The Definite Integral
The definite integral \(\int_a^b f(x)\,dx\) is the number the Riemann sums from the last lesson were closing in on: the limit as the number of rectangles goes to infinity. Geometrically it measures net signed area — area above the x-axis counts positive, area below counts negative.
By the end you'll be able to state the definite integral as a limit of Riemann sums, and evaluate simple integrals geometrically by adding above-axis area and subtracting below-axis area.
Predict: what is \(\int_{-2}^{2} x\,dx\)? Drag a to −2 and b to 2 (the green handles on the axis, or the sliders) and watch the two shaded triangles cancel.
Blue shading above the axis counts positive; amber shading below counts negative — the readout's colors match the regions. Tick “show rectangles” to see the Riemann-sum skeleton underneath: crank n up and the sum closes in on the exact net area.
Above axis: +2.00 · Below axis: -0.50 · Net signed area = 1.50
The definite integral turns “add up infinitely many infinitely thin rectangles” into a single, exact number — the net signed area between a curve and the x-axis.
The definite integral of \(f\) from \(a\) to \(b\) is defined as the limit of Riemann sums as the number of rectangles \(n\) goes to infinity (so each width \(\Delta x\) shrinks to zero): $$\int_a^b f(x)\,dx \;=\; \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x, \qquad \Delta x = \frac{b-a}{n}.$$ Here \(a\) and \(b\) are the limits of integration, \(f\) is the integrand, and the result is a single number — not a function. When this limit exists, \(f\) is integrable on \([a, b]\), and every function continuous on \([a, b]\) is integrable there.
Picture the region trapped between the curve and the x-axis over \([a, b]\). The integral is net signed area: area above the axis counts positive, area below the axis counts negative — exactly the blue and amber regions in the explorer above. So \(\int_a^b f(x)\,dx = \) (area above) − (area below), which is not the total geometric area. A curve symmetric about the axis can integrate to exactly zero even though it clearly encloses real region — the two halves cancel. A negative answer isn't an error; it just means more area sat below the axis.
Evaluate \(\int_{-2}^{2} x\,dx\) geometrically. On \([-2, 0]\) the line \(y = x\) lies below the axis, forming a triangle of area \(\tfrac{1}{2}\cdot 2\cdot 2 = 2\) that counts as \(-2\). On \([0, 2]\) it lies above the axis — an equal triangle counting as \(+2\). Net signed area: \(-2 + 2 = \) 0. The positive and negative pieces cancel exactly, even though the graph encloses 4 units of real geometric area.
Evaluate \(\int_{-1}^{3} x\,dx\) geometrically. Below the axis on \([-1, 0]\): a triangle of area \(\tfrac{1}{2}\cdot 1\cdot 1 = 0.5\), counting as \(-0.5\). Above the axis on \([0, 3]\): a triangle of area \(\tfrac{1}{2}\cdot 3\cdot 3 = 4.5\), counting as \(+4.5\). Now finish it: net signed area \(= -0.5 + 4.5 = \) ____
Reveal the answer
\(-0.5 + 4.5 = \) 4. The big above-axis triangle dominates the small below-axis one. Check it live: pick f(x) = x in the explorer, drag a to −1 and b to 3 — the readout shows +4.50 above, −0.50 below, net 4.00.
More info — why bother with the limit if it's “just area”?
For lines and circles you can read areas off with triangle and circle formulas — that's what the worked examples exploit. But for a region under \(y = x^2 - 1\) there is no elementary geometry formula, and that's where the definition earns its keep: any finite Riemann sum from the previous lesson is an approximation with visible gaps and overshoots, while the limit \(n \to \infty\) squeezes those errors to zero and defines the exact area. Turn on the rectangles in the explorer with f(x) = x² − 1 and step \(n\) from 4 to 64 — the Riemann-sum readout converges to the net-area readout. That the limit lands on one definite value for every continuous function is the payoff; the OpenStax section in Dive deeper works through this convergence carefully.
Check your understanding
Evaluate \(\int_{-3}^{1} x\,dx\) geometrically.
You compute a definite integral and get a negative number. What does that tell you?
On \([a, b]\), the graph of \(f\) encloses 3 units of area above the x-axis and 5 units below it. What are \(\int_a^b f(x)\,dx\) and the total geometric area, in that order?
For a continuous \(f\) on \([a,b]\), let \(R_n\) be a Riemann sum built from \(n\) rectangles of width \(\Delta x = (b-a)/n\). How is \(\int_a^b f(x)\,dx\) related to the \(R_n\)?
Recap
- Definition: \(\int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x\) with \(\Delta x = (b-a)/n\) — the limit of Riemann sums as the number of rectangles goes to infinity.
- The result is a single number: the net signed area — area above the x-axis counts positive, area below counts negative.
- Net signed area ≠ total geometric area: for total area, integrate \(|f(x)|\) or split the interval at the roots of \(f\).
- A negative integral is not an error — it means more area sat below the axis. Every function continuous on \([a,b]\) is integrable there.
Dive deeper
- OpenStax Calculus Volume 1 — 5.2 The Definite Integral Define the definite integral as a limit of Riemann sums and explain net signed area
Sources
- The Definite Integral as a Limit of Riemann Sums