The Definite Integral

A pump fills a water tank, and a gauge on the pipe shows the flow rate — litres per minute — wobbling up and down as the pump works. Question: how many litres actually went into the tank between 9:00 and 9:30? The gauge never tells you a total, only a rate. Yet the total is hiding in plain sight: chop the half hour into slivers of time, multiply each sliver's flow rate by its duration to get litres-in-that-sliver, and add them all up. That is a Riemann sum — and as the slivers thin, the sum settles on the exact total.

The same trick answers "how far did the car travel?" from a speedometer reading, "how much charge flowed?" from a current meter, "how much rain fell?" from a rainfall-rate sensor. Whenever you know how fast something is accumulating and want how much accumulated, you are computing the same object. It deserves a name and a symbol:

\int_{a}^{b} f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^{*})\,\Delta x.

This is the definite integral of f from a (the lower limit) to b (the upper limit). Unlike the indefinite integral — a whole family of functions — this is one honest number. Geometrically it is the signed area trapped between the curve and the x-axis; physically it is an accumulated total. Those are the same thing: on a graph of flow rate against time, litres are literally area, because \text{rate} \times \text{time} is \text{height} \times \text{width}.

Signed area: below the axis counts negative

Why "signed" area? Each rectangle in the sum contributes f(x_i^{*})\,\Delta x. The width \Delta x > 0 always, but the height f(x_i^{*}) carries the sign of f. So where the curve dips below the axis, its rectangles have negative height and subtract from the total.

And that is exactly what the accumulation story demands. If the "flow rate" goes negative — the tank is draining, the car is reversing — those stretches should reduce the total, not add to it. The definite integral is therefore (area above the axis) − (area below). For example \int_0^{2\pi} \sin x\,dx = 0: the positive hump on [0, \pi] is cancelled exactly by the negative hump on [\pi, 2\pi]. Pump water in for a while, drain the same amount back out, and the tank ends where it started — even though plenty of water moved through the pipe.

Worked example 1 — integrals you can do with school geometry

Before any clever theorems arrive, some definite integrals surrender to a ruler. If the region under the curve is a rectangle, triangle or trapezium, the integral is its area (with a sign), and you learned those formulas years ago.

A constant. Evaluate \int_1^5 3\,dx. The graph of f(x) = 3 is a horizontal line; between x = 1 and x = 5 it roofs a rectangle of width 4 and height 3:

\int_1^5 3\,dx = 3 \times (5 - 1) = 12.

In general \int_a^b c\,dx = c\,(b - a) — worth memorising, it is the sanity check for everything else.

A slope. Evaluate \int_0^6 x\,dx. The line y = x makes a right triangle with the axis: base 6, height 6, so

\int_0^6 x\,dx = \tfrac{1}{2}\cdot 6 \cdot 6 = 18.

A slanted roof. Evaluate \int_0^4 (x + 2)\,dx. The region is a trapezium — or, sliced horizontally, a rectangle with a triangle perched on top. Step through the figure:

\int_0^4 (x+2)\,dx = \underbrace{4 \times 2}_{\text{rectangle}} + \underbrace{\tfrac{1}{2}\cdot 4 \cdot 4}_{\text{triangle}} = 16.

Notice what just happened: we evaluated three definite integrals exactly, straight from the definition's geometric meaning, without a single antiderivative. The Fundamental Theorem will soon give a faster route — but the integral's meaning lives here, in area, not in that shortcut.

Worked example 2 — when the areas fight each other

Now a curve that changes sign. Evaluate \int_{-3}^{2} x\,dx.

Step 1 — find where the sign flips. The line y = x crosses the axis at x = 0, so split the interval there: [-3, 0] (curve below the axis) and [0, 2] (curve above).

Step 2 — measure each triangle. On [-3, 0] the region is a triangle of base 3 and height 3 — ordinary area \tfrac{9}{2} — but it hangs below the axis, so it contributes -\tfrac{9}{2}. On [0, 2] the triangle has base and height 2: it contributes +2.

Step 3 — let them cancel.

\int_{-3}^{2} x\,dx = -\tfrac{9}{2} + 2 = -\tfrac{5}{2}.

The answer is negative: more of the region lies below the axis than above. And a pleasing special case falls out for free — on a symmetric interval the two triangles are twins, so \int_{-a}^{a} x\,dx = 0 for every a. Perfect cancellation, no computation required.

Two properties, derived from the definition — line by line

Additivity over adjacent intervals

Claim: \displaystyle\int_a^b f + \int_b^c f = \int_a^c f. The picture is obvious — two adjacent areas glue into one — and the sum definition makes it exact.

Step 1 — tile each piece. Approximate \int_a^b by a Riemann sum over a partition of [a, b], and \int_b^c by one over [b, c], both using the common breakpoint b.

\sum_{[a,b]} f(x_i^{*})\,\Delta x \;+\; \sum_{[b,c]} f(x_j^{*})\,\Delta x.

Step 2 — concatenate the partitions. Laid end to end, the two partitions form a single partition of [a, c] (with b as an interior breakpoint), and the two sums become one sum over all the strips:

= \sum_{[a,c]} f(x_k^{*})\,\Delta x.

Step 3 — pass to the limit. Refining both partitions (\Delta x \to 0), each side converges to its integral:

\int_a^b f\,dx + \int_b^c f\,dx = \int_a^c f\,dx.

Linearity

Step 1 — start from one sum. Apply the definition to \int_a^b (\alpha f + \beta g):

\int_a^b (\alpha f + \beta g)\,dx = \lim_{n\to\infty}\sum_{i=1}^n \big(\alpha f(x_i^{*}) + \beta g(x_i^{*})\big)\,\Delta x.

Step 2 — split the finite sum (a finite sum of a sum is the sum of the sums, and constants factor out):

= \lim_{n\to\infty}\left(\alpha\sum_{i=1}^n f(x_i^{*})\,\Delta x + \beta\sum_{i=1}^n g(x_i^{*})\,\Delta x\right).

Step 3 — split the limit. Both limits exist (continuity), so the limit of the sum is the sum of the limits:

= \alpha\int_a^b f\,dx + \beta\int_a^b g\,dx.

Set c = a in additivity: \int_a^b f + \int_b^a f = \int_a^a f. A degenerate interval [a, a] has width zero, so every Riemann sum is zero — giving \int_a^a f = 0. Substituting back,

\int_a^b f\,dx + \int_b^a f\,dx = 0 \quad\Longrightarrow\quad \int_b^a f\,dx = -\int_a^b f\,dx.

So flipping the limits flips the sign — both the "empty interval" rule and the "reverse the direction" rule fall out of additivity for free, no new assumption needed.

If m \le f(x) \le M for all x in [a, b], then every rectangle's height lies between m and M, so every Riemann sum lies between m\sum\Delta x = m(b-a) and M(b-a). Passing to the limit,

m\,(b - a) \;\le\; \int_a^b f(x)\,dx \;\le\; M\,(b - a).

This bounding inequality traps an unknown integral between two simple rectangle areas — and it is the seed of the squeeze argument behind the Fundamental Theorem.

Let f, g be continuous on the relevant intervals and \alpha, \beta constants. Then:

Empty interval: \displaystyle\int_a^a f\,dx = 0.
Reversal: \displaystyle\int_b^a f\,dx = -\int_a^b f\,dx.
Additivity: \displaystyle\int_a^b f + \int_b^c f = \int_a^c f.
Linearity: \displaystyle\int_a^b (\alpha f + \beta g)\,dx = \alpha\!\int_a^b f + \beta\!\int_a^b g.
Monotonicity: if f \le g on [a,b] then \displaystyle\int_a^b f \le \int_a^b g.
Bounding: if m \le f \le M on [a,b] then m(b-a) \le \int_a^b f \le M(b-a).

Worked example 3 — the properties doing real work

The properties are not decoration; they let you compute integrals you were never told. Suppose all you know about some continuous f is

\int_0^8 f(x)\,dx = 10 \qquad\text{and}\qquad \int_0^3 f(x)\,dx = 4.

Split the interval. Additivity with the breakpoint 3 says \int_0^3 f + \int_3^8 f = \int_0^8 f, so

\int_3^8 f(x)\,dx = 10 - 4 = 6.

Reverse the limits. Then immediately \int_8^3 f(x)\,dx = -6 — same region, opposite direction of travel, opposite sign.

Rescale with linearity. And \int_3^8 \big(2f(x) + 1\big)\,dx = 2\cdot 6 + 1\cdot(8-3) = 17, using our rectangle rule \int_3^8 1\,dx = 5 for the constant part. Three numbers we were never given, all conjured from two facts and the properties. Exam questions adore this move — but so do physicists and engineers, who often know a few measured totals and need the pieces in between.

"Area under the curve" is signed. If a question asks for the geometric area enclosed and the curve dips below the axis, the raw integral under-counts — the below-axis parts arrive with minus signs. For genuine area you must split at the axis crossings and add absolute values: \int_{-3}^{2} x\,dx = -\tfrac{5}{2}, but the shaded area in that picture is \tfrac{9}{2} + 2 = \tfrac{13}{2}.
Displacement is not distance. Integrating a velocity that changes sign gives the net change in position (displacement). A car that drives 30 km east then 30 km west has displacement 0 — but its odometer, which integrates |v|, honestly reports 60 km. Same trip, two different integrals.
Backwards limits flip the sign. Nothing stops you writing \int_5^2 f\,dx — but it equals -\int_2^5 f\,dx. If an answer's sign looks wrong, check which limit is on top before checking anything else.

Swallow a tablet and the drug's concentration in your blood rises, peaks, then decays as your liver clears it. Pharmacologists plot concentration against time and compute the AUC — literally "Area Under the Curve", a definite integral of concentration over time. It measures your total exposure to the drug: a tall short spike and a low long plateau can deliver exactly the same AUC, just as two very different flow rates can fill the same tank.

Regulators use AUC to decide whether a cheap generic tablet is equivalent to the branded original (their AUCs must match closely), and oncologists dose some chemotherapy drugs by prescribing a target AUC rather than a target amount. So when this idea feels abstract, remember: somewhere right now, a hospital computer is evaluating a definite integral to work out how much medicine a patient should get.

Signed area, live

Time to watch the definition breathe. The curve is f(x) = \sin x — think of it as a pump's flow rate that smoothly swings from filling to draining. Slide the upper limit b and watch the shaded region grow. Where the curve is above the axis the shading is one colour and the integral climbs; once b passes \pi the curve drops below the axis, the new shading switches colour, and the running value of \int_0^{b}\sin x\,dx = 1 - \cos b falls — negative area cancelling the positive.

Three moments deserve a pause. At b = \pi the total peaks at exactly 2 — the full positive hump. At b = 2\pi the total is back to 0: perfect cancellation, the tank refilled and re-drained. And notice the readout is a function of where you stop — an integral with a moving upper limit is itself a new function of b, an idea that is about to become the hero of the Fundamental Theorem.