Answer
$5.074212$
Work Step by Step
$y = x^{3/2}$ then $y' = \frac{1}{3} x^{-2/3}$
$L = \int^{6}_{1} f(x) dx$ where $f(x) = \sqrt{1+\frac{1}{9} x^{-4/3}}$
Since $n = 10, \Delta x = \frac{6-1}{10} = \frac{1}{2}$
Now$L = \approx S_{10} = \frac{1/2}{3} [f(1) + 4f(1.5) + 2f(2) + 4f(2.5) + 2f(3) + 4f(3.5) + 2f(4) + 4f(4.5) + 2f(5) + 4f(5.5) + f(6)] \approx 5.074212$
