NapkinCalc

Calculus 2 — Integrals & Series

The Riemann sum — area by brute force

Area under f(x) = x² from 0 to 1: slice it into 100 rectangles of width 0.01, add them up. The exact answer (you'll see why below) is 1/3:

R100:=Sum("(i/100)2∗0.01","i",1,100)R_{100} := Sum("(i/100)^2 * 0.01", "i", 1, 100) = 0.33830.3383 100 rectangles: 0.33835 — a touch high
R1000:=Sum("(i/1000)2∗0.001","i",1,1000)R_{1000} := Sum("(i/1000)^2 * 0.001", "i", 1, 1000) = 0.33380.3338 1000 rectangles: closing in
✓ pass abs(R1000−13)<0.001\mathrm{abs}\left(R_{1000} - \frac{1}{3}\right) < 0.001 converging on exactly 1/3
x^2
00.20.40.60.8100.20.40.60.81

the region being measured — area under x² from 0 to 1