The integral: adding up infinitely many slivers

A derivative pulls a curve apart to ask how fast it changes. The integral does the opposite — it builds a quantity back up by adding tiny pieces. And the most natural thing those pieces add up to is area. Once you can find the area under a curve, you can find a total distance, a total cost, a total anything that comes from a rate. That single move turns out to be one of the two great pillars of calculus.

Area is a way of totalling things

Suppose you draw a graph of a car's speed against time. Speed is a rate: kilometres per hour. Now ask a plain question — how far did the car travel?

If the speed were a flat line at 60 km/h for 2 hours, you'd answer instantly: \( 60 \times 2 = 120 \) km. But look at what you just did on the graph. The flat line and the time axis enclose a rectangle, 2 wide and 60 tall, and its area is exactly that 120. The distance is the area under the speed curve.

This is the big idea in one sentence: the area under a rate is a total. Area under speed is distance. Area under a flow rate is a volume. Area under power is energy. Geometry has quietly become a tool for adding things up.

The trouble: curves don't have an area formula

Real graphs bend. The region under a curve like \( y = x^2 \) is not a rectangle, not a triangle, not any shape with a formula in the back of a geometry book. So we do what mathematicians always do with a hard shape: we approximate it with easy shapes, then make the approximation better and better.

Here is the plan. Take the stretch of the \( x \)-axis we care about — say from \( x = a \) to \( x = b \) — and chop it into \( n \) equal strips. Each strip has the same width, which we call \( \Delta x \) (read "delta x", meaning a small change in \( x \)):

\[ \Delta x = \frac{b - a}{n} \]

Because the \( n \) strips share the whole interval equally, each one is exactly \( \tfrac{1}{n} \) of it. That fraction is worth feeling directly — slide it below to see what one strip out of four looks like, and imagine the staircase that four such strips would make under a curve.

So with four strips, each takes up one quarter of the interval; with ten strips, each is a tenth; with a hundred, a hundredth. The more strips, the thinner each one — and that thinness is the whole secret, as you'll see.

Standing a rectangle on each strip

A strip is thin, but it still has a curvy top, so its area is still awkward. So we cheat — gently. Over each strip we stand a rectangle. Its width is \( \Delta x \), and for its height we just read the curve at one chosen point in the strip:

• Left rule — use the function value at the strip's left edge.
• Right rule — use the value at the right edge.
• Midpoint rule — use the value at the middle.

Each rectangle's area is just height times width, \( f(x_i)\,\Delta x \). Add up all \( n \) of them and you get an estimate of the total area:

\[ \text{area} \approx \sum_{i=1}^{n} f(x_i)\,\Delta x \]

That sum has a name: a Riemann sum, after Bernhard Riemann. It is only an estimate. If the curve is rising, left rectangles sit under the curve and undershoot, while right rectangles poke above it and overshoot. The midpoint rule splits the difference and usually lands closest.

Letting the slices get infinitely thin

Now the magic move. Keep the interval the same but let \( n \) grow — more strips, thinner strips, \( \Delta x \to 0 \). The staircase of rectangles presses tighter and tighter against the curve. The gaps where rectangles overshoot or undershoot shrink toward nothing. The left estimate creeps up, the right estimate creeps down, and they squeeze toward a single number.

That number — the limit the Riemann sums settle on as the slices become infinitely thin — is the exact area. We call it the definite integral:

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

Read the new symbols slowly, because each one is a leftover from the sum it replaced:

• \( \displaystyle\int \) is an elongated letter S — it stands for "sum." It is just \( \Sigma \) stretched out for the infinite case.
• \( f(x) \) is the height of the curve, exactly the rectangle height from before.
• \( dx \) is the width — the \( \Delta x \) after it has shrunk to an infinitely thin sliver.
• The little \( a \) and \( b \) say where to start and stop.

So \( \int_a^b f(x)\,dx \) literally reads "add up height times tiny width, all the way from \( a \) to \( b \)." The notation is a sentence describing the very process we just built.

Time to watch it happen. Below is the area under \( y = x^2 \) from \( 0 \) to \( 3 \); the true answer is \( 9 \). Switch between left, midpoint, and right rectangles, then push the count from 2 up toward 32. Watch the left sum climb toward 9 from below, the right sum fall toward 9 from above, and the staircase melt into the curve.

Notice three things the widget makes obvious: left rectangles always undershoot here, right rectangles always overshoot, and the midpoint rule is uncannily close even with just a handful of rectangles. All three rules chase the same destination — the integral, \( 9 \).

Doing it by hand, small and slow

The widget does the adding for you, but you should grind through it once yourself so the formula stops being abstract. We'll estimate \( \int_0^3 x^2\,dx \) — which we already know equals \( 9 \) — using just three rectangles.

Now keep the same three strips but read the height in the middle of each instead of at the left edge. Watch how much closer we land.

Two rules, three rectangles each, two very different answers: 5 versus 8.75. The lesson isn't that one rule is "right" — it's that every rule converges to the integral as the rectangles get thinner, and a smarter sample point gets you there faster.

Putting it together

You now have the whole arc of the idea. Area under a rate is a total. A curvy region has no formula, so we approximate it with rectangles — \( n \) equal strips of width \( \Delta x \), each \( \tfrac{1}{n} \) of the interval, each topped by a height read off the curve. Add the rectangles to get a Riemann sum, an estimate. Then let the strips become infinitely thin, and the estimates collapse onto one exact value, the definite integral \( \int_a^b f(x)\,dx \).

What we have not done is find a quick way to compute that exact number — we leaned on knowing the answer was 9. The shortcut exists, and it is breathtaking: it ties integrals straight back to derivatives. That is the Fundamental Theorem, and it is next.

Practice

Try each one yourself, then reveal the full solution.