The Definite Integral

A Riemann sum converges, for continuous f, to a single number as the strips thin. That number is the definite integral — and we staple the endpoints to the integral sign to name it:

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

Here a is the lower limit and b the upper limit of integration. Unlike the indefinite integral — a family of functions — this is one honest number: the signed area trapped between the curve and the x-axis.

Signed area: below the axis counts negative

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 area.

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].

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).

Signed area, live

The curve is f(x) = \sin x. 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. At b = 2\pi the total is back to 0.