Riemann sums and definite integrals

A pump adds and removes water

A tank begins with 15 litres. For the first two minutes a pump adds water at 3 litres per minute. For the next minute it removes water at 1 litre per minute. The changes are 3·2=6 litres and (−1)·1=−1 litre: a net gain of 5 litres and a final content of 20 litres.

Each product is rate times duration. A negative rate means removal; it does not mean a negative duration. On a time graph, the positive contribution is above the zero-rate axis and the negative contribution is below it. Their signed sum answers the net-change question.

For a changing rate, one sampled value times a duration is usually an approximation. The lesson explains how to refine that approximation and justify a single limiting value. It does not infer a rate between measured observations without a model.

Preparation: review function domains and actual values, multiplication of fractions, and rectangle area using the linked lessons. First calculate the four rectangles and the signed example; return to the quantified existence argument afterward. Limits and continuity are explained here; use them on your return through the existence argument.

Specify every interval and its sample

Fix real a<b and a bounded real-valued function f defined at every input of the closed interval [a,b]. Bounded means that one finite B satisfies |f(x)|≤B for every allowed input. A partition P is a finite increasing list a=x₀<x₁<⋯<xₙ=b; its n cells are [xᵢ₋₁,xᵢ] and their widths Δxᵢ=xᵢ−xᵢ₋₁ are positive.

Choose one tag ξᵢ in each closed cell. Endpoints are permitted tags, even when the same boundary input is used in two neighbouring cells. Its sampled height is f(ξᵢ). The cell contribution is f(ξᵢ)Δxᵢ; its sign comes from the height.

The tagged Riemann sum is S(f;P,ξ)=Σᵢ₌₁ⁿ f(ξᵢ)Δxᵢ. Here Σ means add the listed terms, and i only labels a cell. If the widths differ, keep the individual widths inside the sum; a common-width multiplier is then unavailable.

The mesh ||P|| is the largest width, maxᵢΔxᵢ. It measures the widest unresolved interval. The number of cells alone does not say that every interval is small.

Use the same partition to compare three choices

For equal widths Δx=(b−a)/n, the left tags are a+(i−1)Δx, right tags a+iΔx, and midpoint tags a+(i−1/2)Δx. Their sums are Lₙ, Rₙ and Mₙ respectively. These names specify the sample location; they do not promise an exact integral.

For f(x)=x on [0,2] with four widths of 1/2, the left heights are 0, 1/2, 1 and 3/2. Thus L₄=(0+1/2+1+3/2)/2=3/2. Right heights give R₄=5/2. Midpoint heights 1/4, 3/4, 5/4 and 7/4 give M₄=2.

Because x is increasing, each left rectangle is no higher than the curve anywhere in its cell, while each right rectangle is no lower. This comparison concerns signed heights and remains valid if some heights are negative. For a decreasing function the left and right bounds swap.

A midpoint height need not bound all heights in its cell. On one cell [0,1], (x−1/2)² has midpoint height zero but positive heights away from the midpoint; the negative of that function reverses the comparison. We prove their nonzero integrals after introducing the integral definition. Exactness for a straight line is a special case, not a general rule.

Require agreement for all fine partitions and tags

The bounded function f is Riemann integrable on [a,b] if there is one real number I such that, for every error tolerance ε>0, some δ>0 makes |S(f;P,ξ)−I|<ε for every partition with ||P||<δ and every permitted set of tags. We write I=∫ₐᵇ f(x) dx. The symbol dx records the input variable; the definition is a limit of ordinary weighted sums.

A limit here is an all-input guarantee, not an instruction to set a width to zero. The tolerance ε measures output error; δ controls the maximum input width. Their units may differ. The same δ must work for all the allowed partitions and tags after ε is chosen.

The integral is unique: if I and J both qualified and I≠J, choose ε=|I−J|/3 and a partition finer than both radii. One sum would be within ε of each, forcing |I−J|<2ε, a contradiction. Finely dividing a finite interval supplies such a partition.

Finding convergence along one preferred list of samples is weaker. A constant sample rule can conceal variation between sample points, and a large cell can survive while the number of other cells grows. Neither establishes this definition.

Bracket every tag choice

On cell i let mᵢ be the infimum, its greatest lower bound, and Hᵢ the supremum, its least upper bound, of all function values there. A bound need not be attained. Define lower sum ℓ(f;P)=ΣmᵢΔxᵢ and upper sum u(f;P)=ΣHᵢΔxᵢ. Multiplying mᵢ≤f(ξᵢ)≤Hᵢ by each positive width and adding proves ℓ≤S≤u.

Adding partition points makes every new cell smaller. Its lower bound cannot decrease and its upper bound cannot increase, so refinement raises ℓ and lowers u. Given two partitions P and Q, their combined list of points refines both. Therefore ℓ(f;P)≤u(f;Q), even if neither original partition refines the other.

All lower sums are bounded above and all upper sums are bounded below by finite multiples of b−a. Completeness of the real numbers supplies A=sup_P ℓ(f;P) and C=inf_P u(f;P), with A≤C. If partitions with u−ℓ arbitrarily small exist, then 0≤C−A≤u−ℓ forces A=C=I. Every lower and upper sum brackets this I.

Why does one small-gap partition control arbitrary fine partitions? Fix P with r interior boundaries and gap below ε/2. A different partition Q of mesh d has at most 2r cells meeting those boundaries, of total width at most 2 rd. On these cells |f|≤B bounds their combined gap by 4 Brd. Every remaining Q cell lies inside an old P cell; its gap contributes no more than the old gap in total. Thus u(f;Q)−ℓ(f;Q)≤u(f;P)−ℓ(f;P)+4 Brd.

Choose d<ε/[8 max(1,B) max(1,r)]. The gap for Q is below ε, and both its sum and I lie between its bounds, so |S−I|<ε. Conversely, the all-tag definition gives a small gap: on any sufficiently fine partition choose tags approximating each infimum and supremum so their total errors are each below ε/4, and require both tagged sums within ε/4 of I. The triangle inequality gives u−ℓ<ε. This proves the small-gap criterion, not merely a picture of it.

Prove the square integral using finite identities

For x² on [0,1] with n equal cells, increasing heights give ℓ=Lₙ and u=Rₙ. The exact finite sums are Lₙ=(n−1)(2n−1)/(6n²) and Rₙ=(n+1)(2n+1)/(6n²). Their gap is 1/n, so the small-gap criterion establishes integrability.

The identities have a finite algebraic proof. Pairing the integers 0,…,n−1 in reverse order gives Σj=n(n−1)/2. Adding (j+1)³−j³=3j²+3j+1 for j=0,…,n−1 makes the left side telescope to n³; substitution gives Σj²=n(n−1)(2n−1)/6. Divide by n³ for Lₙ; adding the final height and removing the first gives Rₙ=Lₙ+1/n.

Both formulas approach 1/3, so the bracketed integral is ∫₀¹x² dx=1/3. More explicitly, Rₙ=1/3+1/(2n)+1/(6n²) and Lₙ=1/3−1/(2n)+1/(6n²). Given any positive tolerance, sufficiently large n makes both errors small.

For midpoint tags, expanding (j+1/2)² and using those same two finite identities yields Mₙ=1/3−1/(12n²). At n=4, L₄=7/32, R₄=15/32 and M₄=21/64. The midpoint error is 1/192. Its faster convergence here follows from this quadratic identity, not from a universal midpoint promise. The earlier centred square h(x)=(x−1/2)² also has cell oscillation at most its width: |h(x)−h(y)|=|x−y||x+y−1|≤|x−y|. Thus its lower–upper gap is at most 1/n, proving existence. Expanding its midpoint terms with the same finite identities gives Mₙ(h)=1/12−1/(12n²), so its integral is 1/12 although its one-cell midpoint sum is zero. Negating every term gives integral −1/12 for −h. Midpoint sums therefore cannot generally be upper or lower bounds.

State conditions that justify convergence

A Lipschitz condition is a finite constant K≥0 with |f(x)−f(y)|≤K|x−y| on the whole interval. Each cell then has oscillation Hᵢ−mᵢ≤KΔxᵢ. Hence u−ℓ≤KΣ(Δxᵢ)²≤K(b−a)||P||. Equal partitions make this gap vanish, proving existence; every tagged sum has error at most this bound. If K=0 the function is constant and every sum is exact.

For midpoint sums the Lipschitz bound improves to K(b−a)||P||/2. On each cell the distance from its midpoint is at most half the width, so the function differs from the sampled constant by at most KΔxᵢ/2. Bracketing that cell integral and adding the bounds proves the estimate. It does not assert quadratic convergence for all Lipschitz functions.

A bounded increasing function has Hᵢ−mᵢ=f(xᵢ)−f(xᵢ₋₁). Thus u−ℓ≤||P||Σ[f(xᵢ)−f(xᵢ₋₁)]=||P||[f(b)−f(a)]. Decreasing functions give the absolute endpoint difference instead. This proves integrability even when a monotone function has jumps.

Continuity at c means that |f(x)−f(c)| becomes arbitrarily small for all domain inputs sufficiently close to c. On a closed finite interval, continuity also supplies boundedness and one common closeness radius for all input pairs. Here is the bridge: an open-neighbourhood cover of [a,b] has a finite subcover. Otherwise repeatedly bisect an interval lacking a finite subcover; completeness gives a common point of the nested intervals, and one open neighbourhood around that point contains a sufficiently late interval, a contradiction.

For each c choose a continuity radius r_c giving errors below η/2; the half-radius neighbourhoods cover [a,b]. Take a finite subcover and δ equal to half the smallest selected radius. If |x−y|<δ, a selected half-neighbourhood containing x puts both x and y within its full radius, giving |f(x)−f(y)|<η. Applying the same finite-cover argument with error 1 bounds all values by a finite maximum of |f(c)|+1.

With η=ε/[2(b−a)], cells of mesh below δ have oscillation at most η; their total gap is at most ε/2. The criterion proves every continuous real function on [a,b] integrable. The finite-cover argument is local justification for the closed-interval condition; it does not assert this result on an unbounded interval or across a missing input.

Keep net change separate from total area

For f(x)=x on [−1,1], the negative and positive triangular regions have areas 1/2 each. Their signed contributions cancel, so ∫₋₁¹x dx=0. The geometric area between the graph and the axis is 1. Zero net accumulation does not mean that no region or activity exists.

For an integrable f, |f| is also integrable: ||f(x)|−|f(y)||≤|f(x)−f(y)| makes its cell oscillation no larger, so the same small-gap criterion applies. The geometric area is ∫ₐᵇ|f(x)| dx. The triangle inequality for every finite sum then gives |∫f|≤∫|f|.

Units multiply. A rate in litres per minute multiplied by minutes gives litres; velocity in metres per second multiplied by seconds gives signed displacement in metres. Integrating the magnitude of velocity gives distance travelled for that specified continuous-time model. It is a different question from signed displacement.

If both graph coordinates measure length in metres, a height-times-width contribution measures square metres. A plotted rate-time rectangle instead measures the product of its named units. The diagram's pixel area and stretched axes are not a measuring instrument.

Bounded does not always mean integrable

The pump rate 3 on 0≤t<2 and −1 on 2≤t≤3 is decreasing and bounded. Its integral is 5: partitions aligned at 2, with tags inside each stage, give the two constant contributions, while their lower–upper gap is at most 4||P|| and tends to zero. Coarse right tags at 1,2,3 give sum 1; midpoint tags at 1/2,3/2,5/2 give 5. A boundary value can matter in a finite sum without changing the integral.

Changing finitely many values of an integrable bounded function preserves its integral. If f and g differ at k points with |f−g|≤D, at most 2k cells can sample those points, so |S(f)−S(g)|≤2 kD||P||. This bound goes to zero and the all-tag definition transfers. A function zero except for height 5 at x=1/2 on [0,1] therefore has integral 0; its finite endpoint samples may still contribute.

In contrast, define d(x)=1 for rational x and 0 for irrational x on [0,1]. The symbol ℚ denotes rational numbers; x∈ℚ means x is rational and x∉ℚ means it is not. Every interval of positive width contains both kinds. A rational lies between its endpoints by choosing a sufficiently large denominator; adding a sufficiently small nonzero rational multiple of √2 to a rational inside the interval supplies an irrational there. To recall why √2 is irrational, a reduced fraction p/q satisfying p²=2q² would force p even and then q even, contradicting that it is reduced.

Every lower sum for d is 0 and every upper sum is 1. Even for arbitrarily fine partitions, rational tags give sum 1 and irrational tags give sum 0. They cannot approach a common value, so this bounded function is not Riemann integrable. Its picture shows two tag choices, not a falsely continuous curve or a complete rendering of rational membership.

The formula 1/x on [0,1] has a separate problem: it is undefined at 0 and unbounded near 0. It does not satisfy this lesson's bounded real-function hypotheses. Any treatment using an improper limit requires a separate definition; a finite sum alone does not fix those problems.

Refine the largest unresolved piece

Keep the cell [0,1/2] and divide only [1/2,1] into more and more cells. The cell count grows, but the mesh remains 1/2. For x², choosing tag 0 or tag 1/2 in the large cell changes its contribution by (1/4−0)·(1/2)=1/8; keeping every other tag the same preserves that difference.

For any proposed common value, two sums differing by 1/8 cannot both be within ε=1/32 of it, because their difference would then be less than 1/16. Increasing the cell count in this way never supplies the required all-tag guarantee. This does not contradict integrability of x²: the partition family is failing to make its mesh tend to zero.

For a known Lipschitz constant K>0, require ||P||<ε/[K(b−a)] for the conservative strict error target. For a monotone function with nonzero endpoint difference V, ||P||<ε/V suffices. Equality in a nonstrict error bound cannot by itself promise a strictly smaller error.

Use elementary properties with their hypotheses

If f and g are integrable on the same interval and α,β are fixed real numbers, then αf+βg is integrable with integral α∫f+β∫g. Each tagged sum obeys that identity exactly. For a chosen ε, ask the f and g errors to be less than ε/[2(|α|+|β|+1)] and use the smaller mesh radius; the weighted error is below ε, including zero coefficients.

For a<c<b, splitting at c gives ∫ₐᵇf=∫ₐᶜf+∫ᶜᵇf. A small-gap partition restricts to each subinterval after c is inserted, establishing those integrals. Partitions including c add exactly; an arbitrary fine partition can be split at c, and its one changed contribution differs by at most 2B||P||. That vanishes, proving the same whole-interval value.

By convention ∫ₐᵃf=0 and ∫ᵦᵃf=−∫ₐᵇf when the forward integral exists. Reversing bounds changes orientation, not the positive widths in the original forward definition. These rules describe accumulation directly; no antiderivative or fundamental theorem is used here.

Predict before revealing the exact records

Choose a supplied function, partition and tag rule. Predict the sign and the comparison with the proved integral or lower–upper bracket. Reveal the exact widths, heights, products and sum; compare left, right and midpoint cases on the same partition.

In the envelope cases, compare both bounds as the largest cell shrinks. In the mesh cases, inspect the large cell rather than only counting rectangles. For the rational indicator, two admissible tag classes remain incompatible however many cells you draw.

The finite explorer checks exact rational arithmetic and bounded illustrative cases. Its curves and rectangles illustrate the proofs; they do not replace the quantified definition. Return to the small-gap and error-bound sections when a numerical estimate needs a justified tolerance.

Worked examples

Try it yourself

1. A constant rate is −2 litres per minute for 3 minutes. Find its integral and explain the sign.

   Show the answer and reasoning

   −6 litres.

   Every sum is (−2)ΣΔt=−2·3. The negative sign records removal, while durations remain positive.

2. Calculate the left sum for x on [0,2] with four equal cells.

   Show the answer and reasoning

   3/2.

   Width 1/2 times heights 0,1/2,1,3/2 gives (1/2)·3=3/2; increasing heights make this a lower bound.

3. Find the right sum for x² on [0,1] with four equal cells and its error from 1/3.

   Show the answer and reasoning

   15/32; error 13/96.

   Heights 1/16,1/4,9/16,1 sum to 15/8; multiply by 1/4. Subtract 1/3 to obtain 13/96.

4. Find the four-cell midpoint sum for x² on [0,1] and decide whether it is exact.

   Show the answer and reasoning

   21/64, lower than 1/3 by 1/192.

   The tags 1/8,3/8,5/8,7/8 give heights 1,9,25,49 divided by 64. Their total times 1/4 is 84/256. The gap follows by a common denominator.

5. A partition has widths 1/8,1/8,3/4. What is its mesh? Does its three-cell count guarantee a small error?

   Show the answer and reasoning

   Mesh 3/4; the count alone supplies no error guarantee.

   Mesh is the maximum width, not their average 1/3. An error estimate needs a function hypothesis and control of that largest width.

6. Why is (f(ξ₁)+f(ξ₂))/2 not generally the sum on cells of widths 1/4 and 3/4?

   Show the answer and reasoning

   The required sum is f(ξ₁)/4+3f(ξ₂)/4.

   Each sample is weighted by its own width. Equal weighting is justified only when the stated cell widths are equal.

7. For x on [−1,1], is a zero integral the same as zero geometric area?

   Show the answer and reasoning

   No: integral 0, geometric area 1.

   The two triangles each have area 1/2. Their signed contributions cancel, while integrating the magnitude adds them.

8. A bounded increasing f has f(0)=−2 and f(3)=4. What error bound follows for any tags with mesh 1/10?

   Show the answer and reasoning

   At most 3/5.

   The endpoint variation is 4−(−2)=6. The lower–upper gap is at most 6·(1/10)=3/5, which brackets both the sum and integral.

9. Does an error bound≤1/100 by itself prove the error is<1/100?

   Show the answer and reasoning

   No.

   Equality remains possible in a nonstrict bound. Use a strictly smaller mesh bound or another estimate that makes the final comparison strict.

10. Must changing one finite value of an integrable bounded function change its integral?

    Show the answer and reasoning

    No; its integral stays the same.

    If the change magnitude is at most D, only at most two cells can use that input as a tag, bounding the sum difference by 2D||P||→0. Finite sums may change.

11. If ∫₀²f=3 and ∫₂⁵f=−1, find ∫₀⁵f and ∫₅⁰f.

    Show the answer and reasoning

    2 and −2 respectively.

    Add adjacent signed integrals:3+(−1)=2. Reversing the bounds changes the sign of that result.

12. A bounded function has lower sum 0 and upper sum 1 on every partition of[0,1]. Can it be Riemann integrable?

    Show the answer and reasoning

    No.

    The lower–upper gap never becomes small. Approximating both bounds with permitted tags gives incompatible all-tag sums.

13. If a Lipschitz constant is K=0 on a nonempty interval, what happens to the sums?

    Show the answer and reasoning

    All tags and partitions give the same exact constant-height sum.

    The inequality |f(x)−f(y)|≤0 forces f to be constant. Every sum is that height times b−a; no division by K is needed.

What to carry forward

• A tagged sum adds sampled heights multiplied by their individual positive widths.
• Left, right and midpoint describe sample choices; bounds require the function's stated behaviour.
• A bounded Riemann integral requires one common value for every sufficiently small mesh and every permitted tag choice.
• Lower and upper sums squeeze all tags; arbitrarily small gaps are equivalent to integrability.
• Continuity on a closed finite interval, a Lipschitz condition, and bounded monotonicity supply justified existence conditions.
• Signed accumulation can cancel; geometric area and total travel use the magnitude instead.
• Growing the cell count, matching a finite table or choosing only favourable tags is not the all-input limit.