Run differentiation backwards, add up infinitely many thin strips, and watch the two halves of calculus snap together.
In the last stage you learned to differentiate: given a position, you found a speed; given a curve, you found its slope. This lesson asks the reverse question — given the speed, can you recover the distance? Given a curve, can you find the area beneath it? The answer is integration, and it is built by slicing a region into countless thin strips and adding them up. The astonishing part — the Fundamental Theorem of Calculus — is that this area is just a difference of an antiderivative, so the two operations you are about to meet, differentiation and integration, are inverses of each other. Here is the keystone of the whole calculus arch.
Suppose a car drives at a steady 60 miles per hour for 2 hours. How far did it go? You multiply: 60 × 2 = 120 miles. On a graph of speed against time, that product is exactly the area of the rectangle under the speed line — height (the speed) times width (the time).
But what if the speed keeps changing? In Stage 36 you ran this the other way: from a position function you took a derivative to get the speed. Integration runs that arrow backwards. If you know the speed at every instant and you want the total distance, you have to add up "speed × tiny bit of time" over and over — and that sum, in the limit, is the integral.
Differentiation turns distance into speed (and a curve into its slope). Integration turns speed back into distance (and a curve into the area beneath it). They are two directions on the same road.
To find the area under a curve y = f(x) from x = a to x = b, we do the only thing we know how to do exactly — find rectangle areas — and we do it many times. Cut [a, b] into n thin strips, each of width Δx = (b − a)/n. On each strip pick a sample point xᵢ, build a rectangle of height f(xᵢ), and add the areas:
Riemann sum = f(x₁)·Δx + f(x₂)·Δx + ⋯ + f(xₙ)·Δx = Σ f(xᵢ)·Δx
The thinner the strips — the larger the n — the better the staircase of rectangles hugs the curve, and the closer the sum comes to the true area. That limiting value, written with a long stretched-S, is the definite integral:
∫ab f(x) dx = limn→∞ Σ f(xᵢ)·Δx
The ∫ is a stretched "S" for sum; the dx is the ghost of the width Δx after it has shrunk to nothing.
Take f(x) = x² on [0, 3]. With just n = 5 midpoint rectangles the sum is 8.91; with n = 20 it is 8.994; with n = 60 it is 8.999. It is marching toward the exact value 9.
The limit of those strip areas is the definite integral — but with one important twist. A rectangle whose top dips below the x-axis has a negative height f(xᵢ), so it contributes a negative area. The integral therefore measures net (signed) area: area above the axis counts as positive, area below counts as negative, and they can cancel.
The integral is not the same as the total amount of ink under the curve. A piece below the axis subtracts. If a function is below the axis as much as it is above, the net integral can even be 0 while the actual geometric area is large.
Before we can compute anything quickly, four properties — each obvious once you picture the strips — make integrals easy to handle.
| Property | In symbols | Why |
|---|---|---|
| Zero width | ∫aa f = 0 | no strip has any width |
| Flip the limits | ∫ab f = −∫ba f | walking backwards reverses sign |
| Split the interval | ∫ab f = ∫ac f + ∫cb f | two slabs make the whole |
| Constant pulls out | ∫ab k·f = k·∫ab f | every strip scales by k |
For f(x) = x² − 2x, the split rule says ∫₀³ = ∫₀² + ∫₂³. The first piece is −1.33 (below the axis), the second is +1.33 (above), and they sum to 0 — exactly the net you saw in the slider above.
Here is the move that makes everything practical. An antiderivative of f is any function F whose derivative is f: that is, F′ = f. Differentiation read backwards. Since the derivative of x² is 2x, we know x² is an antiderivative of 2x; running the power rule in reverse, an antiderivative of x² is x³/3.
But the derivative forgets constants: (x³/3 + 7)′ and (x³/3 − 4)′ are both x². So an antiderivative is never unique — you can always add a constant. We write the whole family as the indefinite integral, and we must not forget the + C:
∫ x² dx = x³/3 + C
The indefinite integral asks for the whole family of antiderivatives, so it always carries a + C; dropping it is the single most common slip. And do not confuse the function f with its antiderivative F — the next theorem lives or dies on getting the right one (F′ = f).
Now the two strands tie together. We built the definite integral as a limit of strip areas — slow, infinite, geometric. We built the antiderivative as differentiation in reverse — fast, algebraic. The Fundamental Theorem of Calculus says these are the same thing: to find the area, just take any antiderivative and subtract its values at the endpoints.
∫ab f(x) dx = F(b) − F(a) (where F′ = f)
No strips, no limits — just a difference of two numbers. The + C cancels in the subtraction, which is why the definite integral has a single, definite value. Differentiation and integration genuinely undo one another.
∫₀³ x² dx = [ x³/3 ]₀³ = 3³/3 − 0³/3 = 9 − 0 = 9. The same 9 the amber strips were crawling toward — now found in one step.
Two everyday questions are now one calculation. Area under a curve is the integral, as we have seen. And distance from a varying speed is the integral of the speed: if v(t) is the speed (in metres per second) and you travel from t = 0 to t = 4 seconds, the distance is the area under the speed curve — and the units agree, metres-per-second × seconds = metres.
A car starts from rest with speed v(t) = t m/s. Distance over the first 4 seconds is ∫₀⁴ t dt = [ t²/2 ]₀⁴ = 16/2 − 0 = 8 metres. If instead v(t) = 2t + 1, the distance over the first 3 seconds is ∫₀³ (2t + 1) dt = [ t² + t ]₀³ = 9 + 3 = 12 metres.
| Idea | What it says |
|---|---|
| Riemann sum | Σ f(xᵢ)·Δx — add thin strips; thinner strips → better |
| Definite integral | ∫ab f dx = the limit of those sums = net (signed) area |
| Signed area | below the axis counts negative — net ≠ total |
| Antiderivative | F with F′ = f; the indefinite integral ∫ f dx = F + C |
| Fundamental Theorem | ∫ab f = F(b) − F(a) — area is a difference of an antiderivative |
Six questions to lock it in. Tap the answer you think is right.
This lesson is a first, honest look at the core of AP Calculus AB: Riemann sums, the definite integral as a limit and as signed area, antiderivatives and the indefinite integral, and the Fundamental Theorem of Calculus (it also touches HSF-IF, interpreting functions). The widgets are computed, not cosmetic — every strip sum equals the program's Riemann sum and every quoted integral comes from an exact antiderivative — so a learner can verify that the staircase really does converge and that the FTC really does reproduce the area. From here the road forks toward techniques of integration, differential equations, and ultimately the rigorous limits of real analysis — a door opening, not a curtain falling.