What the definite integral means geometrically

Primary Tool

2D Graphing

The graphing tool makes the connection between the curve and the area under it immediate. Plotting $f (x) = x^{2}$ over $[0, 3]$ lets you see exactly what region the integral $\int_{0}^{3} x^{2} d x = 9$ measures.

The geometric picture

The definite integral $\int_{a}^{b} f (x) d x$ is the signed area of the region between the graph of $f$ and the $x$ -axis, from $x = a$ to $x = b$ .

Signed means areas above the axis count as positive and areas below the axis count as negative. If the curve dips below the axis in part of $[a, b]$ , those strips subtract from the total.

For example, $\int_{0}^{3} x^{2} d x = 9$ means the region between the parabola $y = x^{2}$ and the $x$ -axis from $0$ to $3$ has area exactly $9$ square units.

Where the integral comes from: Riemann sums

To make area precise, divide $[a, b]$ into $n$ subintervals of width $Δ x = (b - a) / n$ . On each subinterval pick a sample point $x_{i}^{*}$ and build a rectangle of height $f (x_{i}^{*})$ and width $Δ x$ .

The sum of rectangle areas $\sum_{i = 1}^{n} f (x_{i}^{*}) Δ x$ is a Riemann sum. As $n \to \infty$ and the rectangles get thinner, the Riemann sum converges to the integral — provided $f$ is integrable on $[a, b]$ .

1Choose a partition: split $[0, 3]$ into $n$ equal pieces of width $Δ x = 3/ n$ .
2Pick sample points: use the right endpoint $x_{i} = i \cdot Δ x$ for $i = 1, \dots, n$ .
3Compute the sum: $\sum_{i = 1}^{n} (x_{i})^{2} \cdot Δ x = \sum_{i = 1}^{n} (\frac{3 i}{n})^{2} \cdot \frac{3}{n}$ .
4Simplify using the formula $\sum_{i = 1}^{n} i^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$ .
5Take the limit as $n \to \infty$ . The leading term gives $\frac{27}{3} = 9$ .

The Fundamental Theorem of Calculus

Computing limits of Riemann sums directly is tedious. The Fundamental Theorem of Calculus (FTC) gives a much faster route.

If $F$ is any antiderivative of $f$ — meaning $F^{'} (x) = f (x)$ — then $\int_{a}^{b} f (x) d x = F (b) - F (a)$ .

For $f (x) = x^{2}$ , an antiderivative is $F (x) = x^{3} /3$ because $(x^{3} /3)^{'} = x^{2}$ . So $\int_{0}^{3} x^{2} d x = F (3) - F (0) = 27/3 - 0 = 9$ . The Riemann sum limit and the FTC give the same answer, which is the content of the theorem.

Signed area in practice

Consider $\int_{0}^{2 π} sin (x) d x$ . On $[0, π]$ the sine curve is positive, contributing area $+ 2$ . On $[π, 2 π]$ it is negative, contributing area $- 2$ . The total integral is $0$ , even though the curve sweeps out nonzero area on each half.

If you want the total unsigned area swept, integrate the absolute value: $\int_{0}^{2 π} ∣ sin (x) ∣ d x = 4$ .

This distinction matters whenever a function changes sign over the interval of integration.

How to read the notation

In $\int_{a}^{b} f (x) d x$ : $a$ and $b$ are the lower and upper limits of integration. $f (x)$ is the integrand. The $d x$ signals that $x$ is the variable of integration and $Δ x \to 0$ in the limit.

The elongated $\int$ symbol is Leibniz's notation for an infinite sum of infinitesimal strips. The $d x$ is the infinitesimal width. The product $f (x) d x$ is the area of one infinitely thin strip.

Common Pitfall

The definite integral is not the same as finding an antiderivative. An antiderivative is a function; a definite integral is a number. The FTC connects them, but they are different objects.

Try a Variation

Sketch $f (x) = x^{2} - 4$ and identify where it crosses the $x$ -axis. Then compute $\int_{0}^{3} (x^{2} - 4) d x$ using the FTC and check whether the sign of the result matches your sketch.

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