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6.3 Riemann Sums, Summation Notation, and Definite Integral Notation

Syllabus
2020
Topic
6.3
Level
—

Learning objectives

—LIM-5.B—Interpret the limiting case of the Riemann sum as a definite integral• LIM-5.B Interpret the limiting case of the Riemann sum as a definite integral.• LIM-5.B.1 The limit of an approximating Riemann sum can be interpreted as a definite integral.• LIM-5.B.2 A Riemann sum, which requires a partition of an interval I, is the sum of products, each of which is the value of the function at a point in a subinterval multiplied by the length of that subinterval of the partition.• Enduring understanding LIM-5: Definite integrals can be approximated using geometric and numerical methods.—LIM-5.C—Represent the limiting case of the Riemann sum as a definite integral• LIM-5.C Represent the limiting case of the Riemann sum as a definite integral.• LIM-5.C.1 The definite integral of a continuous function f over the interval [a, b], denoted by ∫ₐᵇ f(x)dx, is the limit of Riemann sums as the widths of the subintervals approach 0. That is, ∫ₐᵇ f(x)dx = lim(max Δxᵢ→0) Σ(i=1 to n) f(xᵢ*)Δxᵢ, where n is the number of subintervals, Δxᵢ is the width of the ith subinterval, and xᵢ* is a value in the ith subinterval.• LIM-5.C.2 A definite integral can be translated into the limit of a related Riemann sum, and the limit of a Riemann sum can be written as a definite integral.• Enduring understanding LIM-5: Definite integrals can be approximated using geometric and numerical methods.

From Riemann Sums to an Exact Accumulation

A Riemann sum approximates accumulated change on [a,b][a,b] by adding contributions from a partition a=x0<x1<⋯<xn=ba=x_0<x_1<\cdots<x_n=b. In subinterval ii, the product f(xi∗)Δxif(x_i^*)\Delta x_i uses a sampled function value as the rectangle height and Δxi=xi−xi−1\Delta x_i=x_i-x_{i-1} as its width.

\int_a^b f(x),dx=\lim_{\max \Delta x_i\to 0}\sum_{i=1}^{n} f(x_i^*)\Delta x_i

Refining the partition makes every rectangle narrower, so the sum follows the changing function more closely. The condition max⁡Δxi→0\max \Delta x_i\to0 matters: it prevents even one subinterval from remaining wide. For a continuous function, different valid choices of sample points approach the same definite integral.

For a positive rate r(t)r(t) on [0,4][0,4], a sum ∑r(ti∗)Δti\sum r(t_i^*)\Delta t_i adds approximate changes over short time intervals. As the largest time width approaches zero, the approximation approaches ∫04r(t) dt\int_0^4 r(t)\,dt, the exact accumulated change over those four time units.

The integral is a limit of signed contributions, not automatically ordinary geometric area. Terms are negative where f(xi∗)<0f(x_i^*)<0, so regions below the axis reduce the accumulated value.

Translate Between a Riemann Limit and an Integral

To translate notation, match four pieces: the interval [a,b][a,b], the width Δx\Delta x, the sample point xi∗x_i^*, and the function evaluated there. For nn equal subintervals, Δx=(b−a)/n\Delta x=(b-a)/n; a right-endpoint sample is xi=a+iΔxx_i=a+i\Delta x.

In lim⁡n→∞∑i=1n(1+3i/n)2(3/n)\lim_{n\to\infty}\sum_{i=1}^n(1+3i/n)^2(3/n), the width is 3/n3/n, so b−a=3b-a=3. The sample point is 1+3i/n1+3i/n, so a=1a=1 and b=4b=4. The squared factor is f(xi)f(x_i). Therefore the limit is ∫14x2 dx\int_1^4 x^2\,dx.

For ∫021+x3 dx\int_0^2\sqrt{1+x^3}\,dx, use Δx=2/n\Delta x=2/n and right endpoints xi=2i/nx_i=2i/n. An equivalent limit is lim⁡n→∞∑i=1n1+(2i/n)3(2/n)\lim_{n\to\infty}\sum_{i=1}^n\sqrt{1+(2i/n)^3}(2/n). The factor outside the function is the subinterval width.

Do not infer the upper endpoint from Δx\Delta x alone; combine the width with the sample-point formula. Also distinguish i/ni/n, which locates a sample point, from the separate width factor that multiplies every function value.