1. Rates and areas¶

Calculus began as two separate puzzles. The first: a stone is falling, faster every moment; how fast is it going at one exact instant, when in an instant it covers no distance at all? The second: a shape has a curved edge; how much area does it hold, when no number of squares fits it exactly? In this chapter you will attack both puzzles with nothing more than arithmetic, tables and patience. You will squeeze an instantaneous speed out of shorter and shorter averages, trap a curved area between rectangles, follow Archimedes as he finds the area of a parabolic segment exactly, and end with the surprise that turned two puzzles into one subject: each question is the other one run backwards.

In this chapter

You will learn:

the two questions that calculus answers: the rate at an instant, and the area under a curve
how average speeds over shorter and shorter time intervals home in on a speed at one instant
how to trap an area between lower and upper rectangle sums, and why the trap closes
how Archimedes found the area of a parabolic segment exactly, with a complete proof by exhaustion
why the two questions are inverse to each other, a first look at the fundamental theorem of calculus

You need: school arithmetic, fractions and the graph of \(y = x^2\). The algebra book's Chapter 6 explains functions and graphs if they are new. Gentler versions on the Topics shelf: Fast and slow, Comparing speeds fairly, Average speed for a trip, Speed-time graphs, Falling, on Earth and on Mars, Area under speed-time graphs and Shrinking the interval.

Levels: 1.1 to 1.3 are beginner; 1.4 is intermediate; 1.5 is beginner to intermediate.

1.1 Two old questions¶

Here are two questions that sound innocent.

Question 1 (the rate question). A car's speedometer says 60 kilometres per hour. What does that number mean? It does not mean the car will travel 60 kilometres in the next hour; it might stop at a light in ten seconds. It does not mean it travelled 60 kilometres in the last hour. It claims to be the speed right now. But "right now" is a single instant, and in a single instant the car moves no distance in no time. Speed is distance divided by time, and \(0 \div 0\) means nothing. So what is the speedometer measuring?

Question 2 (the area question). How much area lies under the curve \(y = x^2\) between \(x = 0\) and \(x = 1\)? The region has two straight sides, a straight bottom and a curved top. Areas of rectangles and triangles are easy: length times width, half base times height. But no finite number of rectangles fits under a curve exactly. So what does "the area" of a curved region even mean, and how could you compute it?

These look unrelated. One is about motion and time, the other about shapes and space. Yet both have the same difficulty at their heart: the answer is something we can only approach, by a process that never stops. Speeds over shorter and shorter time intervals; rectangles thinner and thinner. Calculus is the mathematics of what those endless processes are heading towards.

Two words before we start.

A rate is how fast one quantity changes compared with another: metres per second, litres per minute, dollars per year. Speed is the rate at which distance changes with time.
A function is a rule that gives exactly one output for each input. We write \(s(t)\) for "the distance at time \(t\)", read "s of t". The graph of a function is the picture of all its input-output pairs.

The plan of this chapter is to answer both questions numerically, as well as we like, using only arithmetic. Then we will find, almost by accident, that the two answers are tied together. The exact, rigorous version of every idea here (limits, the derivative, the integral) takes the next eight chapters. Here we get the feel of it.

1.2 Average speed and instantaneous speed¶

Drop a stone from a high cliff. Since Galileo's experiments in the early 1600s (with balls rolling down slopes, which slow the fall enough to time it), we know that, ignoring air resistance, the distance fallen grows with the square of the time. Near the Earth's surface, with \(t\) in seconds and distance in metres,

\[ s(t) = 4.9\,t^2 . \]

The 4.9 is half of \(g = 9.8\) metres per second per second, the Earth's gravitational acceleration. (Why "half" will be clear in section 1.5.) Here is the table:

time \(t\) (s)	0	1	2	3	4	5
distance \(s(t)\) (m)	0	4.9	19.6	44.1	78.4	122.5

The average speed over a time interval is the distance covered divided by the time taken:

\[ \text{average speed from } t = a \text{ to } t = b \;=\; \frac{s(b) - s(a)}{b - a}. \]

From \(t = 1\) to \(t = 2\) the stone falls \(19.6 - 4.9 = 14.7\) metres in 1 second, an average of 14.7 m/s. From \(t = 2\) to \(t = 3\) it falls \(44.1 - 19.6 = 24.5\) metres: 24.5 m/s. The stone is speeding up, so each second's average is bigger than the last.

Now the rate question. How fast is the stone moving at exactly \(t = 1\)? We cannot divide by a zero time interval, but we can take a short interval starting at \(t = 1\) and shrink it. Write \(h\) for the length of the interval, so it runs from \(t = 1\) to \(t = 1 + h\).

interval length \(h\) (s)	\(s(1+h)\) (m)	distance \(s(1+h) - s(1)\)	average speed (m/s)
1	19.6	14.7	14.7
0.5	11.025	6.125	12.25
0.1	5.929	1.029	10.29
0.01	4.99849	0.04949	9.849
0.001	4.9098049	0.0049049	9.8049

The averages are settling down. They are heading towards 9.8 m/s. Nobody has divided by zero; we have watched a sequence of honest averages and seen where they are going.

Algebra shows why they go to 9.8. For any \(h \neq 0\),

\[ \frac{s(1+h) - s(1)}{h} = \frac{4.9(1 + 2h + h^2) - 4.9}{h} = \frac{4.9(2h + h^2)}{h} = 9.8 + 4.9h . \]

Dividing by \(h\) was allowed, because \(h\) is not zero. The result \(9.8 + 4.9h\) is a plain expression, and as \(h\) gets smaller it gets closer and closer to \(9.8\). We can make it as close to \(9.8\) as we please: to be within \(0.001\), take \(h\) below \(0.001/4.9 \approx 0.0002\). So we define the speed at the instant \(t = 1\) to be \(9.8\) m/s. That is the instantaneous speed: the value the average speeds approach as the interval shrinks to nothing.

Check. Intervals that end at \(t = 1\) instead of starting there give \(\frac{s(1) - s(1-h)}{h} = 9.8 - 4.9h\), which approaches 9.8 from below. Both sides agree.

The same algebra at any time \(a\) gives

\[ \frac{s(a+h) - s(a)}{h} = \frac{4.9(2ah + h^2)}{h} = 9.8a + 4.9h , \]

which approaches \(9.8a\). So the speed at time \(t\) is \(v(t) = 9.8\,t\): 9.8 m/s after one second, 19.6 m/s after two, 29.4 m/s after three. The speed grows by 9.8 m/s every second, which is exactly what "acceleration of 9.8 metres per second per second" means.

Worked example: falling on Mars. Mars is smaller than Earth, and a stone there accelerates at only about \(3.71\) m/s\(^2\). The fall law is \(s(t) = 1.855\,t^2\) (half of 3.71). The average speed from \(t = a\) to \(t = a + h\) is

\[ \frac{1.855\big((a+h)^2 - a^2\big)}{h} = 1.855(2a + h) = 3.71a + 1.855h, \]

which approaches \(3.71a\). After 2 seconds a stone on Mars is falling at \(7.42\) m/s and has fallen \(1.855 \times 4 = 7.42\) metres. On Earth the same 2 seconds give 19.6 m/s and 19.6 metres. Check: \(3.71 \times 2 = 7.42\) and \(1.855 \times 2^2 = 7.42\); that both numbers are equal at \(t = 2\) is a coincidence of this time (speed \(= 2s/t\) for this law, so they agree only when \(t = 2\)).

What we did, in one sentence¶

To find a rate at an instant, take the average rate over a short interval, simplify while the interval is not zero, then see what the result approaches as the interval shrinks. Chapter 4 names the answer the derivative, and Chapter 2 makes "approaches" precise.

The picture: secants and tangents¶

On a graph of \(s\) against \(t\), the average speed from \(t = a\) to \(t = b\) is the slope (rise over run) of the straight line joining the two points \((a, s(a))\) and \((b, s(b))\). Such a line, cutting the curve in two points, is called a secant (from the Latin for "cutting"). As \(b\) slides towards \(a\), the secants swing round and settle on one line that just touches the curve at \((a, s(a))\): the tangent line. So the instantaneous speed is the slope of the tangent. The rate question and the tangent question (what is the slope of a curve at a point?) are the same question in two costumes. Mathematicians of the 1600s, Fermat and Descartes among them, attacked it as a question about tangents to curves.

1.3 Areas by rectangles¶

Now the area question. Let \(A\) be the area under \(y = x^2\) from \(x = 0\) to \(x = 1\): the region above the \(x\)-axis, below the curve, between the vertical lines \(x = 0\) and \(x = 1\).

We know areas of rectangles, so trap \(A\) between two staircases. Cut \([0, 1]\) into \(n\) equal pieces of width \(1/n\). On each piece, the curve rises from left to right, because \(x^2\) grows as \(x\) grows (for \(x \ge 0\)). So:

a rectangle whose height is the value at the right end of the piece reaches above the curve, and
a rectangle whose height is the value at the left end stays below it.

Adding the tall rectangles gives an upper sum \(U_n\) that is too big; adding the short ones gives a lower sum \(L_n\) that is too small:

\[ L_n \le A \le U_n . \]

With \(n = 4\) the pieces are \([0, \tfrac14], [\tfrac14, \tfrac12], [\tfrac12, \tfrac34], [\tfrac34, 1]\). The right ends are \(\tfrac14, \tfrac12, \tfrac34, 1\), with squares \(\tfrac1{16}, \tfrac4{16}, \tfrac9{16}, \tfrac{16}{16}\). Each rectangle has width \(\tfrac14\), so

\[ U_4 = \tfrac14\left(\tfrac{1 + 4 + 9 + 16}{16}\right) = \frac{30}{64} = 0.46875 . \]

The left ends are \(0, \tfrac14, \tfrac12, \tfrac34\), so \(L_4 = \tfrac14 \cdot \tfrac{0 + 1 + 4 + 9}{16} = \tfrac{14}{64} = 0.21875\). The area lies between 0.21875 and 0.46875. Not very precise yet, but it is a fact, not a guess.

More rectangles tighten the trap.

\(n\)	lower sum \(L_n\)	upper sum \(U_n\)	gap \(U_n - L_n\)
4	0.21875	0.46875	0.25
10	0.285	0.385	0.1
100	0.32835	0.33835	0.01
1000	0.3328335	0.3338335	0.001

The two columns close in on \(0.333\ldots\), which suggests \(A = \tfrac13\). To be sure we need a formula.

A formula for the sums¶

With \(n\) pieces, the right ends are \(\tfrac1n, \tfrac2n, \ldots, \tfrac nn\), so

\[ U_n = \frac1n\left(\frac{1^2}{n^2} + \frac{2^2}{n^2} + \cdots + \frac{n^2}{n^2}\right) = \frac{1^2 + 2^2 + \cdots + n^2}{n^3}. \]

We need the sum of the first \(n\) squares.

Claim. \(1^2 + 2^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6}\).

Proof (telescoping). For every whole number \(k\), expanding gives \((k+1)^3 - k^3 = 3k^2 + 3k + 1\). Write this line down for \(k = 1, 2, \ldots, n\) and add all \(n\) lines. On the left almost everything cancels: \(2^3 - 1^3\) plus \(3^3 - 2^3\) plus ... plus \((n+1)^3 - n^3\) leaves only \((n+1)^3 - 1\). On the right we get \(3S + 3\cdot\frac{n(n+1)}{2} + n\), where \(S\) is the sum of squares we want and \(\frac{n(n+1)}2\) is the sum \(1 + 2 + \cdots + n\). So

\[ 3S = (n+1)^3 - 1 - \tfrac32 n(n+1) - n = (n+1)\left[(n+1)^2 - \tfrac32 n - 1\right] = (n+1)\cdot\frac{2n^2 + n}{2}, \]

and \(S = \frac{n(n+1)(2n+1)}{6}\). (In the middle step we used \((n+1)^3 - 1 - n = (n+1)^3 - (n+1)\).) \(\square\)

Check: \(n = 4\) gives \(\frac{4 \cdot 5 \cdot 9}{6} = 30 = 1 + 4 + 9 + 16\).

So

\[ U_n = \frac{(n+1)(2n+1)}{6n^2} = \frac13 + \frac1{2n} + \frac1{6n^2}, \qquad L_n = \frac{(n-1)(2n-1)}{6n^2} = \frac13 - \frac1{2n} + \frac1{6n^2}. \]

(For \(L_n\) the left ends are \(\tfrac0n, \ldots, \tfrac{n-1}n\), so we need the sum of the first \(n-1\) squares: replace \(n\) by \(n - 1\) in the claim.)

Now the argument is airtight. For every \(n\), \(L_n < \tfrac13 < U_n\): the upper sum is \(\tfrac13\) plus positive terms, and the lower sum is \(\tfrac13\) minus \(\tfrac1{2n} - \tfrac1{6n^2} = \tfrac{3n - 1}{6n^2} > 0\). And the area \(A\) is trapped in the same interval. Both \(A\) and \(\tfrac13\) lie between \(L_n\) and \(U_n\), whose gap is \(U_n - L_n = \tfrac1n\). So \(A\) and \(\tfrac13\) differ by less than \(\tfrac1n\), for every \(n\). The only non-negative number smaller than \(\tfrac1n\) for every \(n\) is \(0\). Therefore

\[ A = \frac13 . \]

That last step is worth a second look, because it is the logical pattern of the whole of calculus. We never added up "infinitely many infinitely thin rectangles". We showed that any other answer would be wrong by some definite amount, and then made the rectangles thin enough to rule that amount out. Archimedes argued the same way, as you will see next.

Scaling. The same method on \([0, b]\) gives area \(\tfrac{b^3}3\) (try it: each width is \(\tfrac bn\) and each height carries a factor \(b^2\), so every sum is multiplied by \(b^3\)). So the area under \(y = x^2\) from \(0\) to \(2\) is \(\tfrac83\), and from \(1\) to \(2\) it is \(\tfrac83 - \tfrac13 = \tfrac73\).

Area means accumulated quantity¶

Areas under graphs are not only geometry. If a car moves at a constant 20 m/s for 30 s, it covers \(20 \times 30 = 600\) m, which is the area of a rectangle 30 wide and 20 high under the speed-time graph. If the speed changes, cut time into short pieces on which the speed is nearly constant, and add up speed times time on each piece: that is a rectangle sum under the speed graph. So

\[ \text{distance travelled} = \text{area under the speed-time graph}. \]

For the falling stone, \(v(t) = 9.8t\) is a straight line, and the area under it from \(0\) to \(3\) is a triangle: \(\tfrac12 \times 3 \times 29.4 = 44.1\). And the table in 1.2 says \(s(3) = 44.1\). Keep this in mind for section 1.5.

Try it in Python. A few lines compute rectangle sums for any function.

def sums(f, a, b, n):
    w = (b - a) / n
    left  = sum(f(a + k * w) for k in range(n)) * w
    right = sum(f(a + (k + 1) * w) for k in range(n)) * w
    return left, right

print(sums(lambda x: x * x, 0, 1, 1000))   # (0.3328335, 0.3338335) up to rounding

1.4 Archimedes and the parabola¶

Level: intermediate.

About 2250 years ago, Archimedes of Syracuse found the exact area of a region bounded by a parabola and a straight line, with no algebra, no coordinates and no formula for sums of squares. He wrote it up in a work now called Quadrature of the Parabola ("quadrature" means finding a square, or any rectilinear figure, of the same area). His result:

Theorem (Archimedes). The area of a parabolic segment is \(\tfrac43\) of the area of the triangle inscribed in it with the same base and the same height.

A parabolic segment is the region between a parabola and a chord (a straight line segment joining two points of the parabola). The inscribed triangle has the chord as its base, and its third vertex is the point of the parabola where the tangent line is parallel to the chord; that is the point of the arc farthest from the chord.

We will prove the theorem in modern coordinates, but following Archimedes' plan step by step. The plan has two parts: fill the segment with triangles whose total area we can add up, then show by a double contradiction that the segment's area can be neither more nor less than that total. This style of proof is called the method of exhaustion, because the triangles "exhaust" the region.

Step 1: the area of a triangle on the parabola¶

Use the parabola \(y = x^2\). Take three points on it with \(x\)-coordinates \(p < q < r\).

Lemma 1. The triangle with vertices \((p, p^2)\), \((q, q^2)\), \((r, r^2)\) has area \(\tfrac12 (q - p)(r - q)(r - p)\).

Proof. The area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is half the absolute value of \((x_2 - x_1)(y_3 - y_1) - (x_3 - x_1)(y_2 - y_1)\) (this is the "shoelace" or cross-product formula). Here

\[ (q - p)(r^2 - p^2) - (r - p)(q^2 - p^2) = (q - p)(r - p)\big[(r + p) - (q + p)\big] = (q - p)(r - p)(r - q), \]

using \(r^2 - p^2 = (r - p)(r + p)\) and \(q^2 - p^2 = (q - p)(q + p)\). All three factors are positive. \(\square\)

Now take a chord from \(x = a\) to \(x = b\), of width \(w = b - a\). The chord's slope is \(\frac{b^2 - a^2}{b - a} = a + b\). Where is the tangent parallel to it? The slope of \(y = x^2\) at \(x = m\) is \(2m\) (by the algebra of 1.2: \(\frac{(m+h)^2 - m^2}{h} = 2m + h \to 2m\)). So \(2m = a + b\), and \(m = \frac{a+b}2\), the midpoint. Lemma 1 with \(p = a\), \(q = m\), \(r = b\) gives the inscribed triangle's area:

\[ T(w) = \tfrac12 \cdot \tfrac w2 \cdot \tfrac w2 \cdot w = \frac{w^3}{8}. \]

It depends only on the width. That is the key fact.

Step 2: the triangles at each stage¶

The inscribed triangle on the chord \(AB\) (call its area \(T\)) leaves two smaller segments, on the chords \(AC\) and \(CB\), each of width \(\tfrac w2\). Each of those has its own inscribed triangle, of area \(\frac{(w/2)^3}8 = \tfrac18 T\). So the two new triangles together add \(\tfrac28 T = \tfrac14 T\). They leave four segments of width \(\tfrac w4\), whose four triangles add \(4 \cdot \frac{(w/4)^3}{8} = \tfrac1{16} T\), and so on. At stage \(k\) there are \(2^k\) new triangles with total area \(\tfrac1{4^k} T\).

After stages \(0, 1, \ldots, n\) the triangles cover

\[ P_n = T\left(1 + \frac14 + \frac1{16} + \cdots + \frac1{4^n}\right) = \frac43 T - \frac13 \cdot \frac{T}{4^n}. \]

The last equality is the finite geometric sum: if \(G = 1 + \tfrac14 + \cdots + \tfrac1{4^n}\) then \(G - \tfrac14 G = 1 - \tfrac1{4^{n+1}}\), so \(G = \tfrac43\left(1 - \tfrac1{4^{n+1}}\right) = \tfrac43 - \tfrac13 \cdot \tfrac1{4^n}\). Archimedes proved exactly this identity, in words: each term plus a third of itself equals a third of the term before.

So the triangles always cover less than \(\tfrac43 T\), but they get as close to \(\tfrac43 T\) as we like.

Step 3: the leftover shrinks¶

We still need to know that the triangles really fill the segment. Here Archimedes used a beautiful observation.

Lemma 2. The inscribed triangle covers more than half of its segment.

Proof. Draw the tangent line at \(C\); it is parallel to the chord \(AB\). Draw the lines through \(A\) and \(B\) parallel to the axis (vertical lines, in our picture) until they meet the tangent. Together with the chord, these make a parallelogram with base \(AB\) whose opposite side lies on the tangent. The triangle \(ABC\) has the same base and the same height as this parallelogram, so its area is exactly half the parallelogram. The parabola is convex (it bends one way): between \(A\) and \(B\) it lies below the chord and above the tangent line. So the whole segment sits inside the parallelogram, and it does not fill it (the parabola touches the tangent only at \(C\)). Hence segment \(<\) parallelogram \(= 2 \times\) triangle. \(\square\)

Now let \(S\) be the area of the whole segment, and let \(R_n = S - P_n\) be the leftover: the total area of the \(2^{n+1}\) small segments not yet covered after stage \(n\). Stage \(n+1\) puts a triangle into each of those small segments, and by Lemma 2 each triangle takes more than half of its segment. So \(R_{n+1} < \tfrac12 R_n\), and

\[ R_n < \frac{R_0}{2^n} < \frac{S}{2^n} \qquad (n \ge 1). \]

The leftover can be made smaller than any amount you name.

Step 4: the double contradiction¶

Proof of the theorem. Suppose first that \(S > \tfrac43 T\). The difference \(d = S - \tfrac43 T\) is a positive number. Choose \(n\) so large that the leftover \(R_n < d\). Then \(P_n = S - R_n > S - d = \tfrac43 T\). But Step 2 showed \(P_n < \tfrac43 T\) for every \(n\). Contradiction.

Suppose next that \(S < \tfrac43 T\). Now \(e = \tfrac43 T - S\) is positive. Choose \(n\) so large that \(\tfrac13 \cdot \tfrac T{4^n} < e\). Then \(P_n = \tfrac43 T - \tfrac13 \cdot \tfrac T{4^n} > \tfrac43 T - e = S\). But the triangles lie inside the segment and do not overlap, so \(P_n \le S\). Contradiction.

Both possibilities are impossible, so \(S = \tfrac43 T\). \(\square\)

What it gives, in modern terms¶

The segment on a chord of width \(w\) has area \(\tfrac43 \cdot \tfrac{w^3}8 = \tfrac{w^3}{6}\).

Worked example. Recover the area under \(y = x^2\) from \(0\) to \(1\). The chord from \((0, 0)\) to \((1, 1)\) has width \(1\), so the segment between the chord and the parabola has area \(\tfrac16\). The region under the chord (a triangle with base 1 and height 1) has area \(\tfrac12\). The region under the parabola is the triangle minus the segment: \(\tfrac12 - \tfrac16 = \tfrac13\). Check: this matches the rectangle sums of 1.3.

Worked example. The line \(y = 4\) cuts \(y = x^2\) at \(x = \pm 2\), width \(w = 4\). The segment has area \(\tfrac{64}{6} = \tfrac{32}3\). The inscribed triangle has apex at \((0, 0)\), base 4, height 4, area 8, and indeed \(\tfrac43 \times 8 = \tfrac{32}3\). Check another way: the rectangle \([-2, 2] \times [0, 4]\) has area 16, and the area under the parabola on \([-2, 2]\) is \(2 \times \tfrac83 = \tfrac{16}3\) (by the scaling remark in 1.3); the segment is \(16 - \tfrac{16}3 = \tfrac{32}3\).

A note on history¶

Archimedes lived in Syracuse, in Sicily, and died in 212 BC when the Romans took the city. The Quadrature of the Parabola contains two proofs of this theorem: the triangle-filling proof above, and a "mechanical" one, balancing thin slices of the segment against a triangle on an imaginary lever. In a letter to Eratosthenes, known as The Method, Archimedes explained that he found results by such balancing arguments first and then proved them rigorously by exhaustion. The Method was lost for centuries; its text survives in a palimpsest (a prayer book written over older erased pages) that the Danish scholar Johan Ludvig Heiberg identified in 1906. The method of exhaustion itself goes back to Eudoxus of Cnidus, a century before Archimedes; Euclid's Elements, Book XII, uses it to prove that the areas of circles are as the squares of their diameters.

What Archimedes did not have was a general method. Each curve needed its own clever dissection. The step from clever dissections to a method that works for every curve took another nineteen centuries.

1.5 The surprise: the two questions are inverse¶

We now have two procedures.

Rate. From a distance function \(s(t)\), find the speed: average over a shrinking interval.
Area. From a speed function \(v(t)\), find the distance: add up rectangles under its graph.

Put the falling stone through both. The rate procedure turned the distance \(s(t) = 4.9t^2\) into the speed \(v(t) = 9.8t\). The area procedure turns the speed \(9.8t\) back into the distance: the area under the line from \(0\) to \(t\) is a triangle with base \(t\) and height \(9.8t\), area \(\tfrac12 \cdot t \cdot 9.8t = 4.9t^2\). We went round a circle and came home. (This is also where the "half" in \(4.9 = \tfrac12 g\) comes from: the distance is the area of a triangle, not of a rectangle.)

Is that a coincidence of this example? Try the parabola. Let \(A(x)\) be the area under \(y = t^2\) from \(t = 0\) to \(t = x\) (we use \(t\) as the variable on the axis so that \(x\) is free to be the right end). Section 1.3 found \(A(x) = \tfrac{x^3}3\). Now ask the rate question about \(A\): how fast does the area grow as \(x\) moves?

\[ \frac{A(x+h) - A(x)}{h} = \frac{(x+h)^3 - x^3}{3h} = \frac{3x^2 h + 3x h^2 + h^3}{3h} = x^2 + xh + \tfrac13 h^2 , \]

which approaches \(x^2\) as \(h\) shrinks. The rate at which the area grows is the height of the curve. The area under \(x^2\) grows at rate \(x^2\).

The picture explains why this must happen for every curve, not just these two. When \(x\) moves to \(x + h\), the area gains a thin strip. The strip has width \(h\) and height about \(f(x)\), the height of the curve there, so its area is about \(f(x) \cdot h\). Divide by \(h\): the rate of growth of the area is about \(f(x)\), and the approximation gets better as the strip gets thinner. In symbols, if \(A(x)\) is the area under \(y = f(t)\) from a fixed starting point to \(x\), then

\[ \text{rate of change of } A \text{ at } x = f(x). \]

So finding areas is undoing rates. To find the area under a curve \(y = f(x)\), you do not need rectangles at all: find a function whose rate of change is \(f\), and read off the area from it. For example, the rate of change of \(\tfrac{x^4}4\) is \(x^3\) (expand \((x+h)^4\) and try it, it is exercise 6's shortcut), so the area under \(y = x^3\) from \(0\) to \(1\) should be \(\tfrac14\). The sum-of-cubes formula confirms it in exercise 6, the hard way.

This is a preview of the fundamental theorem of calculus, which Chapter 9 states precisely and proves. It is the reason calculus is one subject and not two. Archimedes needed a fresh piece of genius for each curved area; after the fundamental theorem, a student can find the area under \(y = x^7\) in one line, because the area question has been turned into the (usually easier) rate question run backwards.

History in brief¶

Rates and areas were studied for centuries before anyone tied them together. In the 1300s, scholars at Merton College, Oxford, stated the "mean speed theorem": a body accelerating uniformly from rest covers the same distance as a body moving at its average speed for the same time. Nicole Oresme, in Paris, gave a geometric proof in the mid-1300s by drawing speed as a height over a time axis, so that distance became an area: the area of a triangle equals the area of a rectangle of half the height. That is our falling stone, three centuries early. Galileo's Two New Sciences (1638) used the same picture for falling bodies.

In the 1600s, Fermat, Descartes and others found tangents to many curves, and Cavalieri, Fermat, Wallis and others found areas under curves \(y = x^n\). Isaac Barrow, Newton's predecessor as Lucasian professor at Cambridge, published a geometric theorem in his Lectiones Geometricae (1670) that, read today, contains the inverse relation. But it was Isaac Newton (from the mid-1660s) and Gottfried Wilhelm Leibniz (from the mid-1670s), working independently, who saw it as the centre of a general method with its own rules and notation. Chapter 9 tells that story, including the dispute over who came first.

Check yourself¶

1. A stone falls with \(s(t) = 4.9t^2\). What is its average speed between \(t = 2\) and \(t = 3\), and between \(t = 3\) and \(t = 4\)?

From the table: \(\frac{44.1 - 19.6}{1} = 24.5\) m/s, and \(\frac{78.4 - 44.1}{1} = 34.3\) m/s. Each second the average goes up by \(9.8\) m/s, the acceleration.

2. Find the instantaneous speed of the stone at \(t = 2\) by computing the average speed from \(t = 2\) to \(t = 2 + h\) and letting \(h\) shrink.

\(\frac{4.9(2+h)^2 - 4.9 \cdot 4}{h} = \frac{4.9(4h + h^2)}{h} = 19.6 + 4.9h\), which approaches \(19.6\) m/s. Check: the formula \(v(t) = 9.8t\) gives \(9.8 \times 2 = 19.6\).

3. On Mars, \(s(t) = 1.855t^2\). How long does a stone take to fall 100 m, and how fast is it going when it lands? Compare with Earth.

Mars: \(1.855t^2 = 100\) gives \(t = \sqrt{100/1.855} \approx 7.34\) s, and the speed is \(3.71t \approx 27.2\) m/s. Earth: \(t = \sqrt{100/4.9} \approx 4.52\) s and speed \(9.8t \approx 44.3\) m/s. Ignoring air (a fair approximation on Mars, whose air is thin, much less so on Earth), the Martian stone takes over half as long again and lands at about 60 percent of the speed.

4. Find the lower and upper sums \(L_n\) and \(U_n\) for \(y = x\) on \([0, 1]\) with \(n\) equal pieces, and show they trap the area \(\tfrac12\).

Right ends \(\tfrac kn\), \(k = 1, \ldots, n\): \(U_n = \frac1n \cdot \frac{1 + 2 + \cdots + n}{n} = \frac{n(n+1)}{2n^2} = \frac12 + \frac1{2n}\). Left ends \(\tfrac kn\), \(k = 0, \ldots, n-1\): \(L_n = \frac{(n-1)n}{2n^2} = \frac12 - \frac1{2n}\). So \(L_n < \tfrac12 < U_n\) with gap \(\tfrac1n\), and the area (a triangle, \(\tfrac12 \cdot 1 \cdot 1\)) is \(\tfrac12\). Check: \(n = 4\) gives \(0.375\) and \(0.625\).

5. Compute the upper sum \(U_4\) for \(y = x^2\) on \([0, 2]\) and compare with the exact area.

Width \(\tfrac12\), right ends \(\tfrac12, 1, \tfrac32, 2\) with squares \(0.25, 1, 2.25, 4\). \(U_4 = \tfrac12(7.5) = 3.75\). The exact area is \(\tfrac{2^3}3 = \tfrac83 \approx 2.667\), so four rectangles overestimate by about \(1.08\).

6. Use \(1^3 + 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2\) to show the area under \(y = x^3\) on \([0, 1]\) is \(\tfrac14\).

\(U_n = \frac1n \sum_{k=1}^n \frac{k^3}{n^3} = \frac{n^2(n+1)^2}{4n^4} = \frac14\left(1 + \frac1n\right)^2\), and \(L_n = \frac14\left(1 - \frac1n\right)^2\) (replace \(n\) by \(n - 1\) in the sum). Both are within \(\tfrac1n\) of \(\tfrac14\): \(U_n - \tfrac14 = \tfrac1{2n} + \tfrac1{4n^2} \le \tfrac3{4n}\) and \(\tfrac14 - L_n = \tfrac1{2n} - \tfrac1{4n^2} < \tfrac1{2n}\). The area lies between \(L_n\) and \(U_n\) for every \(n\), so it equals \(\tfrac14\). (Check: \(U_{10} = 0.3025\), \(U_{100} = 0.255025\).) The shortcut of 1.5: the rate of change of \(\tfrac{x^4}{4}\) is \(\frac{(x+h)^4 - x^4}{4h} = x^3 + \tfrac32 x^2 h + x h^2 + \tfrac14 h^3 \to x^3\).

7. A car's speed is measured every 10 s: 0, 8, 14, 18, 20 m/s at \(t = 0, 10, 20, 30, 40\). The speed rises all the time. Between what limits must the distance in those 40 s lie?

Lower sum (speed at the start of each 10 s piece): \(10(0 + 8 + 14 + 18) = 400\) m. Upper sum (speed at the end): \(10(8 + 14 + 18 + 20) = 600\) m. Because the speed only increases, the true distance is between 400 m and 600 m. More frequent readings would narrow the gap.

8. Find the area of the parabolic segment cut from \(y = x^2\) by the chord from \((1, 1)\) to \((3, 9)\), in two ways.

By Archimedes: width \(w = 2\), triangle \(\tfrac{w^3}8 = 1\), segment \(\tfrac43\). (Lemma 1 with \(p = 1, q = 2, r = 3\): \(\tfrac12 \cdot 1 \cdot 1 \cdot 2 = 1\).) By areas: under the chord is a trapezoid of width 2 and heights 1 and 9, area \(10\); under the parabola from 1 to 3 is \(\tfrac{27}3 - \tfrac13 = \tfrac{26}3\). Segment \(= 10 - \tfrac{26}3 = \tfrac43\).

9. Show that \(1 + \tfrac14 + \tfrac1{16} + \cdots + \tfrac1{4^n}\) is less than \(\tfrac43\) for every \(n\) but gets as close as you like.

Call the sum \(G\). Then \(\tfrac14 G = \tfrac14 + \cdots + \tfrac1{4^{n+1}}\), and subtracting, \(\tfrac34 G = 1 - \tfrac1{4^{n+1}}\), so \(G = \tfrac43 - \tfrac13 \cdot \tfrac1{4^n}\). The subtracted term is positive, so \(G < \tfrac43\), and it is below any positive number you name once \(n\) is large (it shrinks by a factor of 4 each step).

10. Let \(A(x)\) be the area under \(y = 2t + 1\) from \(t = 0\) to \(t = x\). Find \(A(x)\) by geometry and check that its rate of change is \(2x + 1\).

The region is a trapezoid with parallel sides \(1\) (at \(t = 0\)) and \(2x + 1\) (at \(t = x\)) and width \(x\): \(A(x) = \tfrac x2 (1 + 2x + 1) = x^2 + x\). Rate: \(\frac{(x+h)^2 + (x+h) - x^2 - x}{h} = 2x + h + 1 \to 2x + 1\), the height of the line, as section 1.5 predicts.

11. Why does Lemma 2 (the triangle covers more than half its segment) need the parabola to be convex? What fails for a wiggly curve?

The proof put the whole segment inside a parallelogram between the chord and the tangent at \(C\). That needs the curve between \(A\) and \(B\) to stay between those two lines. A curve that bends both ways can cross the tangent line, poke out of the parallelogram, and then the triangle may cover less than half. The parabola never changes its direction of bending, so it stays on one side of every tangent line.

12. A tank fills at a rate of \(r(t) = 3t^2\) litres per minute. Using the pattern of 1.5, guess how much water enters in the first 2 minutes, and confirm with rectangles.

We need a function whose rate of change is \(3t^2\): from 1.5, \(t^3\) works (\(\frac{(t+h)^3 - t^3}{h} \to 3t^2\)). So the volume is \(2^3 - 0^3 = 8\) litres. Rectangles: the area under \(3t^2\) on \([0, 2]\) is \(3 \times \tfrac{2^3}3 = 8\) by the scaling remark in 1.3. Both agree.

What you learned¶

The rate question (speed at an instant) is answered by averaging over a shrinking interval and seeing what the averages approach; geometrically it is the slope of the tangent line.
The area question is answered by trapping the area between lower and upper rectangle sums whose gap shrinks to zero; this gave \(\tfrac13\) for \(y = x^2\) on \([0, 1]\).
The logic behind both is "any other answer would be wrong by a definite amount, and we can rule that amount out".
Archimedes found the area of a parabolic segment, \(\tfrac43\) of its triangle, by exhaustion and a double contradiction, about two thousand years before calculus.
Distance is the area under the speed graph, and speed is the rate of change of distance.
In general, the rate of change of the area under a curve is the height of the curve: the two questions are inverse, which is the fundamental theorem of calculus in embryo.

Next: Chapter 2 makes "approaches" exact: limits, the laws they obey, and the epsilon-delta definition.