= lim(n->; ∞) ( 1/n)∑(i: 1->n) √( 1-(i/n)^2)
=∫(0-& gt; √( 1-x^2) dx
=π/4
let
x=sinu
dx=cosu du
x=0,u=0
x= 1,u=π/2
∫(0->; √( 1-x^2) dx
=∫(0->; π/2) (cosu) du
=( 1/2)∫(0->; π/2) ( 1+cos2u) du
=( 1/2)[u+( 1/2)sin2u]|(0-& gt; π/2)