Riemann Sums, Sigma Notation and the Definite Integral

Riemann sums estimate an area. Use more subintervals and the estimate improves. This page follows that idea to the end: take the limit as the number of subintervals goes to infinity. The result is the definite integral, the central object of the rest of the course.

Sigma notation for a Riemann sum

Writing out a sum with 100 terms is slow, so we use sigma notation. The symbol $\sum$ (capital sigma) means "add up":

$$\sum_{i=1}^{4} (2i + 1) = 3 + 5 + 7 + 9 = 24.$$

The index $i$ starts at the number below the sigma and goes up by 1 to the number on top. Substitute each value of $i$ into the expression and add.

Now split $[a, b]$ into $n$ subintervals of equal width. Each has width

$$\Delta x = \frac{b - a}{n},$$

and the right ends are $x_1 = a + \Delta x$, $x_2 = a + 2\Delta x$, and in general $x_i = a + i\,\Delta x$, up to $x_n = b$. A right Riemann sum is then

$$\sum_{i=1}^{n} f(x_i)\,\Delta x = \sum_{i=1}^{n} f\left(a + \frac{(b - a)i}{n}\right)\frac{b - a}{n}.$$

The limit: the definite integral

As $n$ grows, the rectangles get thinner and hug the curve more closely, as in the figure. For a continuous function the sums approach a single number, whichever endpoints you use (left, right or midpoint). That number is the definite integral:

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

Each piece of the symbol echoes the sum. The integral sign, ∫, is a stretched S, for "sum". The function $f(x)$ is the integrand, the height. The numbers $a$ and $b$ are the limits of integration, the ends of the interval. And $dx$ stands for the vanishing width $\Delta x$; it also names the variable. Never drop it.

The letter is a placeholder: $\int_2^7 g(t)\,dt$ and $\int_2^7 g(u)\,du$ are the same number. The definite integral is a number, not a function of $t$.

Translating a limit of sums into an integral

The exam often shows a limit of sums and asks which integral it equals. Match three things:

1. Width. The factor $\frac{c}{n}$ is $\Delta x$, so the interval has length $c = b - a$.
2. Points. The expression $a + \frac{ci}{n}$ is $x_i$; its constant $a$ is the left end.
3. Height. Whatever is done to $x_i$ is the function $f$.

A careful point: the same limit also equals $\int_0^3 (2 + x)\,dx$, read with points $\frac{3i}{n}$ on $[0, 3]$ and the 2 moved into the function. Both are right. What is wrong is to mix them: $\int_2^5 (2 + x)\,dx$ counts the 2 twice.

Translating an integral into a limit of sums

What a definite integral means

• If $f(x) \ge 0$ on $[a, b]$, $\int_a^b f(x)\,dx$ is the area of the region under the graph. When that region is a triangle, trapezoid or part of a circle, geometry gives the value: the upper half of a circle of radius 2 encloses $\frac{1}{2}\pi(2)^2 = 2\pi$.
• If $f$ is negative, each product $f(x_i)\Delta x$ is negative, so the integral is negative: $\int_1^4 (-2)\,dx$$= -6$.
• If $f$ is a rate, the integral is the accumulated change, in (rate units) times (input units). If $P(t)$ is power in kilowatts and $t$ is in hours, $\int_0^{24} P(t)\,dt$ is the energy used over the day, in kilowatt-hours.

With a calculator

Every graphing calculator, and the Desmos calculator built into the exam, computes definite integrals numerically. On a calculator question, write the integral as your setup, such as $\int_0^2 e^{-x^2}\,dx$, then the value correct to three decimal places: $0.882$. A number with no integral written beside it shows no setup, and the setup earns its own point.