Answer
$1.132123$
Work Step by Step
let
$y$ = $x^{-1}$
$y'$ = $-x^{-2}$
$1+(y')^{2}$ = $1+\frac{1}{x^{4}}$
so, the arc length over $[1,2]$ is
$\int_1^2{\sqrt {1+\frac{1}{x^{4}}}}dx$
let f(x) = ${\sqrt {1+\frac{1}{x^{4}}}}$ with n = $8$
$Δx$ = $\frac{2-1}{8}$
Simpson's rule
$\int_1^2{\sqrt {1+\frac{1}{x^{4}}}}dx$ $\approx$ $\frac{Δx}{3}[f(x_{0})+4{\Sigma}_{i=1}^{4}f(x_{2i-1}+2{\Sigma}_{i=1}^{3}f(x_{2i})+f(x_{8})]$ = $1.132123$
