqrt x) \sqrt {1+\dfrac{1}{4x}} dx$ or, $=\delta \int_0^2 \sqrt {x+\dfrac{1}{4}} dx$ Suppose $a= x+\dfrac{1}{4}$ So, $\delta \int_{1/4}^{9/4} a^{1/2} da= \dfrac{2 \delta} {3}(\dfrac{27}{8}-\dfrac{1}{8}$ or, $=\dfrac{13 \delta}{6}$