Riemann Sums
How much area sits under the curve y = f(x) between
x = a and x = b? When the boundary is
curved there is no area formula from geometry class. The fix is wonderfully blunt:
tile the region with thin rectangles, add up their areas, and let the
rectangles get thinner.
Chop [a, b] into n equal strips, each of
width
\Delta x = \frac{b - a}{n}.
On each strip stand a rectangle whose height is f sampled at some
point x_i^{*} inside that strip. The total tiled area is a
Riemann sum:
S_n = \sum_{i=1}^{n} f(x_i^{*})\,\Delta x.
Where you sample inside each strip gives the sum its name: the
left, right, or midpoint rule.
Left, right, and midpoint, side by side
With strip i running from x_{i-1} to
x_i where x_i = a + i\,\Delta x:
-
Left sum — sample the left edge,
x_i^{*} = x_{i-1}:
L_n = \sum_{i=1}^{n} f(x_{i-1})\,\Delta x = \big(f(x_0) + f(x_1) + \dots + f(x_{n-1})\big)\,\Delta x.
-
Right sum — sample the right edge,
x_i^{*} = x_{i}:
R_n = \sum_{i=1}^{n} f(x_{i})\,\Delta x = \big(f(x_1) + f(x_2) + \dots + f(x_{n})\big)\,\Delta x.
-
Midpoint sum — sample the centre,
x_i^{*} = \tfrac{x_{i-1}+x_i}{2}, usually the most accurate of the
three for a given n.
A sum we can do by hand — line by line
Take f(x) = x on [0, 1]. Geometry already
knows the answer — the region is a right triangle with legs 1 and
1, so the area is \tfrac12. Let us watch
the right Riemann sum march to exactly that as n \to \infty.
Step 1 — set up the grid. With a = 0,
b = 1, the width is \Delta x = \tfrac{1}{n}
and the right edge of strip i sits at
x_i = \tfrac{i}{n}.
Step 2 — write the right sum. Since f(x) = x, the
height at x_i is just x_i = \tfrac{i}{n}:
R_n = \sum_{i=1}^{n} f\!\left(\frac{i}{n}\right)\frac{1}{n} = \sum_{i=1}^{n} \frac{i}{n}\cdot\frac{1}{n}.
Step 3 — pull the constants out of the sum. The factor
\tfrac{1}{n^2} does not depend on i:
R_n = \frac{1}{n^2}\sum_{i=1}^{n} i.
Step 4 — use Gauss's sum
\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2}:
R_n = \frac{1}{n^2}\cdot\frac{n(n+1)}{2} = \frac{n(n+1)}{2n^2}.
Step 5 — simplify. Cancel one factor of n:
R_n = \frac{n+1}{2n} = \frac{1}{2} + \frac{1}{2n}.
Step 6 — take the limit. As n \to \infty the
crumb \tfrac{1}{2n} \to 0, so
\lim_{n\to\infty} R_n = \frac{1}{2}.
The Riemann sum converges to \tfrac12 — exactly the triangle's area.
The +\tfrac{1}{2n} is the overshoot of the right rule (each rectangle
pokes above the line); it shrinks away as the strips thin.
For a continuous
f on a closed interval, every choice of sample points
gives sums converging to the same limit as the mesh
\Delta x \to 0. Continuity controls how much f
can wiggle within a strip: over a tiny width the spread between the largest and smallest
values is uniformly small, so left, right, and midpoint all get squeezed together. That
common limit is what we will christen the
definite integral.
Let f be continuous on [a, b], and
partition the interval into n equal strips of width
\Delta x = \tfrac{b-a}{n}. Then, for any choice of sample points
x_i^{*} in the strips, the Riemann sums converge:
\lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^{*})\,\Delta x \quad\text{exists and is independent of the sample points.}
That common limit is the (signed) area under f over
[a, b]. In particular the left, right, and midpoint sums all share
the same limit.
Watch the rectangles converge
The curve is f(x) = x^2 on [0, 2], whose
true area is \int_0^2 x^2\,dx = \tfrac{8}{3} \approx 2.667. Slide
n up and the rectangles thin; flip the rule between left, right, and
midpoint sampling. The live readout shows the sum closing in on the true area — fastest for the
midpoint rule.