Unit 5, lesson 49

Partitions and Riemann Sums

How to cut an interval into pieces of any widths, build a Riemann sum from any choice of sample points, and trap every such sum between lower and upper Darboux sums.

  • Theory part 97
  • Practice part 98
  • Definite Integrals

Theory, part 97

Practice, part 98

Equal strips are convenient, but nothing about area requires them. A surveyor measuring a field along a river might take readings closer together where the bank bends and farther apart where it runs straight. Allowing uneven pieces and any sample point in each piece gives the general sums that the definite integral is built from.

Key terms

Partition

Definition A partition of a closed interval [a,b][a, b] with a<ba < b is a finite set of points

P={x0,x1,…,xn},a=x0<x1<⋯<xn=b.P = \{ x_0, x_1, \ldots, x_n \}, \qquad a = x_0 < x_1 < \cdots < x_n = b.

It splits [a,b][a, b] into the nn subintervals [xi−1,xi][x_{i-1}, x_i] with widths Δxi=xi−xi−1>0\Delta x_i = x_i - x_{i-1} > 0, and these widths add up to b−ab - a.

In plain words A partition is a list of cut points that starts at aa, ends at bb and goes strictly upward. The pieces need not have the same width.

Example The set P={0,0.5,1.5,2,3}P = \{0, 0.5, 1.5, 2, 3\} is a partition of [0,3][0, 3] into four subintervals with widths 0.50.5, 11, 0.50.5 and 11, which add to 33.

Mesh of a partition

Definition The mesh, also called the norm, of a partition PP is the width of its widest subinterval,

∥P∥=max⁡1≤i≤nΔxi.\lVert P \rVert = \max_{1 \le i \le n} \Delta x_i.

In plain words The mesh measures how coarse the partition is. A small mesh guarantees that every piece is thin, which a large number of pieces alone does not.

Example For P={0,0.5,1.5,2,3}P = \{0, 0.5, 1.5, 2, 3\} the widths are 0.5,1,0.5,10.5, 1, 0.5, 1, so ∥P∥=1\lVert P \rVert = 1. For the partition of [0,1][0, 1] into nn equal parts the mesh is 1n\frac{1}{n}. The partition {0,0.001,0.002,1}\{0, 0.001, 0.002, 1\} has three pieces of which two are tiny, yet its mesh is 0.9980.998.

Tagged partitions and arbitrary sample points

Definition A tagged partition of [a,b][a, b] is a partition P={x0,…,xn}P = \{x_0, \ldots, x_n\} together with a choice of sample points, or tags, xi∗x_i^* with

xi−1≤xi∗≤xi(i=1,…,n).x_{i-1} \le x_i^* \le x_i \qquad (i = 1, \ldots, n).

Each tag may be any point of its subinterval, endpoints included. Left endpoints xi∗=xi−1x_i^* = x_{i-1}, right endpoints xi∗=xix_i^* = x_i and midpoints xi∗=xi−1+xi2x_i^* = \frac{x_{i-1} + x_i}{2} are common choices, but none is required.

In plain words In every piece you pick one point where you will read the height of the function. You are free to pick it anywhere in that piece.

Example On P={0,0.5,1.5,2,3}P = \{0, 0.5, 1.5, 2, 3\} the midpoints are the tags 0.25,1,1.75,2.50.25, 1, 1.75, 2.5. Choosing 0,1.5,1.5,30, 1.5, 1.5, 3 instead is also allowed, since 1.51.5 is an endpoint of both the second and third pieces.

Riemann sum

Definition Let ff be defined on [a,b][a, b] and let PP be a partition with tags xi∗x_i^*. The Riemann sum of ff for this tagged partition is

∑i=1nf(xi∗) Δxi.\sum_{i=1}^{n} f(x_i^*) \, \Delta x_i.

No sign condition on ff is needed. Terms where f(xi∗)<0f(x_i^*) < 0 are negative.

In plain words Multiply each sampled height by the width of its piece and add. When ff is nonnegative this is the total area of a row of rectangles, and when ff dips below the axis those rectangles count negatively.

Example Take f(x)=x2f(x) = x^2 on [0,3][0, 3] with P={0,0.5,1.5,2,3}P = \{0, 0.5, 1.5, 2, 3\} and midpoint tags.

PieceWidthTagf(tag)f(\text{tag})Product
[0,0.5][0, 0.5]0.50.50.250.250.06250.06250.031250.03125
[0.5,1.5][0.5, 1.5]11111111
[1.5,2][1.5, 2]0.50.51.751.753.06253.06251.531251.53125
[2,3][2, 3]112.52.56.256.256.256.25

The Riemann sum is 8.81258.8125. The exact area under x2x^2 on [0,3][0, 3] is 99, so even this coarse sum is close.

Lower and upper Darboux sums

Definition Let ff be bounded on [a,b][a, b] and let PP be a partition. On each subinterval put

mi=inf⁡xi−1≤x≤xif(x),Mi=sup⁡xi−1≤x≤xif(x).m_i = \inf_{x_{i-1} \le x \le x_i} f(x), \qquad M_i = \sup_{x_{i-1} \le x \le x_i} f(x).

The lower and upper Darboux sums are

L(f,P)=∑i=1nmi Δxi,U(f,P)=∑i=1nMi Δxi.L(f, P) = \sum_{i=1}^{n} m_i \, \Delta x_i, \qquad U(f, P) = \sum_{i=1}^{n} M_i \, \Delta x_i.

For every choice of tags, mi≤f(xi∗)≤Mim_i \le f(x_i^*) \le M_i, so every Riemann sum for PP lies between L(f,P)L(f, P) and U(f,P)U(f, P). When ff is continuous, mim_i and MiM_i are the minimum and maximum of ff on the piece.

In plain words The lower sum uses the lowest height on each piece and the upper sum the highest. Together they bracket every Riemann sum you could build on that partition.

Example For f(x)=x2f(x) = x^2 on P={0,0.5,1.5,2,3}P = \{0, 0.5, 1.5, 2, 3\}, ff is increasing, so mi=f(xi−1)m_i = f(x_{i-1}) and Mi=f(xi)M_i = f(x_i).

L(f,P)=0⋅0.5+0.25⋅1+2.25⋅0.5+4⋅1=5.375,L(f, P) = 0 \cdot 0.5 + 0.25 \cdot 1 + 2.25 \cdot 0.5 + 4 \cdot 1 = 5.375,U(f,P)=0.25⋅0.5+2.25⋅1+4⋅0.5+9⋅1=13.375.U(f, P) = 0.25 \cdot 0.5 + 2.25 \cdot 1 + 4 \cdot 0.5 + 9 \cdot 1 = 13.375.

The midpoint sum 8.81258.8125 sits between them, as it must.

Refinement

Definition A partition P′P' of [a,b][a, b] is a refinement of PP if P⊆P′P \subseteq P'. For bounded ff,

L(f,P)≤L(f,P′)≤U(f,P′)≤U(f,P).L(f, P) \le L(f, P') \le U(f, P') \le U(f, P).

Consequently, for any two partitions PP and QQ, L(f,P)≤U(f,Q)L(f, P) \le U(f, Q), which follows by comparing both with the common refinement P∪QP \cup Q.

In plain words Adding cut points can only raise the lower sum and lower the upper sum. Every lower sum is below every upper sum, no matter which partitions they come from.

Example Add the point 11 to the partition above, so P′={0,0.5,1,1.5,2,3}P' = \{0, 0.5, 1, 1.5, 2, 3\}. For f(x)=x2f(x) = x^2,

L(f,P′)=0⋅0.5+0.25⋅0.5+1⋅0.5+2.25⋅0.5+4⋅1=5.75,L(f, P') = 0 \cdot 0.5 + 0.25 \cdot 0.5 + 1 \cdot 0.5 + 2.25 \cdot 0.5 + 4 \cdot 1 = 5.75,U(f,P′)=0.25⋅0.5+1⋅0.5+2.25⋅0.5+4⋅0.5+9⋅1=12.75.U(f, P') = 0.25 \cdot 0.5 + 1 \cdot 0.5 + 2.25 \cdot 0.5 + 4 \cdot 0.5 + 9 \cdot 1 = 12.75.

The lower sum rose from 5.3755.375 and the upper sum fell from 13.37513.375.

Why the choice of sample points stops mattering

Definition Suppose ff is bounded on [a,b][a, b] and there is a sequence of partitions PkP_k with U(f,Pk)−L(f,Pk)→0U(f, P_k) - L(f, P_k) \to 0 and both sums tending to a number II. Then the Riemann sums on PkP_k tend to II as well, however the tags are chosen, and the choice may change from one kk to the next. The reason is that each such sum lies between L(f,Pk)L(f, P_k) and U(f,Pk)U(f, P_k), so the squeeze theorem applies.

In plain words Once the lower and upper sums squeeze together, any sum you build in between is squeezed too. Left, right, midpoint or a random point all give the same limit.

Example Let f(x)=xf(x) = x on [0,1][0, 1] with nn equal pieces. The lower sum uses left endpoints and the upper sum right endpoints,

Ln=1n2∑i=1n(i−1)=n−12n,Un=1n2∑i=1ni=n+12n.L_n = \frac{1}{n^2} \sum_{i=1}^{n} (i - 1) = \frac{n - 1}{2n}, \qquad U_n = \frac{1}{n^2} \sum_{i=1}^{n} i = \frac{n + 1}{2n}.

Both tend to 12\frac{1}{2}, so every Riemann sum on these partitions, with tags placed anywhere, tends to 12\frac{1}{2}.

Check yourself

  1. Find the mesh of the partition {1,1.2,2,2.5,4}\{1, 1.2, 2, 2.5, 4\} of [1,4][1, 4].
Show the answer

The widths are 0.2,0.8,0.5,1.50.2, 0.8, 0.5, 1.5, so the mesh is 1.51.5.

  1. For f(x)=2x−1f(x) = 2x - 1 on [0,2][0, 2] with the partition {0,1,2}\{0, 1, 2\}, compute the Riemann sum with right endpoint tags and with left endpoint tags.
Show the answer

Right endpoints give f(1)⋅1+f(2)⋅1=1+3=4f(1) \cdot 1 + f(2) \cdot 1 = 1 + 3 = 4. Left endpoints give f(0)⋅1+f(1)⋅1=−1+1=0f(0) \cdot 1 + f(1) \cdot 1 = -1 + 1 = 0. Since ff is increasing these are the upper and lower Darboux sums.

  1. For a bounded ff and a partition PP, can a Riemann sum on PP ever be larger than U(f,P)U(f, P)?
Show the answer

No. Each tag satisfies f(xi∗)≤Mif(x_i^*) \le M_i, and multiplying by Δxi>0\Delta x_i > 0 and adding gives a Riemann sum at most U(f,P)U(f, P).

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