Interpret a numerical integral in context

A wind turbine's energy

A wind turbine's power in kilowatts is read every 2 hours. Energy is power times time, so the energy produced is the area under the power graph, in kilowatt-hours.

The power only rises, so the left sum is too small and the right sum is too large. They are \(2(12 + 18 + 25 + 31) = 172\) and \(2(18 + 25 + 31 + 34) = 216\), so the energy is between 172 and 216 kilowatt-hours.

Read every hour. The sums become 184 and 206. The gap between the sums is the total rise, 22, times the width of a rectangle, so halving the width halves the range.

An estimate with a range

The teal rectangles make the lower sum, and with the faint extra height on each one they make the upper sum. Below, the teal bar runs from one sum to the other, and the chalk tick inside it is the true value.

Move the slider to add readings. The bar narrows, but the true value stays inside it, so the estimate halfway between the sums becomes more reliable.

A good report gives the estimate with its unit and its range, and no more digits than the range supports.

Notation

1. The function inside the integral is a rate, and the integral is the amount it accumulates, here the energy. Kilowatts times hours gives kilowatt-hours.

   \[\begin{aligned} &\int_0^8 \cy{P(t)}\,dt \\ &\text{kilowatts}\times\text{hours} \\ &\quad= \text{kilowatt-hours} \end{aligned}\]

2. If the rate only rises or only falls, one sum is a lower estimate and the other an upper estimate. The integral lies between them.

   \[172 \le \cb{\int_0^8 P(t)\,dt} \le 216\]

3. The estimate is halfway between the two sums, which is the trapezoidal sum. The true value can be anywhere in the range.

   \[\tfrac12\,(172 + 216) = \cy{194}\]

4. Round the estimate to the precision that the data support. A digit much smaller than the width of the range has no meaning.

When it fails

A report of 194.0 kilowatt-hours claims that the value is known to a tenth, and the small orange mark shows how narrow that claim is.

The teal bar runs from 172 to 216, so the ones and tenths digits have no meaning. Report about 190 kilowatt-hours, or give the range.

Example 1: Traffic on a road

A sensor reads the traffic rate, in vehicles per minute, every 10 minutes for an hour. Use the trapezoidal rule to estimate the number of vehicles in the hour.\[\small\begin{array}{c|ccccccc} t & 0 & 10 & 20 & 30 & 40 & 50 & 60 \\ \hline r(t) & 20 & 26 & 30 & 34 & 32 & 28 & 22 \end{array}\]

1. The integral of the rate is the number of vehicles, because vehicles per minute times minutes gives vehicles.

2. Each width is 10 minutes. Count each interior reading twice, because it is a side of two trapezoids, and each end reading once.

   \[\begin{aligned} T &= \frac{\cg{10}}{2}\Big[\cy{20} + 2(\cy{26} + \cy{30} \\ &\qquad + \cy{34} + \cy{32} + \cy{28}) + \cy{22}\Big] \\ &= 1710 \end{aligned}\]

3. The data do not show how the rate changes between readings, so the last digit of 1710 is not reliable. Report about 1700 vehicles.

Example 2: A pump

A pump delivers water at 30, 34, 38 and 40 liters per second at \(t = 0, 2, 4\) and 6 minutes. The rate only rises. Estimate the volume delivered in 6 minutes.

1. The rate only rises, so the left sum is a lower estimate and the right sum an upper one. Each rectangle is 2 minutes wide.

2. A tempting wrong move: Multiply each sum of heights by the width, 2.

   \[\begin{aligned} L &= 2(30 + 34 + 38) = 204 \\ R &= 2(34 + 38 + 40) = 224 \end{aligned}\]

   The pump delivers at least 30 liters every second, so 6 minutes give at least 10,800 liters, and 204 liters is much too small. The rate is in liters per second, so the width must be in seconds.

3. The rate is per second, so convert the width: 2 minutes is 120 seconds. The sums of the heights, 102 and 112, do not change.

   \[\begin{aligned} L &= \cg{120}(102) = 12\,240 \\ R &= \cg{120}(112) = 13\,440 \end{aligned}\]

4. The volume is between 12,240 and 13,440 liters, and the estimate is the average of the two sums.

   \[\tfrac12(12\,240 + 13\,440) = 12\,840\]

5. The range is 1,200 liters wide, so the last two digits of the estimate have no meaning. Report about 12,800 liters.

Practice

A phone's data rate is 8, 12, 14 and 15 megabytes per second at \(t = 0, 10, 20\) and 30 seconds. The rate only rises. Give a lower and an upper estimate of the data transferred, with units, and one estimate with its range.\[\begin{array}{c|cccc} t \text{ (s)} & 0 & 10 & 20 & 30 \\ \hline r(t) & 8 & 12 & 14 & 15 \end{array}\]

The rate rises, so the left rectangles stay below the line and give the lower estimate. The right rectangles give the upper one.

Questions

The figure shows a pipe's leak rate, in liters per minute, at \(t = 0, 3, 6, 9\) and 12 minutes: 10, 7, 5, 4 and 3. The rate only falls.

1. The rectangles of the left sum are shown. Which statement is correct?

   • The left sum is an upper estimate of the volume that leaks.
   • The left sum is a lower estimate of the volume that leaks.
   • The left sum is exact, because the readings are exact.
   Answer

   The left sum is an upper estimate of the volume that leaks.

2. What is the unit of the area of each rectangle in the figure?

   • Liters
   • Liters per minute
   • Minutes
   Answer

   Liters

3. A gas flows at 12, 16, 20 and 18 cubic meters per hour at \(t = 0, 2, 4\) and 6 hours. What is the trapezoidal estimate of the volume?

   • 102 cubic meters
   • 102 cubic meters per hour
   • 204 cubic meters
   Answer

   102 cubic meters

Summary

You interpret a numerical integral by giving the estimate with its unit and saying how far it can be trusted.

The unit comes from the rate times the time, and the left and right sums bracket the true value.

You use this when a problem asks what an estimate means, or how many digits to report.

This is one of five skills in topic 6.2 of Which & Why: AP Calculus.