The definite integral

The derivative was a limit of slopes. The integral is a limit of sums, and the parallel is exact: both take a quantity you can compute for a finite step, and ask what it approaches as the step shrinks to nothing.

Slice the region under a curve into $n$ rectangles of width $\Delta x = \tfrac{b-a}{n}$, add their areas, and let $n$ grow:

$$\int_a^b f(x)\,dx = \lim_{n \to \infty}\sum_{i=1}^{n} f(x_i)\,\Delta x$$

That is the definition. Everything else in this unit is a way of avoiding having to use it.

Watch it converge

Three things are worth doing with that before reading on.

Drag $n$ up on the left rule. The error shrinks, and roughly halves each time you double $n$. Slow, steady, predictable.

Switch to right. On this increasing function the left rule undershoots and the right rule overshoots, so the true value is trapped between them. At $n = 4$ they give $1.75$ and $3.75$ — and their average, $2.75$, is exactly what the trapezoid rule gives. That is not a resemblance: averaging the two rules is algebraically identical to averaging each strip’s two edges.

Switch to midpoint and trapezoid at the same $n$. Both are dramatically better. At $n = 4$ the left rule is out by $0.92$, the trapezoid by $0.083$, and the midpoint by $0.042$ — a factor of twenty for no extra work.

The exact value is $\tfrac83$, and the widget’s error readout is measured against exactly that.

Which rules over- and under-estimate

This is a standard exam question, and it is decided by two properties of the function, not by memorising four cases.

For left and right it is about monotonicity: on an increasing function, the left edge of each strip is the lowest point in it, so every rectangle sits below the curve.

For trapezoid and midpoint it is about concavity: a chord across a concave-up curve lies above it, so the trapezoid overshoots. The midpoint rectangle’s error is subtler — the bit it misses on one side is more than made up on the other — and it lands on the opposite side, which is why midpoint and trapezoid errors have opposite signs.

It is not only area

Area is the picture, but it is not the meaning. A definite integral is accumulated change, and the units come from the product $f(x)\,dx$:

This is why the integral sign turns up throughout physics, and why the units question — “what does this integral represent?” — is answered by multiplying the units of the integrand by the units of the variable.

Signed area. Where $f$ is negative the product $f(x)\Delta x$ is negative, so the integral subtracts. A velocity that goes negative contributes backwards displacement, exactly as it should.