From Rectangle Sums to a Definite Integral
A tagged Riemann sum records the heights of a function at selected sample points and weights each height by the width of its subinterval. It is a finite sum, so it can be computed directly. The definite integral is defined when these sums approach one common number as the subintervals become uniformly small.
The requirement that the limit be common is essential. The partitions need not have equal widths, and the sample points need not be chosen by a particular rule. The value of the integral must emerge regardless of these choices.
Partitions, Tags, and Mesh
Let \(a<b\). A partition \(P\) of \([a,b]\) is a finite increasing list of points
Its \(i\)th subinterval is \([x_{i-1},x_i]\), and its width is \(\Delta x_i=x_i-x_{i-1}\). A tagged partition includes a sample point \(\xi_i\) in each subinterval, so \(x_{i-1}\leq \xi_i\leq x_i\). The mesh of \(P\), written \(\|P\|\), is the width of its widest subinterval:
For a function \(f:[a,b]\to\mathbb{R}\), the tagged Riemann sum associated with these choices is
Each term is a rectangle's signed height times its width. If \(f(\xi_i)\) is negative, that term is negative; the sum records signed accumulation, not just geometric area.
In this definition, “every choice” includes nonuniform partitions and every allowed choice of sample points. More precisely, once the mesh is sufficiently small, every tagged sum must be as close to \(I\) as desired. The next tutorial gives the quantified epsilon-delta form of this limit condition.
The boundedness condition is part of the Riemann-integrability definition used here. It ensures, in particular, that the function's values on the interval cannot grow without bound. The definition does not say that every bounded function is integrable: integrability also requires the common tagged-sum limit to exist.
Why the Limit Must Be Common
One sequence of sums gives information about one sequence of choices. It cannot, by itself, establish the definition, which requires the same limiting value for all partitions and tags whose mesh tends to zero. When the common limit does exist, its value is unique.
To see why, suppose two real numbers \(I\) and \(J\) both satisfy the common-limit requirement. If they were different, their separation \(d=|I-J|\) would be positive. For sufficiently small mesh, every tagged sum would have to be within \(d/3\) of \(I\), and also within \(d/3\) of \(J\). Choosing any tagged partition with mesh that small would give a sum \(S\) satisfying
This contradicts \(|I-J|=d\). Therefore \(I=J\). The common limit, when it exists, is a single well-defined number.
Worked Examples
Worked Example: A Constant Function
Let \(f(x)=3\) on \([1,4]\). Take any partition \(1=x_0<x_1<\cdots<x_n=4\) and any tags \(\xi_i\in[x_{i-1},x_i]\). Since the function has the same value at every tag, its sum is
The widths telescope to the interval length, so every tagged sum equals \(9\), not merely sums from fine partitions. Consequently, the common limit exists and \(\int_1^4 3\,dx=9\). This is checked directly: changing the partition or any tag leaves the sum equal to \(9\).
Worked Example: The Identity Function on an Arbitrary Partition
Let \(f(x)=x\) on \([0,2]\). Consider any partition \(0=x_0<x_1<\cdots<x_n=2\), with any tag \(\xi_i\) in each subinterval. Write \(\Delta x_i=x_i-x_{i-1}\). Since \(x_{i-1}\leq \xi_i\leq x_i\), there is a number \(\theta_i\in[0,1]\) such that \(\xi_i=x_{i-1}+\theta_i\Delta x_i\). Thus
For each subinterval, expanding the difference of squares gives
Summing these identities and using \(x_0=0\), \(x_n=2\), we obtain
Substituting into the tagged sum yields
Because \(0\leq\theta_i\leq1\), we have \(\left|\theta_i-\frac12\right|\leq\frac12\). Also, \(\Delta x_i\leq\|P\|\), so
This bound applies to every partition and every set of tags. As the mesh tends to zero, the difference from \(2\) tends to zero, so \(\int_0^2 x\,dx=2\). For a direct check of the finite-sum formula, take the partition \(0<1<2\) with tags \(0.5\) and \(1.5\). Its sum is \(0.5(1)+1.5(1)=2\), as the bound permits.
Worked Example: An Affine Function on an Interval
Let \(f(x)=2x+1\) on \([1,3]\). For any tagged partition, the sum is
The identity-function calculation above, with endpoints \(1\) and \(3\), gives the exact expression
where each \(\theta_i\in[0,1]\). The widths sum to \(3-1=2\), so
As before, \(\left|\theta_i-\frac12\right|\leq\frac12\). Therefore
This error bound tends to zero with the mesh, whatever the tags, so \(\int_1^3(2x+1)\,dx=10\). As a finite check, use the partition \(1<2<3\) and tags \(1.5\) and \(2.5\). The sum is \((2(1.5)+1)(1)+(2(2.5)+1)(1)=4+6=10\), consistent with the limit.
A Common Mistake: Treating the Mesh as Every Width
The mesh is the largest subinterval width; it is not a replacement for each individual width in the tagged sum. For example, partition \([0,2]\) at \(0.5\), and choose tags \(0.25\) and \(1.5\) for \(f(x)=x\). The widths are \(0.5\) and \(1.5\), and the correct sum is
A mistaken calculation might multiply both sampled heights by the mesh \(1.5\), obtaining \(0.25(1.5)+1.5(1.5)=2.625\). That is not the tagged Riemann sum: each height must be multiplied by the width of its own subinterval. The mesh is used to describe how fine the whole partition is, not as a common width unless the partition is uniform.
What the Definition Establishes
The definite integral is not defined as the value of one particular rectangle sum. It is the common limiting value of all tagged sums as the mesh tends to zero. For that reason, a calculation using equal-width partitions or a favorite tag rule may be useful evidence, but it establishes integrability only if the limit is shown to be independent of those choices.
In applications, \(f(\xi_i)\) can represent a rate and \(\Delta x_i\) a short time or distance interval. Their product approximates an accumulated amount on that subinterval. The definite integral captures the limiting total accumulation when the partition is refined, provided the function meets the definition's boundedness and common-limit requirements.
Check Your Understanding
Use the definitions and examples above to answer these questions.
- What information does a tagged partition add to an ordinary partition?
- How is the mesh of a partition defined, and why is it not necessarily the width of every subinterval?
- In the definition of the Riemann integral, what must be true of the tagged sums as the mesh tends to zero?
- Why does finding a limit along one sequence of uniform partitions not, by itself, establish the common tagged-sum limit?
- For a constant function \(f(x)=c\) on \([a,b]\), explain why every tagged sum equals \(c(b-a)\).
