Area by slices
Cut the region under a curve into n strips, stand a rectangle on each, add up the rectangles. Too tall in places, too short in others. Now make n bigger. The errors shrink and the sum heads for a definite number: the exact area, which calculus writes as an integral.
Mission. Keep x² on [0, 1]. Predict what doubling n does to the gap between the left and right sums. Then read the table at n = 4, 8 and 16.
Answer
It halves: 0.25, 0.125, 0.0625. The gap is the first and last rectangles swapping, (f(1) − f(0))Δx = 1/n. Stretch: the midpoint error for this curve is exactly −1/(12n²), so doubling n quarters it. Check the error column with midpoints selected: n = 8 gives −1.302 × 10⁻³ and n = 16 gives −3.255 × 10⁻⁴.
What's going on
Each rectangle has width Δx = (b − a)/n and a height taken from the curve: at the strip's left edge, its right edge, or its middle. For an increasing curve like x², left rectangles all sit under the curve and right rectangles all poke over it, so the true area is trapped between the two sums. As n grows the trap tightens, because the difference between the left and right sums is just the first and last rectangles swapping, which is f(b)Δx − f(a)Δx, and Δx → 0.
For x² on [0, 1] the right sum can be worked out exactly: Δx = 1/n and the heights are (k/n)², so the sum is (1² + 2² + … + n²)/n³ = n(n + 1)(2n + 1)/(6n³). Expand the top and divide: 1/3 + 1/(2n) + 1/(6n²). As n → ∞ this is 1/3, and the 1/(2n) term is nearly all of the error you see shrinking in the table. Archimedes did essentially this calculation for the parabola two thousand years before calculus had a name.
Midpoint rectangles are much better than either endpoint, because each one overshoots on one side of the midpoint and undershoots on the other, and the two errors nearly cancel. Watch the error column: halve Δx and the midpoint error drops by about four, not two.
Sheet: Area under a curve, Archimedes' way. See also Pi by darts, another way to measure an area by counting.