Answer
$7.634$
Work Step by Step
Integrate the integral to calculate the arc length as follows:
$l= \int_{A}^{B} \sqrt {1+(\dfrac{dy}{dx})^2}$
or, $ =\int_{1}^{8} \sqrt {1+\dfrac{4}{9y^{2/3}}} dy$
or, $=\dfrac{1}{3} \int_{1}^{8} (\sqrt {9y^{2/3} +4}) (y^{-1/3} ) dy $
Suppose $a =9y^{2/3} +4 $ and $ da= \dfrac{6}{y^{1/3}}dy$
$ l= \dfrac{1}{18}\int_{13}^{40} a^{1/2} da =\dfrac{1}{27} [(40)^{3/2}- (13)^{3/2}] \approx 7.634$
