MISSION · Calculus
The Integral as Accumulated Area
Adding up infinitely many rectangles, and why that undoes a derivative
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The idea in one sentence
An integral adds up infinitely many infinitely thin slices, which is how you total a quantity whose rate keeps changing.
Start with rectangles
How much area sits under a curve between two points? For a straight-sided shape you’d use a formula, but a curve has no formula waiting for it.
So approximate. Slice the region into vertical rectangles, each as tall as the curve at that point, and add their areas. With four rectangles the answer is rough — they overshoot or undershoot wherever the curve bends. With forty it’s better. With four hundred the gaps are invisible.
That sum is called a Riemann sum, and the integral is what it converges to as the rectangles get infinitely thin:
\[\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\,\Delta x\]The notation records the construction. The \(\int\) is a stretched S for “sum”, \(f(x)\) is a rectangle’s height, and \(dx\) is its vanishing width.
It isn’t only about area
Area is the picture, but the useful reading is accumulation. Whenever you have a rate and want a total, you integrate:
- Speed against time gives distance travelled.
- Flow rate against time gives volume delivered.
- Force against distance gives work done.
If the rate were constant you’d just multiply — 60 mph for 2 hours is 120 miles. When the rate varies, multiplication fails and integration is what replaces it. Each thin rectangle is a moment brief enough that the rate is effectively constant within it; the integral adds those moments back up.
The theorem that makes it possible
Computing that limit by hand for every problem would be unbearable. The Fundamental Theorem of Calculus rescues it by revealing that integration and differentiation are inverse operations:
\[\int_a^b f(x)\,dx = F(b) - F(a) \quad \text{where } F' = f\]Find a function whose derivative is \(f\) — an antiderivative — evaluate it at both ends, subtract. No rectangles required.
That two operations invented for unrelated problems, slopes and areas, turn out to undo one another is the central surprise of calculus. It makes sense in the accumulation reading: the rate at which accumulated area grows, as you widen the region, is exactly the height of the curve at the edge.
Definite and indefinite
A definite integral has limits and produces a number: an area, a distance, a total.
An indefinite integral has no limits and produces a family of functions: \(\int 2x\,dx = x^2 + C\). The \(+ C\) is there because every function differing by a constant has the same derivative, so the antiderivative is only pinned down up to that constant — which cancels anyway when you subtract two ends of a definite integral.
Common mix-ups
- Forgetting \(+ C\) on an indefinite integral.
- Subtracting the wrong way round. Top limit minus bottom limit.
- Assuming the answer is an area. Below the axis, \(f(x)\) is negative and so is its contribution — the integral gives a signed total.
- Confusing the integral with the derivative. One accumulates, the other measures a rate.
Check yourself
Evaluate \(\int_0^3 2x\,dx\), and confirm it with the triangle it describes.
An antiderivative of \(2x\) is \(x^2\), so the value is \(9 - 0 = 9\). The region is a triangle with base 3 and height 6, and \(\frac{1}{2}(3)(6) = 9\).
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