How do thin pieces add up to a total?

The opening function $x^3/4-x$ is positive on part of $[-2,3]$ and negative on another part. With ten left-endpoint panels the sum is $0.703125$, while the integral is $25/16=1.5625$. The geometric area between the graph and the axis is larger still, $57/16$: the integral subtracts what lies below the axis. Increase $n$ and watch the table: the sums approach $25/16$.

Intuition: rate times time, piece by piece

If a speed $f(t)$ were constant, distance would be speed × time. When it changes, chop time into short intervals on which it is nearly constant, use one sampled speed per interval, and add the products. A negative speed (moving backward) subtracts distance. The total is a signed accumulation; refining the chopping makes it exact in the limit.

Definition

Partition $[a,b]$ into $n$ panels of width $\Delta x=(b-a)/n$ with points $x_i=a+i\,\Delta x$. A Riemann sum chooses a sample $\xi_i$ in each panel:

The left, right and midpoint sums use $\xi_i=x_i$, $x_{i+1}$ or the centre; the trapezoid sum averages the two end heights. For continuous $f$, all of them converge to the same limit, the definite integral

By the same formula, reversing the limits reverses the sign of $\Delta x$: $\int_b^a f=-\int_a^b f$, and $\int_a^a f=0$.

Mechanics: what the explorer computes

For the polynomial integrands every sum is computed exactly with rational arithmetic, and the integral comes from an exact antiderivative; their difference is the exact error. The positive part, the negative part and the geometric area $\int|f|$ are obtained by splitting at the exact rational roots. For $\sin x$ the values are binary64 and marked $\approx$. Each drawn panel is the panel used in the sum, with the same height and width.

Derivation: signed area and geometric area

Split $[a,b]$ at the points where $f$ changes sign. On a subinterval where $f\ge0$, the integral is the area under the graph; where $f\le0$, it is minus the area above the graph. Hence

For $x^3/4-x$ on $[-2,3]$ the roots are $-2$, $0$ and $2$: the area above is $1+25/16=41/16$, the area below is $1$, so the integral is $25/16$ and the geometric area $57/16$.

Worked example

$f(x)=x^3/4-x$ on $[-2,3]$ with $n=10$ left panels: $\Delta x=1/2$, samples $-2,-1.5,\dots,2.5$. The exact sum is $45/64$. The right sum is $165/64$, the midpoint sum $195/128$ and the trapezoid sum $105/64$. With $n=64$ the midpoint and trapezoid sums are within $0.002$ of $25/16$, while the left and right sums are still off by about $0.15$: the sample point matters for the rate.

Exceptional cases

• Exact cancellation. For $x-1$ on $[0,2]$ the integral is exactly $0$ although the area is $1$.
• Reversed limits. With $b<a$, every $\Delta x$ is negative and the sum changes sign.
• Non-integrable behaviour. A function with a pole inside the interval, such as $1/x$ on $[-1,1]$, has no Riemann integral at all; sample points can land arbitrarily close to the pole and the sums do not settle. The explorer’s integrands are bounded on the whole window.

Executable reference code

The reference computes exact left, right, midpoint and trapezoid sums with Fraction and compares them with the antiderivative. The NumPy version vectorises the sums for increasing $n$ and shows the errors shrinking at the expected rates.

"""Exact Riemann and trapezoid sums with Fractions; signed integral versus geometric area (III.12)."""
from fractions import Fraction as F

f = lambda x: x**3 / 4 - x
antiderivative = lambda x: x**4 / 16 - x**2 / 2


def riemann(f, a, b, n, rule):
    dx = (b - a) / n
    xs = [a + i * dx for i in range(n + 1)]
    if rule == "left":
        return sum(f(x) for x in xs[:-1]) * dx
    if rule == "right":
        return sum(f(x) for x in xs[1:]) * dx
    if rule == "midpoint":
        return sum(f((x0 + x1) / 2) for x0, x1 in zip(xs, xs[1:])) * dx
    return sum((f(x0) + f(x1)) / 2 for x0, x1 in zip(xs, xs[1:])) * dx      # trapezoid


a, b = F(-2), F(3)
exact = antiderivative(b) - antiderivative(a)
assert exact == F(25, 16)
assert [riemann(f, a, b, 10, r) for r in ("left", "right", "midpoint", "trapezoid")] == [F(45, 64), F(165, 64), F(195, 128), F(105, 64)]

# Geometric area: split at the roots -2, 0, 2 and add absolute values.
pieces = [(F(-2), F(0)), (F(0), F(2)), (F(2), F(3))]
area = sum(abs(antiderivative(q) - antiderivative(p)) for p, q in pieces)
assert area == F(57, 16) and area != exact

# Reversing the limits reverses the sign (negative dx).
assert riemann(f, b, a, 10, "left") == -riemann(f, a, b, 10, "right")

"""Vectorised Riemann sums: left/right errors shrink like 1/n, midpoint/trapezoid like 1/n**2 (III.12)."""
import numpy as np

f = lambda x: x**3 / 4 - x
a, b, exact = -2.0, 3.0, 25 / 16
errors = {}
for n in (16, 32, 64, 128):
    x = np.linspace(a, b, n + 1)
    dx = (b - a) / n
    errors[n] = {
        "left": abs(np.sum(f(x[:-1])) * dx - exact),
        "midpoint": abs(np.sum(f((x[:-1] + x[1:]) / 2)) * dx - exact),
        "trapezoid": abs(np.trapezoid(f(x), x) - exact),
    }
assert 1.8 < errors[64]["left"] / errors[128]["left"] < 2.2
assert 3.8 < errors[64]["trapezoid"] / errors[128]["trapezoid"] < 4.2
assert 3.8 < errors[64]["midpoint"] / errors[128]["midpoint"] < 4.2

Explore

On Positive: 4 − x², the integral and the area agree. Switch rules on the same $n$: left and right sums bracket the integral when $f$ is monotone on each panel. Drag $a$ past $b$ and read the sign change. On sin x, compare midpoint and trapezoid errors.

Common mistakes

Treating the integral as area when the graph dips below the axis. Forgetting that $\Delta x$ is negative when $b<a$. Using $n$ panels but $n+1$ heights in a left or right sum (the trapezoid uses $n+1$ points, weighted).

Numerical considerations

Floating-point sums of many small terms accumulate rounding error; summing in exact arithmetic avoids it, and in binary64 the order of summation (or compensated summation) matters for very large $n$. Riemann sums converge slowly — the left and right rules have error proportional to $\Delta x$ — which is why III.15 studies better rules.

Engineering and ML connections

Integrals compute totals from rates: energy from power, distance from velocity, probability from a density (Part IV), expected values as weighted sums. Discretised integrals appear in every simulation and in numerical estimates of areas under ROC curves.

Exercises

Summary and checkpoints

A Riemann sum multiplies sampled heights by panel widths and adds; its limit is the definite integral, a signed accumulation. Area above the axis adds, area below subtracts; reversing the limits reverses the sign. Next, III.13 lets the upper limit move and discovers that the rate of accumulation is the integrand itself.