In fact, this calculation should consider the problem of integral, right? W=S 1+S2+…+Sn- 1, when n is larger and larger.

This parabola can be approximately regarded as consisting of countless black and white triangles? That is, W=S 1+S2+…+Sn- 1, when n is larger and larger? W is actually close to half of the parabola area (the greater the n? Black and white triangles can be almost completely inlaid, coincident and equal)

So w = 1/2 ∫ (-x 2+ 1) dx? (0 = & ltx & lt= 1)

W= 1/2？ (- 1/3*x^3+x)i(0, 1)= 1/2(- 1/3+ 1- 1/3*0-0)= 1/2*2/3= 1/3？

Triangle area = (1-1/N2)/2n+(1-2 2/N2)/2n+(1-3 2/N2)/2n. ....

=((n- 1)-n(n- 1)(2n- 1)/6n^2)/2n

=(n- 1)(4n+ 1)/ 12n^2

= 1/3- 1/4n- 1/ 12n^2

When n becomes larger and larger, it is 1/3.