Visual walkthrough — Riemann sums — left, right, midpoint; formal definition of definite integral

By the end you will have watched ${\int }_0^1{x}^{2}dx$ become $\frac{1}{3}$.

Step 1 — One curve, one region, no formula for it

WHAT. We have a curve. Call the height at horizontal position $x$ by the name $f(x)$ — read it "$f$ of $x$", a machine that eats a number $x$ and gives back the height there. We want the area trapped between this curve and the flat ground (the horizontal axis), from a left wall at $x=a$ to a right wall at $x=b$.

WHY. A rectangle's area is width $\times$ height — dead easy. A curvy top has no such formula. That single obstacle is the reason integration exists. Everything below is one long trick to dodge it.

PICTURE. The shaded region below is what we want the area of. Notice its top edge is curved — that is the whole problem.

Step 2 — Chop the region into vertical strips

WHAT. Slice the region with vertical cuts. We place cut-points along the ground and name them left to right: $a=x_0<x_1<x_2<\cdots <x_n=b.$ Here $x_0$ is just another name for the left wall $a$, $x_n$ another name for the right wall $b$, and the ones in between are the interior cuts. This list of cut-points is called a partition. Each slice sits between two neighbours: strip number $i$ lives on $[x_{i-1},x_i]$.

WHY. A thin strip is almost a rectangle — its top barely curves. So if we can approximate each thin strip by a rectangle, we trade one impossible area for many easy ones.

PICTURE. Below, $n=4$ strips. Read the term-by-term meaning right on the figure: the width of strip $i$ is $\Delta x_i=x_i-x_{i-1}$ (the "$\Delta$" symbol just means "the change in", so $\Delta x_i$ = how far $x$ moves across strip $i$).

Step 3 — Turn each strip into a rectangle (pick a height)

WHAT. A strip has a curved top, so it has no single "height". We choose one: pick a sample point $x_i^{\ast }$ somewhere inside strip $i$, measure the curve there, and use $f(x_i^{\ast })$ as a flat height for the whole strip.

WHY. The moment we commit to one flat height, the strip becomes a genuine rectangle: area $=f(x_i^{\ast })\cdot \Delta x_i$. This is the first area we can actually compute.

PICTURE. Watch the red dot: we read the curve's height at the sample point, then extend that height flat across the strip, creating a rectangle. The top-right corner pokes above the curve on one side and leaves a gap below on the other — that mismatch is our error, and it shrinks as strips get thinner.

Step 4 — Three honest ways to choose the height

WHAT. Where inside the strip should the sample point $x_i^{\ast }$ sit? Three natural answers, each giving a differently-named sum:

WHY. There is no single "correct" spot — that freedom is the point. We want to prove the final answer does not depend on this arbitrary choice. So we must first see the choices differ, then watch them agree in the limit.

PICTURE. The same rising curve, sampled three ways. For a curve that increases, the left rule reads a too-low height (underestimate, red gap above), the right rule reads a too-high height (overestimate, red overshoot), and the midpoint sits between — the overshoot on one side of the centre roughly cancels the gap on the other.

Step 5 — Add up all the rectangles (the Riemann sum)

WHAT. We now have $n$ rectangles. Add their areas. That total is the Riemann sum: $S_n={\sum }_{i=1}^{n}f(x_i^{\ast })\Delta x_i.$

WHY. The whole region's area is (approximately) the sum of the strip areas — that is just "the area of the pieces adds to the area of the whole." The sum symbol $\sum$ is bookkeeping for "add these up as $i$ runs $1,2,\dots ,n$."

PICTURE. Decode the symbols right where they live: $\underset{\text{running total}}{\underbrace{S_n}}=\underset{\begin{array}{c}\text{add over}\\ \text{all strips}\end{array}}{\underbrace{\sum _{i=1}^n}}\underset{\text{height of strip }i}{\underbrace{f(x_i^{\ast })}}\underset{\text{width of strip }i}{\underbrace{\Delta x_i}}.$ Below, all four rectangles are drawn and their areas literally stacked into one bar — that bar's height is $S_4$, our first estimate of the true area.

Step 6 — Slice thinner, and thinner, and thinner…

WHAT. Increase $n$ (more strips), which shrinks every width $\Delta x$. Watch the staircase of rectangle-tops hug the curve tighter and tighter.

WHY. Each rectangle's error is that little sliver between its flat top and the curve. Thinner strips $\Rightarrow$ smaller slivers $\Rightarrow$ smaller total error. We are squeezing the approximation toward the truth.

PICTURE. The same region and curve, drawn with $n=2$, then $n=4$, then $n=8$ right sums. The red "error slivers" visibly shrink. The staircase becomes indistinguishable from the curve.

Step 7 — Take the limit: the definite integral is born

WHAT. As $\Vert P\Vert \to 0$, the sums close in on one number $I$. If that same number $I$ is reached no matter how we chose sample points (left, right, midpoint, anything), we call $f$ integrable and name the number the definite integral: ${\int }_a^{b}f(x)dx={\lim }_{\Vert P\Vert \to 0}{\sum }_{i=1}^{n}f(x_i^{\ast })\Delta x_i.$

WHY "every choice"? If left, right, and midpoint all march to the same limit, the answer can't be an artifact of our arbitrary sampling — it is a real property of $f$. (For a well-behaved curve — see Continuity and Integrability — this always works; for pathological ones it can fail.)

PICTURE. Decoding the reborn notation, symbol by symbol, on the figure: $\underset{\text{a long "S" = Sum}}{\underbrace{\int }}\underset{\text{height}}{\underbrace{f(x)}}\underset{\begin{array}{c}\text{infinitely thin}\\ \text{width, }\lim \Delta x\end{array}}{\underbrace{dx}}$ The red staircase and the black curve have merged — the limit is the area.

Step 8 — Degenerate & edge cases you must not skip

Every scenario the reader might hit — shown, not assumed.

Case A · Empty interval, $a=b$. No width to fill: $\Delta x=\frac{b-a}{n}=0$, every rectangle is a line, total area $=0$. So ${\int }_a^{a}f(x)dx=0$. Makes sense — no region.

Case B · Curve below the ground, $f(x)<0$. The "height" is negative, so $f(x_i^{\ast })\Delta x$ is negative — the rectangle counts as negative area. The integral returns Signed area: above-ground positive, below-ground negative, cancelling where they overlap.

Case C · A straight top, $f$ linear. Here the midpoint rule is exact for any $n$ — no limit needed. The little triangle poking above the sample height on one side is congruent to the gap below on the other, so they cancel perfectly. (Left and right are still only approximate.)

Case D · A "wild" function. If different sample choices give different limits, no single $I$ exists and $f$ is not Riemann integrable. The classic villain is the Dirichlet function (1 on rationals, 0 elsewhere): pick rational samples and every sum is $b-a$; pick irrational samples and every sum is $0$. Two limits ⇒ no integral.

PICTURE. Three panels: (A) the collapsed zero-width case, (B) a curve crossing the ground with one red rectangle below counted negative, (C) a straight line with the two cancelling red triangles at a midpoint rectangle.

The one-picture summary

Left → right, the whole derivation in a single strip: curve → chop into strips → flatten each into a rectangle → sum → shrink widths → limit = the exact area = the integral.

Active Recall