Riemann Sum¶

Definition¶

Let $f\colon [a,b]\to\mathbb{R}$ be a bounded function. Given a partition $P=\{x_0,\ldots,x_n\}$ of $[a,b]$ and a tag $x^*_i\in J_i=[x_{i-1},x_i]$ in each subinterval, the corresponding Riemann sum is

S(f, P)=\sum_{i=1}^n f(x^*_i)(x_i-x_{i-1}),

where the tags may be chosen independently of one another.

Interpretation¶

A Riemann sum approximates the area under the graph of $f$ by rectangles. On each subinterval of a partition it places a rectangle of width $x_i-x_{i-1}$ and height $f(x^*_i)$ , the value of $f$ at the chosen tag, and adds up their areas in a finite sum.

The curve x squared over 2 plus 1 over four subintervals, with a rectangle on each whose height is the function value at the chosen tag, the midpoint of the subinterval.

Here $f(x)=\tfrac{x^2}{2}+1$ on $[0,4]$ with $x_i=i$ $i$ , and the tags $x^*_i$ are the midpoints. The rectangle heights are $f(x^*_i)=1.125,\ 2.125,\ 4.125,\ 7.125$ $1.125, 2.125, 4.125, 7.125$ , so the shaded area is $S(f,P)=14.5$ $14.5$ , close to the area $\int_0^4 f(x)\,\mathrm{d}x=\tfrac{44}{3}\approx14.67$ under the curve.

Each term $f(x^*_i)(x_i-x_{i-1})$ is height times width, and the sum depends on two choices: the cut points and the tags. The tags matter: for this increasing $f$ , the left tags give $1+1.5+3+5.5=11$ and the right tags give $1.5+3+5.5+9=19$ $19$ , which bracket the true area. Where $f$ is negative the terms are negative, so a Riemann sum approximates signed area: parts below the axis count negatively.

Riemann sums are the bridge between sums and integrals: the Riemann integral is defined as their limit, and conversely an integral can be estimated numerically by a Riemann sum with many pieces. A common slip on a regular subdivision of $[a,b]$ into $n$ pieces is to take the width as $\tfrac{1}{n}$ . It is $\tfrac{b-a}{n}$ , which equals $\tfrac{1}{n}$ only when $b-a=1$ $1$ .

Key Results¶

Riemann sums depend on both the partition and the choice of tags $x^*_i\text{.}$
On a regular subdivision, the tag $x^*_i=x_{i-1}$ gives the left Riemann sum and $x^*_i=x_i$ gives the right Riemann sum.
Because $f$ is bounded, each subinterval has a finite supremum and infimum, so we may also form the upper and lower Riemann sums

$\begin{aligned} U(f,P)&=\sum_{i=1}^n\left(\sup_{x\in J_i}f(x)\right)|J_i|,\\ L(f,P)&=\sum_{i=1}^n\left(\inf_{x\in J_i}f(x)\right)|J_i|, \end{aligned}$

which bracket every tagged Riemann sum on the same partition. These exist even when the extreme values are not attained. - As the mesh shrinks, the Riemann sums of an integrable function approach its integral, whatever tags are chosen. A single refinement need not improve a particular tagged sum, but it never raises the upper sum and never lowers the lower sum.

Examples¶

For $f(x)=x$ on $[0,1]$ with the regular subdivision into $4$ pieces, the left Riemann sum is $\tfrac{1}{4}\left(0+\tfrac{1}{4}+\tfrac{1}{2}+\tfrac{3}{4}\right)=\tfrac{3}{8}$ and the right Riemann sum is $\tfrac{1}{4}\left(\tfrac{1}{4}+\tfrac{1}{2}+\tfrac{3}{4}+1\right)=\tfrac{5}{8}\text{;}$ the integral $\tfrac{1}{2}$ lies between them.

Prerequisites¶

Sources¶

Math I: Calculus · Lecture 10

Symbols: element of, integral, real numbers, sum

Last updated: 27 October 2026 11:52