Approximating area with Riemann sums

Day 1: Leonor merges onto the highway

Leonor speeds up along an on-ramp. Her speedometer readings, converted to meters per second, follow \displaystyle v(t) = 10 + \frac{t^2}{4} for 0 \le t \le 8, with t in seconds. Estimate the distance she covers with L_4, R_4 and M_4, and say what the first two guarantee.

On a short time interval the velocity is nearly constant, and distance at constant velocity is velocity times time, the area of a rectangle on the graph. So distance is the area under the velocity graph, and each sum estimates it. The units agree: meters per second times seconds is meters.

Here \displaystyle \Delta t = \frac{8 - 0}{4} = 2, the cut points are 0, 2, 4, 6, 8, and the midpoints are 1, 3, 5, 7. At the cut points v is 10, 11, 14, 19, 26, and at the midpoints it is 10.25, 12.25, 16.25, 22.25.

The left sum uses the readings at 0, 2, 4, 6; the right sum those at 2, 4, 6, 8; the midpoint sum those at 1, 3, 5, 7. Each has the common width 2 factored out:

Since \displaystyle v'(t) = \frac{t}{2} > 0 for 0 < t \le 8, the velocity is increasing (Increasing and decreasing from the derivative), so the left sum is under the distance and the right sum over it: Leonor covers between 108 and 140 meters. The midpoint estimate, 122 meters, sits inside that bracket; Example 3 shows how close it is.

Day 1: Duarte brakes to a stop

Duarte brakes his car evenly: v(t) = 20 - 2.5t meters per second for 0 \le t \le 8. Find L_4, R_4 and M_4 and compare each with the exact stopping distance.

Again \Delta t = 2. The readings at 0, 2, 4, 6, 8 are 20, 15, 10, 5, 0, and at the midpoints 1, 3, 5, 7 they are 17.5, 12.5, 7.5, 2.5.

Here the region is a triangle with base 8 and height 20, so the exact distance is \displaystyle \frac{1}{2}(8)(20) = 80 meters (Area of triangles).

The velocity falls, so the order reverses: each left rectangle uses the largest reading in its piece and overestimates, each right rectangle underestimates, and 60 < 80 < 100. The midpoint sum is exact. On each piece the line crosses the top of the midpoint rectangle at its middle. The gap under the line on the left half and the corner of the rectangle above the line on the right half are right triangles with the same legs, 1 second and 2.5 meters per second, so the rectangle gains exactly what it loses.

Day 1: sigma notation and more rectangles for Leonor

Write L_n, R_n and M_n for Leonor’s \displaystyle v(t) = 10 + \frac{t^2}{4} on [0, 8] in sigma notation. Evaluate them for n = 4, 8, 16, 100 and compare with the exact distance, \displaystyle \frac{368}{3} \approx 122.67 meters. Take that number on trust for now: nothing in this lesson can produce it, and The fundamental theorem of calculus will show how to find it.

With n pieces, \displaystyle \Delta t = \frac{8}{n} and \displaystyle t_k = \frac{8k}{n}, so \displaystyle v(t_k) = 10 + \frac{1}{4}\cdot\frac{64k^2}{n^2} = 10 + \frac{16k^2}{n^2}. The midpoint of piece k is \displaystyle m_k = \frac{8k - 4}{n}, and \displaystyle v(m_k) = 10 + \frac{4(2k - 1)^2}{n^2}. Each term is width times height:

For n = 4 the midpoint terms are \displaystyle 2\left(10 + \frac{(2k-1)^2}{4}\right) = 20.5, 24.5, 32.5, 44.5, which add to 122, as in Example 1. Adding the terms by computer for larger n:

The errors, rounded:

Every error shrinks as n grows. Doubling n roughly halves the left and right errors and cuts the midpoint error by a factor of 4. The bracket closes at a known rate: \displaystyle R_n - L_n = \frac{8}{n}\big(v(8) - v(0)\big) = \frac{128}{n}, which is 1.28 meters at n = 100.

Day 1: reading a sum back into rectangles

Describe the rectangles behind \displaystyle \displaystyle\sum_{k=1}^{6} \frac{1}{3}\sqrt{1 + \frac{k}{3}}: the function, the interval, the number of pieces and the kind of sum. Does it overestimate or underestimate the area, and what is it?

The factor \displaystyle \frac{1}{3} multiplying every term is the width, and there are six terms, so n = 6 and the interval has length \displaystyle 6 \cdot \frac{1}{3} = 2. The heights are \displaystyle \sqrt{x} at \displaystyle x = 1 + \frac{k}{3}, which runs \displaystyle \frac{4}{3}, \frac{5}{3}, \dots, 3 as k goes from 1 to 6. These are the right ends of the pieces of [1, 3], because k = 0, the point x = 1, is left out. So the sum is R_6 for \displaystyle f(x) = \sqrt{x} on [1, 3].

The square root is increasing, so the right sum overestimates the area. Adding the six terms gives about 2.917.

The same six numbers are also R_6 for \displaystyle \sqrt{1 + x} on [0, 2], the same rectangles slid one unit left. A sum fixes the areas of the rectangles, not the picture.

Day 2: Afonso’s bike computer

Afonso’s bike computer logs his velocity at uneven times as he speeds up along a flat road. He pushes harder the whole minute, so his acceleration v'(t) is positive throughout.

Estimate the distance with a left and a right sum, and bracket it.

The pieces have widths 10, 15, 5, 20, 10, and each rectangle takes its own width.

A positive derivative makes v increasing (Increasing and decreasing from the derivative), so on each piece the left reading is the lowest and the right reading the highest. Afonso rode between 375 and 455 meters. The table has no readings at the midpoints of the pieces, so no midpoint sum can be formed. Averaging the two sums, 415, is the idea behind the trapezoidal rule, a later lesson.

Day 2: snowmelt into a reservoir

On a spring day, snowmelt runs into a mountain reservoir at a rate r(t), in thousands of cubic meters per hour, t hours after 8 a.m. A hydrologist’s model has r''(t) < 0 for the whole day.

Estimate the water that ran in over the eight hours with M_4. Is it an overestimate or an underestimate, and do the left and right sums come with a guarantee?

Four pieces of width 2 are [0, 2], [2, 4], [4, 6], [6, 8], with midpoints 1, 3, 5, 7, all in the table:

The units are thousands of cubic meters per hour times hours, so about 37{,}600 cubic meters. Since r'' < 0, the rate is concave down, and a midpoint sum for a concave down function overestimates. The true total is less than 37.6 thousand cubic meters.

The rate rises and then falls, so r is not monotone on [0, 8], and the bracketing result does not apply to the whole interval: neither L_4 nor R_4 is known to be over or under. Concavity decides the midpoint sum where monotonicity decides nothing.

Day 2: a Riemann sum with mixed sample points

For f(x) = 16 - x^2 on [0, 4], take the partition 0, 1, 2, 3.5, 4 and the sample points c_1 = 0.5, c_2 = 2, c_3 = 3, c_4 = 4. Find the Riemann sum, then the smallest and largest Riemann sums possible on this partition.

The widths are 1, 1, 1.5, 0.5. Each sample point lies in its piece; c_2 = 2 and c_4 = 4 are right ends, which is allowed.

Since f'(x) = -2x < 0 for x > 0, f is decreasing on [0, 4], so on each piece its largest value is at the left end and its smallest at the right end. The largest sum uses f(0), f(1), f(2), f(3.5) = 16, 15, 12, 3.75, and the smallest uses f(1), f(2), f(3.5), f(4) = 15, 12, 3.75, 0:

Every choice of sample points gives a sum between 32.625 and 50.875, and 38.25 is one of them. Refining the partition squeezes that range; the one number all Riemann sums close in on is the subject of The definite integral as a limit of Riemann sums.