Module 1 · Lesson 1

Calculus: Instantaneous speed and areas under curves: limits, derivatives, and integrals

You'll take on two puzzles that troubled mathematicians for two thousand years: what speed means at a single instant, and how to find the area of a curved shape. You'll estimate both from tables of numbers and see why they turn out to be the same idea run in opposite directions.

What you will be able to do

Estimate a rate of change and an accumulated total from a table of values, and explain how the two are linked.

0% of this lesson · 9 items · 1h 3m total · 48m without the optional problem set

Contents of this lesson9 items

ReadingWhat Does Your Speedometer Mean?3 min
ReadingFrom Archimedes to Newton and Leibniz: How Calculus Took Shape4 min
ReadingThe Derivative: From Average Rate to Instantaneous Rate4 min
ReadingThe Integral: Adding Up Small Pieces4 min
ReadingThe Fundamental Theorem of Calculus: Rates and Totals Undo Each Other4 min
Guided ReadingWorked Example: Distance and Speed from a Table, Both Ways7 min
Guided ConversationEstimate a Rate and a Total from Your Own Numbers12 min
Problem Set · optionalRates and Totals from Tables15 min
Knowledge CheckInstantaneous speed and areas under curves: limits, derivatives, and integrals10 min

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Reading 3 min Done

What Does Your Speedometer Mean?

Glance at a car's speedometer and it shows one number, say 50 kilometers per hour. That number is meant to describe this moment, not the last hour. But speed is distance divided by time. At a single instant no time passes and the car covers no distance, so the division would be zero divided by zero. What, then, is the speedometer reporting?

Try it with numbers before reading on. Suppose a car's distance from a starting line, in meters, is the square of the time in seconds. In symbols, $d = t^2$ , where $d$ is distance and $t$ is time. At 3 seconds the car is 9 meters out. At 4 seconds it's 16 meters out. Its average speed over that second is the change in distance divided by the change in time: 7 meters in 1 second, or 7 meters per second.

Now shrink the interval. Between 3 and 3.1 seconds the car goes from 9 meters to 9.61 meters. That's 0.61 meters in 0.1 seconds, an average of 6.1 meters per second. Between 3 and 3.01 seconds the average is 6.01 meters per second. The shorter the interval, the closer the averages come to one number:

1 second: 7 meters per second
0.1 second: 6.1 meters per second
0.01 second: 6.01 meters per second

No interval in the list gives exactly 6. Yet the averages are heading there, and 6 meters per second is the natural answer to "how fast at the instant $t = 3$ ?" Calculus makes that answer precise. It defines the speed at an instant as the number that average speeds approach as the time interval shrinks (Strang and Herman 2016a, sec. 3.4). A speedometer, in effect, reports that number.

A second puzzle looks unrelated. Finding the area of a rectangle is easy: multiply width by height. Finding the area of a shape with a curved edge is not. No formula for rectangles fits a curve, and cutting the shape into rectangles always leaves slivers over or under the edge. For about two thousand years, mathematicians attacked such areas one shape at a time, with ingenious arguments that didn't carry over to the next shape.

Calculus is the branch of mathematics that answers both puzzles with one set of ideas. One idea handles rates: how fast something changes at an instant. The other handles totals: how much builds up when many small pieces are added together. The surprise, which took centuries to see, is that each idea undoes the other. Finding a speed and finding a distance turn out to be the same problem run in opposite directions.

That link is why the subject reaches so far. Wherever something changes, whether a position, a price, a population, or a temperature, the same pair of questions comes up: how fast is it changing, and how much has it added up to?

References

Strang, Gilbert, and Edwin "Jed" Herman. 2016a. Calculus Volume 1. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 1

Reading 4 min Done

From Archimedes to Newton and Leibniz: How Calculus Took Shape

Introduction

Calculus is often credited to two men working between the 1660s and the 1680s. The ideas behind it are much older, and they came from several places. This reading traces how methods for areas, series, and tangents developed over two thousand years before Isaac Newton and Gottfried Wilhelm Leibniz turned them into a general method, and how the two men's claims ended in a dispute.

Areas and Series Before the 1600s

Around 225 BCE, the Greek mathematician Archimedes found the area of a parabolic segment, the region cut off from a parabola by a straight line. He showed that it is four-thirds of the triangle with the same base and the same height (Archimedes 1897, Quadrature of the Parabola, prop. 24). His argument fills the curved region with triangles, each new batch having one-quarter the area of the last. In modern terms, he summed $1 + \tfrac14 + \tfrac{1}{16} + \ldots$ and got $\tfrac43$ . Historians count this as the first known summation of an infinite series, a sum with endlessly many terms that settles on a finite value (O'Connor and Robertson 1996).

Archimedes never added infinitely many terms outright. He used the method of exhaustion: fill a region with shapes of known area until the gap is smaller than any amount you name, then show by contradiction that the area can be neither more nor less than the claimed value. The idea that a curved quantity can be pinned down by approximations that get as close as you like is the seed of calculus.

Much later and far away, Madhava of Sangamagrama, in Kerala in southern India, found infinite series for the sine, cosine, and arctangent around 1400 (O'Connor and Robertson 2000b). These series express quantities from trigonometry, the mathematics of angles, as endless sums of simpler terms. Madhava's own writings are lost. His results are known through later Kerala texts, such as Jyesthadeva's Yukti-Bhasa of about 1550. European mathematicians found the same series in the 1600s. Whether any of the Kerala work reached Europe is unknown and contested.

Tangents, Fluxions, and Notation

In the 17th century, Pierre de Fermat found where a curve reaches its highest or lowest points by asking where its tangent, the straight line that just touches the curve, is level. Isaac Barrow treated a tangent as the limit of lines through two nearby points on a curve. He worked toward a link between tangents and areas, though he never stated it outright (O'Connor and Robertson 1996).

Newton wrote a tract on what he called fluxions in October 1666, and didn't publish it at the time. A fluxion was Newton's term for the rate at which a "flowing" quantity changes. Leibniz, working on his own, had settled on his notation by 1675, including the long-S sign still used for areas. He published his methods in 1684 and 1686 (O'Connor and Robertson 1996; O'Connor and Robertson 1998a).

Who	When	Contribution
Archimedes	About 225 BCE	Area of a parabolic segment, by an argument that amounts to summing an infinite series
Madhava of Sangamagrama	About 1400	Infinite series for sine, cosine, and arctangent
Pierre de Fermat	17th century	Highest and lowest points found where the tangent is level
Isaac Barrow	17th century	Tangent as a limit; work toward linking tangents and areas
Isaac Newton	1666	Unpublished tract on fluxions
Gottfried Wilhelm Leibniz	1675, 1684, 1686	Notation, then the first published papers on the method

The Priority Dispute

Newton's work circulated privately, and Leibniz published first. In 1711 Leibniz read an accusation of plagiarism in the journal of the Royal Society in London, and he asked the Society for a retraction. The Society appointed a committee, and its 1713 report found for Newton. The report was written by Newton himself, though it didn't appear under his name (O'Connor and Robertson 1998a; O'Connor and Robertson 2000a).

This priority dispute, a quarrel over who reached a result first and deserves credit for it, was settled in Newton's favor by a body he controlled. Historians now treat the two discoveries as independent: Newton got there first, and Leibniz found his own method several years later (O'Connor and Robertson 2000a).

Conclusion

Calculus grew from many sources: Greek area arguments, Kerala series, and 17th-century work on tangents. Newton and Leibniz each turned these threads into a general method, independently. The bitter dispute over credit was decided by a report Newton wrote, and the question of Kerala influence on Europe remains open.

Key Terms

Infinite series: A sum with endlessly many terms that settles on a finite value.
Method of exhaustion: The Greek technique of filling a region with shapes of known area until the gap is smaller than any amount named, then proving the area by contradiction.
Fluxion: Newton's term for the rate at which a "flowing" quantity changes.
Priority dispute: A quarrel over who reached a result first and deserves credit for it.

References

Archimedes. 1897. The Works of Archimedes. Edited and translated by T. L. Heath. Cambridge: Cambridge University Press.
- Free: Internet Archive
O'Connor, J. J., and E. F. Robertson. 1996. "A History of the Calculus." MacTutor History of Mathematics Archive, University of St Andrews. Last updated February 1996.
- Free: MacTutor: A history of the calculus
O'Connor, J. J., and E. F. Robertson. 1998a. "Gottfried Wilhelm von Leibniz." MacTutor History of Mathematics Archive, University of St Andrews. Last updated October 1998.
- Free: MacTutor: Leibniz
O'Connor, J. J., and E. F. Robertson. 2000a. "Isaac Newton." MacTutor History of Mathematics Archive, University of St Andrews. Last updated January 2000.
- Free: MacTutor: Newton
O'Connor, J. J., and E. F. Robertson. 2000b. "Madhava of Sangamagrama." MacTutor History of Mathematics Archive, University of St Andrews. Last updated November 2000.
- Free: MacTutor: Madhava

Reading 4 min Done

The Derivative: From Average Rate to Instantaneous Rate

Introduction

A weather app says the temperature is rising 2 degrees an hour. A news report says a company's sales are growing fastest right now. Both describe how fast something changes at one moment, and that idea needs care. This reading explains average and instantaneous rates of change, the limit that connects them, the derivative, and how the idea was put on a rigorous footing.

Average Rates

Start with what you can measure. The average rate of change of a quantity over an interval is the change in the quantity divided by the change in time. If a tank holds 20 liters at 2 p.m. and 50 liters at 5 p.m., it gained 30 liters in 3 hours, an average of 10 liters per hour. Speed is the most familiar example: distance covered divided by time taken.

An average rate hides what happens inside the interval. The tank may have filled fast at first and then slowly. To describe a single moment, you need a different idea.

Shrinking the Interval

Take a car whose distance from a starting line, in meters, is the square of the time in seconds: $d = t^2$ , where $d$ is distance and $t$ is time. Here are its average speeds over intervals that start at $t = 3$ and get shorter (Strang and Herman 2016a, sec. 3.4).

Interval	Change in distance	Change in time	Average speed
3 to 4 s	16 − 9 = 7 m	1 s	7 m/s
3 to 3.1 s	9.61 − 9 = 0.61 m	0.1 s	6.1 m/s
3 to 3.01 s	9.0601 − 9 = 0.0601 m	0.01 s	6.01 m/s

The averages close in on 6 meters per second. No interval in the table produces exactly 6, and no interval of any length would. But 6 is the number they approach as the interval shrinks toward zero.

That number is the instantaneous rate of change: the rate at a single moment, defined as the value that average rates approach as the interval shrinks. The approach itself has a name. A limit is the number a sequence of values gets closer and closer to, as close as you like, even if no value in the sequence reaches it. So the car's speed at $t = 3$ is the limit of its average speeds, 6 meters per second.

The Derivative and the Slope

The derivative of a quantity is its instantaneous rate of change: the limit of its average rates as the time interval shrinks toward zero. Mathematicians often write it as $\frac{dy}{dt}$ , read as "the rate of change of $y$ with respect to $t$ ." You won't need the notation. The idea is what matters: a derivative is a rate at an instant, found as the end point of a sequence of averages.

The derivative also has a picture. On a graph of distance against time, an average speed is the slope of the straight line joining two points on the curve. The slope of a line is its steepness: how much it rises for each unit it runs to the right. As the two points move together, those lines settle toward the line that just touches the curve at one point. The derivative is that line's slope. A steep curve means a large rate, a flat curve a rate near zero, and a falling curve a negative rate.

Making Limits Rigorous

Newton and Leibniz used these ideas with great success, but they couldn't say exactly what a quantity that is "almost zero" was. In 1734 the philosopher George Berkeley attacked the lack of rigor in the calculus and disputed its logic, and mathematicians spent much of the following century trying to tighten the reasoning (O'Connor and Robertson 1996).

The firm basis came in the 1800s. Augustin Louis Cauchy's 1821 textbook set out to develop the basic theorems of calculus as rigorously as possible (O'Connor and Robertson 1997). Karl Weierstrass, lecturing in Berlin from the late 1850s, took on the foundations again and set standards of rigor that shaped later mathematics (O'Connor and Robertson 1998b). Their work turned "the number the averages approach" into a precise definition, and that answered the old complaint.

Conclusion

An average rate divides a change by an interval; an instantaneous rate is the limit those averages approach as the interval shrinks. The derivative names that limit, and on a graph it is the slope of the curve at a point. The logic that early critics found unclear was made rigorous in the 19th century.

Key Terms

Average rate of change: The change in a quantity divided by the change in time over an interval.
Instantaneous rate: The rate of change at a single moment, defined as the value that average rates approach as the interval shrinks.
Limit: The number a sequence of values gets closer and closer to, as close as you like, even if no value in the sequence reaches it.
Derivative: A quantity's instantaneous rate of change: the limit of its average rates as the time interval shrinks toward zero.
Slope: The steepness of a line: how much it rises for each unit it runs to the right.

References

O'Connor, J. J., and E. F. Robertson. 1996. "A History of the Calculus." MacTutor History of Mathematics Archive, University of St Andrews. Last updated February 1996.
- Free: MacTutor: A history of the calculus
O'Connor, J. J., and E. F. Robertson. 1997. "Augustin Louis Cauchy." MacTutor History of Mathematics Archive, University of St Andrews. Last updated January 1997.
- Free: MacTutor: Cauchy
O'Connor, J. J., and E. F. Robertson. 1998b. "Karl Theodor Wilhelm Weierstrass." MacTutor History of Mathematics Archive, University of St Andrews. Last updated October 1998.
- Free: MacTutor: Weierstrass
Strang, Gilbert, and Edwin "Jed" Herman. 2016a. Calculus Volume 1. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 1

Reading 4 min Done

The Integral: Adding Up Small Pieces

Introduction

A fitness watch reports how far you ran, though it never measured the distance directly. It sampled your speed again and again and added up the pieces. This reading explains the integral, the calculus idea behind that kind of total: how to estimate a total with rectangles, why thinner rectangles do better, and where the idea is used beyond area.

Totals from a Changing Rate

If you drive at a steady 20 meters per second for 10 seconds, you cover 200 meters: speed times time. When speed changes, that single multiplication fails. The fix is to cut the time into short stretches. Over each stretch, the speed barely changes, so speed times time gives a good estimate for that stretch. Add the stretches and you have an estimate of the whole distance.

This kind of building up has a general name. Accumulation is the total that builds up when a changing rate acts over time, found by adding the contributions of many short stretches. Distance accumulates from speed, water in a tank accumulates from a flow rate, and a bill accumulates from a price per hour.

Rectangles Under a Curve

Accumulation has a picture. Plot speed against time. Over a short stretch, speed times time is the area of a thin rectangle: its height is the speed, its width the time. So the distance traveled is the area under a curve: the area of the region between the graph of a rate and the time axis.

Here is a car whose speed rises, but less and less quickly, with a reading every 10 seconds.

Time (s)	0	10	20	30	40
Speed (m/s)	0	6	10	12	13

A rectangle approximation estimates an area under a curve by covering it with rectangles and adding their areas (Strang and Herman 2016a, sec. 5.1). There are two natural choices for each rectangle's height.

Use the speed at the start of each stretch: $(0 + 6 + 10 + 12) \times 10 = 280$ meters. The speed was rising during every stretch, so this undercounts.
Use the speed at the end of each stretch: $(6 + 10 + 12 + 13) \times 10 = 410$ meters. This overcounts for the same reason.

The true distance lies somewhere between 280 and 410 meters. One estimate is too low and the other too high.

Thinner Rectangles and the Integral

To narrow the gap, take readings more often. With a reading every second, giving 40 one-second stretches instead of 4 ten-second ones, each rectangle hugs the curve more closely, and the low and high estimates move toward each other. The error in each rectangle is the thin sliver between its top and the curve. Thinner rectangles leave thinner slivers, so the total error shrinks. Using smaller and smaller rectangles gives closer and closer approximations, and taking the limit gives the exact area (Strang and Herman 2016a, sec. 5.1).

The integral is the exact total that rectangle estimates approach as the rectangles get thinner and thinner. It is the accumulated amount, with no approximation left. Leibniz chose a symbol for it, $\int$ , an elongated S that suggests a sum (Strang and Herman 2016a, sec. 5.2). He first published his results on these sums in the 1680s under the name calculus summatorius, the calculus of sums (O'Connor and Robertson 1996).

Beyond Area

The integral is defined through area, but it is used for any total built from a changing rate. Engineers total the water that passes through a pipe from a flow rate that rises and falls. Utilities total the energy a building uses from its power draw, which changes through the day. A fitness watch estimating a run's distance from sampled speeds is doing the same thing on a small scale. In each case the method is the same: slice time finely, multiply rate by slice, add, and take the limit.

Conclusion

When a rate changes, you can estimate the total by cutting time into short stretches and adding rate times time. On a graph, that is a rectangle approximation of an area, too low with one choice of heights and too high with another. The integral is the value those estimates approach as the rectangles get thinner.

Key Terms

Accumulation: The total that builds up when a changing rate acts over time, found by adding the contributions of many short stretches.
Area under a curve: The area of the region between the graph of a rate and the time axis.
Rectangle approximation: An estimate of an area under a curve made by covering it with rectangles and adding their areas.
Integral: The exact total that rectangle estimates approach as the rectangles get thinner and thinner.

References

O'Connor, J. J., and E. F. Robertson. 1996. "A History of the Calculus." MacTutor History of Mathematics Archive, University of St Andrews. Last updated February 1996.
- Free: MacTutor: A history of the calculus
Strang, Gilbert, and Edwin "Jed" Herman. 2016a. Calculus Volume 1. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 1

Reading 4 min Done

The Fundamental Theorem of Calculus: Rates and Totals Undo Each Other

Introduction

Finding a speed at an instant and finding the area under a curve look like separate problems, with separate histories. One theorem shows they are the same problem run in opposite directions. This reading explains that theorem, the fundamental theorem of calculus, with a small table, then says where it came from and why it matters.

Two Operations That Undo Each Other

Some pairs of operations reverse each other. Add 5, then subtract 5, and you're back where you started. Double a number, then halve it, and the same thing happens. Inverse operations are a pair of operations in which each one undoes the other.

Calculus has such a pair. Finding a rate takes a quantity, such as distance, and gives how fast it changes. Finding a total takes a rate, such as speed, and gives how much it accumulates. The fundamental theorem of calculus says that these two are inverse operations: accumulate a rate and then find the rate of change of the result, and you get the original rate back (Strang and Herman 2016a, sec. 5.3).

A Table Example

A cyclist rides for four minutes and holds a steady speed within each minute: 5 meters per second in the first minute, 5 in the second, 4 in the third, and 6 in the fourth. Each minute has 60 seconds, so the distance for each minute is speed times 60.

A running total is the sum so far, updated as each new piece is added. Here it is the distance covered since the start.

Minute	Speed (m/s)	Distance that minute (m)	Running total (m)
1	5	300	300
2	5	300	600
3	4	240	840
4	6	360	1,200

Reading down the speed column and building the running total is accumulation: the total that builds up when a changing rate acts over time, found by adding the contributions of many short stretches. Now run it backward. Take the running total and find its rate of change in each minute: $(600 - 300) \div 60 = 5$ , $(840 - 600) \div 60 = 4$ , $(1200 - 840) \div 60 = 6$ . Those are the speeds you started with. Totaling the speeds gave distance; the rates of the distances gave the speeds back.

With speeds that change smoothly, the same holds in the limit. The integral of the speed gives the distance, and the derivative of the distance gives the speed.

Where It Came From

Isaac Barrow worked toward the link between tangents, which measure rates, and areas, which measure totals, but he never stated it outright. Isaac Newton's tract on fluxions of October 1666 contains the first clear statement of the theorem (O'Connor and Robertson 1996). Gottfried Wilhelm Leibniz found and explored the same relationship, and the theorem was developed by both men, among others, in the late 1600s and early 1700s (Strang and Herman 2016a, sec. 5.3).

Why It's a Landmark

Before the theorem, each curved area needed its own argument. Archimedes' result for the parabola took a long, careful proof. The theorem changed the strategy. To find the area under a curve, you can look for a quantity whose rate of change is that curve. If you find one, the area is how much that quantity changes from start to finish, with no rectangles to add.

Here is a small case. A car's speed in meters per second is $2t$ at $t$ seconds, so its speed graph is a straight line. What is the area under that line from 0 to 5 seconds? The distance $t^2$ has $2t$ as its rate of change, since it gains about $2t$ meters each second at time $t$ . So the area is how much $t^2$ changes from 0 to 5: $25 - 0 = 25$ . No rectangles needed. That turns many hard area problems into easier rate problems. It also unites the two ancient questions, the speed at an instant and the area of a curved shape, into one subject. Every later use of calculus in physics, engineering, and economics leans on that link.

Conclusion

The fundamental theorem of calculus says that finding a rate and finding a total are inverse operations. A table of speeds and running totals shows it in miniature. Newton gave the first clear statement in 1666, Leibniz reached it in his own work, and the theorem made curved areas a matter of working backward from rates.

Key Terms

Inverse operations: A pair of operations in which each one undoes the other.
Fundamental theorem of calculus: The result that finding a rate and finding a total are inverse operations, so accumulating a rate and then finding its rate of change gives the original rate back.
Running total: The sum so far, updated as each new piece is added.
Accumulation: The total that builds up when a changing rate acts over time, found by adding the contributions of many short stretches.

References

O'Connor, J. J., and E. F. Robertson. 1996. "A History of the Calculus." MacTutor History of Mathematics Archive, University of St Andrews. Last updated February 1996.
- Free: MacTutor: A history of the calculus
Strang, Gilbert, and Edwin "Jed" Herman. 2016a. Calculus Volume 1. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 1

Guided Reading 7 min Done

Worked Example: Distance and Speed from a Table, Both Ways

Introduction

Calculus answers two questions about anything that changes: how much has built up, and how fast is it changing right now? This worked example answers both for one car, using nothing but a table of numbers and arithmetic. First you'll estimate the distance the car travels from its speeds. Then you'll run the reasoning backward and recover its speed from its distances.

The Problem

A car starts from rest and speeds up steadily. Its speed is recorded once a second for five seconds:

Time (s)	0	1	2	3	4	5
Speed (m/s)	0	2	4	6	8	10

Two questions:

How far does the car travel in the five seconds?
Working from the distances alone, can you recover the car's speed at the instant $t = 3$ seconds, where $t$ stands for time?

The table gives speeds only at whole seconds. Between readings the speed keeps changing, so neither question can be answered by one multiplication.

Working It Through

Step 1: Estimate the distance using the speed at the start of each second

Over a short stretch of time, distance is about speed times time. Each stretch here is 1 second long. If you assume the car holds the speed it had at the start of each second, the five seconds contribute:

0 \times 1 + 2 \times 1 + 4 \times 1 + 6 \times 1 + 8 \times 1 = 20

That gives 20 meters. This is a rectangle approximation: on a graph of speed against time, each term is the area of a rectangle 1 second wide and as tall as the starting speed. The car was speeding up during every second, so its true speed was always at least the starting value. The estimate of 20 meters is too low.

Step 2: Estimate the distance using the speed at the end of each second

Now assume the car holds the speed it reaches at the end of each second:

2 \times 1 + 4 \times 1 + 6 \times 1 + 8 \times 1 + 10 \times 1 = 30

That gives 30 meters. During each second the car was never going faster than its end speed, so this estimate is too high. The true distance is squeezed between the two: at least 20 meters and at most 30 meters.

Figure: A graph of speed (vertical axis, 0 to 10 m/s) against time (horizontal axis, 0 to 5 s), showing a straight line rising from 0 to 10. Five rectangles, each 1 second wide, sit under the line with their tops at the speed at the start of each second (total area 20). Five more rectangles, drawn in a second color, rise above the line with their tops at the speed at the end of each second (total area 30). Alt text: A rising speed line with one set of rectangles fitting below it and another set poking above it, showing the low and high estimates of distance.

Step 3: Average the two estimates

The average of the low and high estimates is $(20 + 30) \div 2 = 25$ meters. Here that average is exact, not just a better guess.

The reason is the shape of the graph. The speed rises at a steady rate, so its graph is a straight line from 0 to 10 meters per second. In each second, the low rectangle misses a small triangle above it, and the high rectangle overshoots by a triangle of the same size. Averaging the two cancels the error. The area under the line is also a triangle with base 5 and height 10, whose area is $\tfrac12 \times 5 \times 10 = 25$ . Both routes give 25 meters.

Step 4: Work backward to the distances

Now turn the problem around. The same reasoning gives the distance covered by the end of each second, as a running total, the sum so far. Over the first second the speed rises from 0 to 2, an average of 1 meter per second, so the car covers 1 meter. Over the first two seconds the average speed is 2, so it covers 4 meters, and so on:

Time (s)	0	1	2	3	4	5
Distance (m)	0	1	4	9	16	25

The pattern is that the distance in meters is the square of the time in seconds: $d = t^2$ , where $d$ is distance. That formula lets you find the distance at any moment, not only at whole seconds. At $t = 3.1$ , for instance, $d = 3.1^2 = 9.61$ meters.

Step 5: Estimate the speed at three seconds from the distances

Pretend you'd lost the speed table and had only the distances. The average speed over an interval is the change in distance divided by the change in time. Start at $t = 3$ and shrink the interval:

From 3 to 4 seconds: $(16 - 9) \div 1 = 7$ meters per second
From 3 to 3.1 seconds: $(9.61 - 9) \div 0.1 = 6.1$ meters per second
From 3 to 3.01 seconds: $(9.0601 - 9) \div 0.01 = 6.01$ meters per second

The averages approach 6 meters per second. That number, the limit of the average speeds as the interval shrinks, is the instantaneous speed at $t = 3$ . Check it against the original table: at 3 seconds the car's speed was 6 meters per second. The backward route recovered the forward data.

Step 6: Name what happened

You did two things that undo each other. Going forward, you totaled the speeds to get distance: accumulation, which in the limit is the integral. Going backward, you took the rate of change of distance to get speed: the instantaneous rate, which is the derivative. Each operation reversed the other. That is the fundamental theorem of calculus, seen in one small table.

Key Considerations

Why 25 meters is exact here. The average of the two rectangle estimates matched the true distance only because the speed rose at a constant rate, so its graph was a straight line. If the speed graph curves, as it does for a car that speeds up quickly and then levels off, the low and high triangles no longer match. Then the rectangle estimates, and their average, only approximate the distance. Thinner rectangles bring them closer, and the exact answer is the limit they approach.

Which estimate is low and which is high. For a speed that is rising, start-of-interval speeds give a low estimate and end-of-interval speeds give a high one. For a falling speed the roles swap. Always ask which way the rate is moving before you say whether an estimate is high or low.

The common mistake: total distance over total time. It's tempting to say that the car's speed is 25 meters divided by 5 seconds, or 5 meters per second. That's the average speed over the whole trip. It isn't the speed at any particular instant you choose. At $t = 3$ the car is going 6 meters per second; at $t = 1$ it's going 2. An instantaneous speed comes only from shrinking the interval around the instant you care about (Strang and Herman 2016a, sec. 3.4).

Measurement error in step 5. Each average in step 5 is exact for this car. With measured data, readings carry small errors, and dividing by a tiny interval magnifies them. In practice, people stop shrinking the interval once the averages settle.

Summary

Totaling the speeds gave the distance, and taking rates of the distances gave the speed back. Here is the full solution, with the check built in:

Part 1: Distance from speed (rectangle estimates)
Method Estimate Verdict
Speed at the start of each second 20 m Too low
Speed at the end of each second 30 m Too high
Average of the two 25 m Exact, because speed rises at a steady rate
Part 2: Speed from distance (shrinking intervals at $t = 3$ )
Interval Average speed
3 to 4 s 7 m/s
3 to 3.1 s 6.1 m/s
3 to 3.01 s 6.01 m/s
Limit 6 m/s, matching the speed table

Method	Estimate	Verdict
Speed at the start of each second	20 m	Too low
Speed at the end of each second	30 m	Too high
Average of the two	25 m	Exact, because speed rises at a steady rate

Interval	Average speed
3 to 4 s	7 m/s
3 to 3.1 s	6.1 m/s
3 to 3.01 s	6.01 m/s
Limit	6 m/s, matching the speed table

Part 1 is accumulation: rate times time, added up, squeezed between a low and a high estimate.
Part 2 is the instantaneous rate: averages over shrinking intervals, approaching a limit.
The match between part 2's limit and the original speed table is the check that both parts are right.

References

Strang, Gilbert, and Edwin "Jed" Herman. 2016a. Calculus Volume 1. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 1

Guided Conversation 12 min Done

Estimate a Rate and a Total from Your Own Numbers

In this conversation you'll pick something in your own life that changes over time, make up a short table for it, and estimate a rate and a total from your numbers, with hints when you want them. You'll leave able to explain, in two sentences of your own, how rates and totals undo each other.

You'll have this conversation with an AI assistant, using your own account. Choose a button to open a new chat with the prompt already filled in, then press send to start. If the chat opens empty, copy the prompt and paste it in.

Run this conversation in whichever assistant you already use:

Open in Claude Open in ChatGPT

Claude desktop app

To use another LLM, simply copy and paste the prompt into its chat window.

Show the full prompt (it lists misreadings to watch for, so skip it if you would rather come to the conversation fresh)

Guided Conversation: Estimate a Rate and a Total from Your Own Numbers (about 12 minutes)

Note to the learner: press send to start. Everything below is facilitator guidance for the AI. It lists misconceptions to watch for, so skip it if you'd rather come to the conversation fresh.

Please facilitate a coached problem session with me. I'm an adult who is comfortable with basic high-school algebra but may be rusty or uneasy with math, exploring what calculus is and whether to study it further. I'm studying how calculus estimates rates of change and accumulated totals. Follow this guidance for the whole conversation.

GOAL
Help me estimate a rate of change and an accumulated total from a table of values, and explain how the two are linked, using a situation I choose.

HOW TO RUN THE CONVERSATION
- Ask one question at a time, then wait for my reply. Keep each of your turns under about 120 words.
- Don't lecture. Explain a point only when I need it to continue, then return to my table.
- Be curious, collegial, and patient. Give hints and ask questions; don't solve the problem for me. Ask for my approach before any hint, give one hint at a time, and never state an answer before I've tried it. Define terms briefly on first use. Welcome disagreement when I give a reason.
- Write math in plain text, for example (18 - 10) / 1 = 8.
- Plain conversation only: don't search the web or create files or documents.
- Aim for about 12 minutes, with most of the time on topics 2 and 3. If my replies are brief, offer one concrete prompt, such as "What would the bathtub hold after one minute?", and move on. If I seem uncertain, shorten the conversation to 5-7 minutes. Always reach the final topic.
- Start now. Open with one or two warm sentences: this is a conversation, not a quiz; trying an estimate matters more than getting it exactly right; I can ask you to clarify anything. Then ask me to name something in my life that changes over time, such as a run, a bank balance, or a filling bathtub.

TOPICS, IN ORDER
1. My table. Help me make up a short table of the amount at four or five equally spaced times, with units. If I'm stuck, offer this one: a bathtub holds 0, 10, 18, 24, and 28 liters at 0, 1, 2, 3, and 4 minutes. Follow-up: is the amount changing at a steady pace or not?
2. A rate, then a sharper rate. Ask me to estimate how fast the amount changes between two of my times. Then ask how I could get a better estimate of the rate at one exact moment. Hint, if needed: use a shorter interval. If my table can't give one, let me invent a reading in between, or use the bathtub, which holds 11t - t^2 liters at t minutes. Follow-up: what number do my averages seem to approach?
3. From rates to a total. Give me a table of rates instead: the bathtub fills at 11, 9, 7, 5, and 3 liters per minute at 0, 1, 2, 3, and 4 minutes. Ask me to estimate how much water arrives in the 4 minutes, and whether my estimate is too high or too low. Hint, if needed: multiply each rate by 1 minute and add. Follow-up: what if I used the rate at the end of each minute instead?
4. Closing. Ask me to explain in two sentences, in my own words, how finding a total and finding a rate undo each other, so I leave with an explanation I can test on a set of practice problems about rates and totals.

CORRECT METHOD AND ANSWER
- An average rate is the change in the amount divided by the change in time, with units such as liters per minute. For the bathtub table, the rates over each minute are 10, 8, 6, and 4 liters per minute.
- The rate at one instant is the number the averages approach as the interval shrinks. For the bathtub at t = 2: from 2 to 3 minutes the average is 6; from 2 to 2.5 it's 6.5; from 2 to 2.1 it's 6.9; from 2 to 2.01 it's 6.99. They approach 7 liters per minute.
- A total from rates is rate times time, added up. Using the rate at the start of each minute: 11 + 9 + 7 + 5 = 32 liters, too high, because the rate is falling. Using the rate at the end: 9 + 7 + 5 + 3 = 24 liters, too low. The true total lies between. Here it is 28 liters, the average of the two, matching the first table at 4 minutes. For a rising rate, the roles swap.
- Totals and rates are inverse operations: totaling the rates gives the amount, and the rate of change of the amount gives the rates back.
- With my own numbers, check my arithmetic, units, and whether I said each estimate is high or low.

MISCONCEPTIONS TO CORRECT GENTLY
When one appears, name the accurate version briefly, then return to my table.
- "Instantaneous speed is total distance over total time": that's the average over the whole period; the rate at an instant is the value averages approach as the interval shrinks.
- "Calculus is about formulas": it's about rates and totals; formulas are shortcuts for working them out.
- "Newton invented calculus alone": Leibniz developed it independently, and the ideas go back much further, to Archimedes and the Kerala school in India.

LIMITS
- No differentiation or integration rules.
- Write math in plain notation only.
- Don't compute my estimates for me; check them after I've tried.

TO FINISH
After my two-sentence explanation, close in one short turn:
- Affirm one specific thing I worked out, in my own words where possible.
- Suggest one or two next steps that fit how the conversation went. Possible steps: try a still shorter interval to see the averages settle; redo the total with the end-of-minute rates and compare; make a table where the rate rises and check which estimate is low; revisit what an average rate means.
- Restate my two-sentence explanation on its own line, labeled "Rates and totals", so I can copy it.

Problem Set 15 minOptional Done

Rates and Totals from Tables

Overview

This optional problem set gives you practice with the two basic moves of calculus: finding a rate of change and finding an accumulated total. An average rate is a change in some amount divided by the time it took. A rate at one instant is the number those averages approach as the time interval shrinks. A total from a rate is rate times time, added up over short stretches.

Work through the problems in order. For each one:

Write down what the problem gives you: amounts or rates.
Divide differences to find rates; add rate times time to find totals.
Compare your work with the answer and method given.

Problems

Problem 1

A train covers 150 km in 2 hours. What is its average speed?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 1Hide the solution

Answer: 75 km/h

Method:

Average speed is distance divided by time.
$150 \div 2 = 75$ , so the average speed is 75 kilometers per hour.

Most common error: Multiplying instead of dividing, $150 \times 2 = 300$ . Check the units: kilometers divided by hours gives kilometers per hour, which is a speed.

Problem 2

An object's distance from a starting point is $d = t^2$ meters at $t$ seconds, where $d$ is distance and $t$ is time. What is its average speed between $t = 2$ and $t = 2.1$ seconds?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 2Hide the solution

Answer: 4.1 m/s

Method:

Find the distance at each time. At $t = 2$ , $d = 2^2 = 4$ meters. At $t = 2.1$ , $d = 2.1^2 = 4.41$ meters.
The change in distance is $4.41 - 4 = 0.41$ meters.
The change in time is $2.1 - 2 = 0.1$ seconds.
Divide: $0.41 \div 0.1 = 4.1$ meters per second.

Most common error: Dividing the distance at 2.1 seconds by 2.1, which gives about 2.1 m/s. That's the average speed since the start, not over the short interval you were asked about. Use the changes in distance and time.

Problem 3

Using the same object as in Problem 2, estimate its speed at the instant $t = 2$ seconds.

Once you've tried the problem, check your work against the solution.Show the solution to Problem 3Hide the solution

Answer: 4 m/s

Method:

Shrink the interval that starts at $t = 2$ and watch the averages.
From 2 to 2.1 seconds: 4.1 m/s, from Problem 2.
From 2 to 2.01 seconds: $2.01^2 = 4.0401$ , so the average is $(4.0401 - 4) \div 0.01 = 4.01$ m/s.
From 2 to 2.001 seconds, the same steps give 4.001 m/s.
The averages approach 4, so the speed at the instant $t = 2$ is 4 meters per second.

Most common error: Stopping at 4.1 and calling it the answer. Each average over an interval overshoots a little here, because the object is speeding up. The instantaneous speed is the number the averages approach, not any one of them.

Problem 4

A tap fills a tub at 3 liters per minute for 4 minutes. How much water is in the tub, if it started empty?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 4Hide the solution

Answer: 12 L

Method:

The rate is steady, so the total is rate times time.
$3 \times 4 = 12$ liters.

Most common error: Dividing, $3 \div 4$ or $4 \div 3$ . A total built from a rate is a product: liters per minute times minutes gives liters.

Problem 5

A cyclist's speed is recorded each second:

Time (s)	0	1	2	3	4
Speed (m/s)	0	1	4	9	16

Estimate the distance traveled over the 4 seconds two ways: first using the speed at the start of each second, then using the speed at the end of each second. Say which estimate is too low and which is too high.

Once you've tried the problem, check your work against the solution.Show the solution to Problem 5Hide the solution

Answer: 14 m using starting speeds (too low); 30 m using ending speeds (too high). The true distance lies between 14 and 30 m.

Method:

Each stretch is 1 second long, so each piece of distance is the speed times 1 second.
Starting speeds: $0 + 1 + 4 + 9 = 14$ meters.
Ending speeds: $1 + 4 + 9 + 16 = 30$ meters.
The speed rises during every second. So the starting speed undercounts each stretch and the ending speed overcounts it. The first estimate is too low and the second too high.

Most common error: Adding all five speeds, $0 + 1 + 4 + 9 + 16 = 30$ , and getting the right number for the wrong reason, or adding them and dividing by 5. There are only four one-second stretches, so each estimate has four terms.

Self-Check

When you're done, check that:

Each rate has units such as km/h, m/s, or L/min, and each total has units such as km, m, or L.
You said whether each rectangle estimate in Problem 5 is high or low, and why.
Your answer to Problem 3 matches the trend in the averages from Problem 2 and the shorter intervals after it.

Knowledge Check 10 min Done

Instantaneous speed and areas under curves: limits, derivatives, and integrals

This ungraded knowledge check assesses your understanding of how calculus measures rates and totals. You'll be asked about how calculus developed from Archimedes to Newton and Leibniz, the derivative as a limit of average rates, the integral as accumulation, and the fundamental theorem of calculus.

Note: Use this to test yourself, review the feedback on any questions you miss, and retry until you feel confident before moving forward.

5 questions · ungraded · retry as often as you like