Diagnosing a Claimed Riemann-Sum Limit

Introduction

A proposed Riemann-sum calculation can look convincing and still be wrong. Its algebra may be accurate while its widths do not match the partition, its tags lie outside their subintervals, or its partitions never become fine. Even a correctly computed limit can be misinterpreted if signed accumulation is confused with geometric area.

A useful diagnosis checks the setup before the arithmetic. We will examine the interval, partition, tags, widths, boundedness, and mesh, then check what the resulting number means.

The structure of a tagged Riemann sum

Let $f$ be a function on a closed interval $[a,b]$, where $a<b$. A partition $P$ of that interval is a finite ordered list of points

a=x_0<x_1<\cdots<x_n=b.

The $i$th subinterval is $[x_{i-1},x_i]$, and its width is $\Delta x_i=x_i-x_{i-1}$. A tag $\xi_i$ for that subinterval must satisfy $x_{i-1}\leq \xi_i\leq x_i$. The tagged Riemann sum is the sum of each function value at its tag multiplied by the corresponding width.

S(f,P,\xi)=\sum_{i=1}^{n} f(\xi_i)\Delta x_i.

Definition: The mesh of a partition $P$ is its largest subinterval width: $$ \|P\|=\max_{1\leq i\leq n}\Delta x_i. $$ A sequence of partitions has mesh tending to zero if $\|P_k\|\to 0$ as $k\to\infty$.

The mesh measures the size of the widest subinterval, not the number of subintervals by itself. A partition with many pieces can still have a large mesh if even one piece is wide. For an equal-width partition of $[a,b]$ into $n$ pieces, each width is $(b-a)/n$, so its mesh tends to zero as $n\to\infty$.

What must be checked before accepting a limit?

For a proper Riemann integral on $[a,b]$, the function must be bounded and Riemann integrable there. Boundedness means there is a finite number $M$ such that $|f(x)|\leq M$ for every $x\in[a,b]$. Boundedness is necessary for proper Riemann integrability, but it is not by itself enough to guarantee integrability.

Riemann integrability can be characterized using upper and lower sums: for every $\varepsilon>0$, there is a partition for which the upper sum minus the lower sum is less than $\varepsilon$. For a bounded Riemann-integrable function, the standard tagged-sum criterion says that sums converge to the definite integral whenever the partition mesh tends to zero, regardless of how valid tags are chosen.

Theorem: If $f$ is Riemann integrable on $[a,b]$, and $P_k$ are partitions with $\|P_k\|\to 0$, then for any choice of tags $\xi_{i,k}\in[x_{i-1,k},x_{i,k}]$, $$ \lim_{k\to\infty}\sum_{i=1}^{n_k} f(\xi_{i,k})\Delta x_{i,k} =\int_a^b f(x)\,dx. $$

This theorem gives a compact audit rule: a claimed limit must come from genuine partitions of the stated interval, valid tags, and the actual widths, with mesh tending to zero. The function must also meet the conditions for a proper Riemann integral.

Why shrinking mesh controls the sums

Here is a justification of the key idea. For a bounded function, the upper and lower sums of a partition use, on each subinterval, the supremum and infimum of the function there. Every tagged sum for that partition lies between those two sums. Integrability means a fixed partition can be chosen with an arbitrarily small upper-minus-lower difference.

To see why that helps for any sufficiently fine partition, fix such a partition $Q$. When another partition $P$ has small enough mesh, each subinterval of $P$ either lies within a single subinterval of $Q$, or crosses one of the finitely many internal breakpoints of $Q$. For the pieces lying within a single interval of $Q$, their total contribution to the upper-minus-lower difference is no greater than the corresponding difference for $Q$. The crossing pieces have total width at most the number of internal breakpoints times $\|P\|$. If $|f|\leq M$, the oscillation of $f$ on any such piece is at most $2M$; consequently their total contribution tends to zero with the mesh.

Thus, first choose $Q$ so the difference is small, and then choose $P$ fine enough that the crossing-piece contribution is small. The upper and lower sums for $P$ are then close. Since every tagged sum and the integral lie between those bounds, the tagged sum is close to the integral too. This explains why fine partitions, together with integrability, control the result.

Worked Example: Verifying a midpoint sum for a linear function

Consider $f(x)=x$ on $[0,2]$. Divide the interval into $n$ equal subintervals and choose each midpoint as its tag. Check the Riemann-sum limit.

The partition points are $x_i=2i/n$, for $i=0,\ldots,n$. They begin at $0$, end at $2$, and increase in order. Each width is

\Delta x_i=\frac{2i}{n}-\frac{2(i-1)}{n}=\frac{2}{n}.

The midpoint tag is

\xi_i=\frac{x_{i-1}+x_i}{2} =\frac{2(i-1)/n+2i/n}{2} =\frac{2i-1}{n}.

This lies between the stated endpoints of its subinterval, as required. The function is bounded on $[0,2]$, and the mesh is $2/n$, which tends to zero. The sum is

\begin{aligned} S_n &=\sum_{i=1}^{n} f(\xi_i)\Delta x_i\\ &=\sum_{i=1}^{n}\frac{2i-1}{n}\frac{2}{n}\\ &=\frac{2}{n^2}\sum_{i=1}^{n}(2i-1)\\ &=\frac{2}{n^2}\,n^2\\ &=2. \end{aligned}

Since $f(x)=x$ is Riemann integrable, the tagged-sum theorem applies. The resulting integral is $2$, which also agrees with the triangular area under the line from $0$ to $2$. Thus the partition, tags, widths, mesh, computation, and interpretation all check out.

Worked Example: Finding an incorrect width

A calculation for $f(x)=x$ on $[0,1]$ uses right endpoints $i/n$, but assigns width $2/n$ to every subinterval. It therefore claims

\sum_{i=1}^{n}\frac{i}{n}\frac{2}{n} =\frac{2}{n^2}\frac{n(n+1)}{2} =\frac{n+1}{n} \longrightarrow 1.

The arithmetic is correct, but the proposed widths fail the partition check. For $n$ subintervals of equal width on $[0,1]$, the widths must add to $1$. Here they add to $n(2/n)=2$, so they cannot be the widths of a partition of $[0,1]$.

The actual width is $1/n$. With the same right-endpoint tags, the valid sum is

\begin{aligned} \sum_{i=1}^{n}\frac{i}{n}\frac{1}{n} &=\frac{1}{n^2}\frac{n(n+1)}{2}\\ &=\frac{n+1}{2n}\\ &\longrightarrow \frac{1}{2}. \end{aligned}

The mesh is $1/n\to0$, so this limit agrees with $\int_0^1 x\,dx=1/2$. The incorrect answer came from widths that did not match the interval, not from a mistake in the sum formula.

Worked Example: Checking signed interpretation

Let $f(x)=x-1$ on $[0,2]$. Divide the interval into $n$ equal pieces and use midpoints. Check the sum and interpret its value.

Each width is $2/n$, and the midpoint of the $i$th piece is $(2i-1)/n$. The tags are within their subintervals, and $f$ is bounded on this interval. The mesh $2/n$ tends to zero. The tagged sum is

\begin{aligned} S_n &=\sum_{i=1}^{n}\left(\frac{2i-1}{n}-1\right)\frac{2}{n}\\ &=\frac{2}{n^2}\sum_{i=1}^{n}(2i-1-n). \end{aligned}

The terms in the last sum cancel in pairs: the term indexed by $n+1-i$ is the negative of the term indexed by $i$. If $n$ is odd, the unpaired middle term is zero. Hence $S_n=0$ for every $n$, and the limit is $0$. This is the signed integral: the contribution below the horizontal axis cancels the contribution above it. The geometric areas of the two triangles add to $2\cdot(1/2)=1$, so geometric area is not the value of this signed integral.

Worked Example: A fixed partition does not establish a limit

For $f(x)=x$ on $[0,1]$, take just the partition $0<1$, choose the tag $0$, and compute the sum $f(0)(1-0)=0$. This is a valid tagged sum for that one partition. It does not show that the integral is $0$, because its mesh is $1$, and repeating the same partition does not make the mesh tend to zero.

For comparison, divide $[0,1]$ into $n$ equal pieces and choose left endpoints. The mesh is $1/n$, and the valid sums are

\begin{aligned} S_n &=\sum_{i=1}^{n}\frac{i-1}{n}\frac{1}{n}\\ &=\frac{1}{n^2}\frac{n(n-1)}{2}\\ &=\frac{n-1}{2n}\\ &\longrightarrow \frac{1}{2}. \end{aligned}

The limiting sum agrees with $\int_0^1 x\,dx=1/2$. A single tagged sum may be far from the integral; it is the shrinking-mesh limit that matters.

A practical diagnostic checklist

When checking a claimed Riemann-sum limit, test its ingredients in this order:

Interval: Do the partition points start at $a$, end at $b$, and increase strictly?
Widths: Is each width $x_i-x_{i-1}$, and do all the widths add to $b-a$?
Tags: Is each tag inside its assigned subinterval?
Function: Is the function bounded and Riemann integrable on the stated closed interval?
Mesh: Does the largest width tend to zero, rather than just some widths or the average width?
Interpretation: Does the answer represent signed accumulation, rather than an unsigned geometric area?

Also check the index range: a sum with $n$ subintervals should have exactly $n$ terms, and its $i$th function value must be multiplied by the width of the same subinterval. A correct-looking formula can conceal a mismatch between the index, tag, and width.

Why these checks matter

Riemann sums connect local function values to a total accumulated quantity. Their reliability depends on matching each value to the portion of the interval it represents. Incorrect widths distort that contribution; invalid tags evaluate the function at the wrong location; a mesh that stays large leaves some part of the interval unresolved. These are structural errors, and simplifying the algebra cannot repair them.

Finally, the integral uses signed contributions. Where the function is below the horizontal axis, its values contribute negatively. A result may therefore be smaller than the total geometric area, or even zero when positive and negative contributions cancel. A sound diagnosis checks both how the limit was formed and what its value means.

Check Your Understanding

For each question, identify which part of a claimed Riemann-sum calculation should be checked.

A partition of $[2,5]$ has subinterval widths that add to $4$. What must be wrong?
In a proposed tagged sum, a tag lies outside the subinterval whose width multiplies its function value. Is this a valid tagged sum?
Why is it not enough for the number of subintervals to grow if the mesh does not tend to zero?
What conditions allow arbitrary valid tags to give sums converging to the definite integral?
A function has equal geometric areas above and below the horizontal axis on an interval. What can be said about its signed integral there, and why?

Diagnosing a Claimed Riemann-Sum Limit

What You'll Learn

Introduction

The structure of a tagged Riemann sum

What must be checked before accepting a limit?

Why shrinking mesh controls the sums

Worked Example: Verifying a midpoint sum for a linear function

Worked Example: Verifying a midpoint sum for a linear function

Worked Example: Finding an incorrect width

Worked Example: Finding an incorrect width

Worked Example: Checking signed interpretation

Worked Example: Checking signed interpretation

Worked Example: A fixed partition does not establish a limit

Worked Example: A fixed partition does not establish a limit

A practical diagnostic checklist

Why these checks matter

Check Your Understanding

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Diagnosing a Claimed Riemann-Sum Limit

What You'll Learn

Introduction

The structure of a tagged Riemann sum

What must be checked before accepting a limit?

Why shrinking mesh controls the sums

Worked Example: Verifying a midpoint sum for a linear function

Worked Example: Verifying a midpoint sum for a linear function

Worked Example: Finding an incorrect width

Worked Example: Finding an incorrect width

Worked Example: Checking signed interpretation

Worked Example: Checking signed interpretation

Worked Example: A fixed partition does not establish a limit

Worked Example: A fixed partition does not establish a limit

A practical diagnostic checklist

Why these checks matter

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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