Introductory calculus · The Lumenary

Lens

MotivationH

Two questions about motion

A stone is dropped from a bridge. Ignoring air resistance, after $t$ seconds it has fallen about $s(t) = 4.9\,t^2$ metres. How fast is it falling at exactly $t = 1$ ?

Speed is distance divided by time — but at a single instant no time passes and no distance is covered, so "distance ÷ time" reads $\tfrac00$ . What we can compute is the average speed over a short interval from $t = 1$ to $t = 1 + h$ :

\frac{s(1 + h) - s(1)}{h} = \frac{4.9(1 + 2h + h^2) - 4.9}{h} = \frac{9.8h + 4.9h^2}{h} = 9.8 + 4.9h \qquad (h \ne 0).

Over $0.1$ s this is $10.29$ m/s; over $0.001$ s it is $9.8049$ m/s. As $h$ shrinks, the averages close in on $9.8$ m/s. That number — the value the averages approach — is the speed at the instant $t = 1$ .

The opposite question is just as old: a car's speedometer is recorded every moment of a journey from Accra to Kumasi. How far did the car travel? If the speed were constant, distance would be speed × time, the area of a rectangle under the speed–time graph. When the speed varies, the distance is the area under the graph — and computing areas under curves again means adding up more and more, thinner and thinner pieces.

Calculus is the mathematics of these two limits: the derivative (instantaneous rate of change) and the integral (accumulated total). Its central surprise, the fundamental theorem of calculus, is that each undoes the other.

IntuitionH

Zooming in and slicing up

Zoom in on the graph of a smooth function, such as $y = x^2$ near the point $(1, 1)$ . The more you magnify, the straighter the curve looks; under a strong enough microscope it is indistinguishable from a straight line, the tangent line. The slope of that line is the derivative: how fast $y$ changes per unit change of $x$ , right at that point.

To find an area, do the opposite: cut the region under the curve into thin vertical strips, replace each strip by a rectangle, and add up the rectangles. Thinner strips give better approximations, and the area is what these sums approach.

Both ideas rest on a single notion: the limit, the value that approximations approach as some quantity ( $h$ , or the strip width) shrinks to $0$ — without ever being $0$ .

DefinitionsH

Definitions

Definition— limit (informal)

We write $\displaystyle\lim_{x \to a} f(x) = L$ ("the limit of $f(x)$ as $x$ tends to $a$ is $L$ ") if $f(x)$ can be made as close to $L$ as we like by taking $x$ close enough to $a$ , but not equal to $a$ . The value $f(a)$ — if it exists at all — plays no part.

The precise version ("for every $\varepsilon > 0$ there is a $\delta > 0$ such that $0 < |x - a| < \delta$ implies $|f(x) - L| < \varepsilon$ ") took mathematicians until the 19th century to find; it is the starting point of real analysis. In this lesson we use the following limit laws, which can be proved from that definition: if $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ , then $f + g \to L + M$ , $f g \to L M$ , $cf \to cL$ , and $f/g \to L/M$ provided $M \ne 0$ ; also $\lim_{x \to a} c = c$ and $\lim_{x \to a} x = a$ . Consequently $\lim_{x \to a} p(x) = p(a)$ for every polynomial $p$ . Limits also respect inequalities: if $f(x) \le g(x)$ near $a$ , then $L \le M$ .

A function is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$ : the value is what the nearby values predict.

Definition— derivative

The derivative of $f$ at $a$ is

f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h},

when this limit exists; then $f$ is differentiable at $a$ . The fraction is the slope of the secant line through $(a, f(a))$ and $(a + h, f(a + h))$ , and $f'(a)$ is the slope of the tangent line $y = f(a) + f'(a)(x - a)$ . The function $x \mapsto f'(x)$ is the derivative function $f'$ , also written $\dfrac{dy}{dx}$ or $\dfrac{d}{dx} f(x)$ .

Definition— Riemann sum and definite integral

Divide $[a, b]$ into $n$ strips of width $\Delta x = \tfrac{b - a}{n}$ and choose a point $x_k^*$ in the $k$ -th strip (its left end, right end or midpoint). The Riemann sum is $\sum_{k} f(x_k^*)\,\Delta x$ . If these sums approach a single number as $n \to \infty$ (for every way of choosing the points), that number is the definite integral $\displaystyle\int_a^b f(x)\,dx$ . Strips where $f < 0$ contribute negatively: the integral is a signed area. Continuous functions on $[a, b]$ always have integrals (a theorem of real analysis).

Definition— antiderivative

A function $F$ is an antiderivative of $f$ on an interval if $F'(x) = f(x)$ for every $x$ in it. For example $\tfrac{x^3}{3}$ , and also $\tfrac{x^3}{3} + 7$ , are antiderivatives of $x^2$ .

Worked examplesH

Worked examples

Example— Limits from tables, and limits by algebra

A table. For $f(x) = \tfrac{\sin x}{x}$ (with $x$ in radians): $f(0.1) \approx 0.99833$ , $f(0.01) \approx 0.9999833$ , $f(-0.01) \approx 0.9999833$ . The table suggests $\lim_{x \to 0} \tfrac{\sin x}{x} = 1$ , which is true — but a table is evidence, not proof (a table can be fooled by a function that misbehaves between the sample points).
Algebra. $\displaystyle\lim_{x \to 1}\frac{x^2 - 1}{x - 1}$ : for $x \ne 1$ , $\tfrac{x^2 - 1}{x - 1} = \tfrac{(x - 1)(x + 1)}{x - 1} = x + 1$ , so the limit is $1 + 1 = 2$ . The function is not even defined at $x = 1$ ; the limit exists anyway.
No limit. $\tfrac1x$ has no limit at $0$ (it grows without bound), and the sign function, $-1$ for $x < 0$ and $1$ for $x > 0$ , has none either: from the left it is near $-1$ , from the right near $1$ .

Example— Derivatives from the definition

For $f(x) = x^2$ and $h \ne 0$ : $\;\dfrac{(a + h)^2 - a^2}{h} = \dfrac{2ah + h^2}{h} = 2a + h \to 2a$ . So $f'(a) = 2a$ : the slope of the parabola doubles as $a$ doubles.

For $f(x) = \tfrac1x$ and $a \ne 0$ : $\;\dfrac{\frac{1}{a + h} - \frac1a}{h} = \dfrac{a - (a + h)}{h\,a(a + h)} = \dfrac{-1}{a(a + h)} \to -\dfrac{1}{a^2}$ .

The tangent line to $y = x^2$ at $a = 3$ has slope $6$ and passes through $(3, 9)$ : $y = 9 + 6(x - 3) = 6x - 9$ .

Example— Finding the turning points

Let $f(x) = x^3 - 3x$ . Then $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$ , which is $0$ at $x = \pm 1$ . For $x < -1$ , $f' > 0$ (rising); for $-1 < x < 1$ , $f' < 0$ (falling); for $x > 1$ , $f' > 0$ again. So $f$ has a local maximum $f(-1) = 2$ and a local minimum $f(1) = -2$ .

Example— An optimisation

A farmer in Malawi has $20$ m of fencing for a rectangular goat pen. With sides $x$ and $10 - x$ , the area is $A(x) = x(10 - x) = 10x - x^2$ for $0 \le x \le 10$ . Then $A'(x) = 10 - 2x = 0$ at $x = 5$ , where $A = 25$ ; at the ends $A(0) = A(10) = 0$ . The best pen is a $5 \times 5$ square.

Explore

Explore: secants, tangents and strips

Shrink $h$ and watch the secant through $P$ and $Q$ turn into the tangent. For $f(x) = x^2$ at $x_0 = 1$ the table of slopes should approach $2$ — and you know exactly why, since the slope is $2 + h$ .

From secant to tangent

The slope of the secant is exactly 2 + h. What happens as h shrinks? Then try fn = abs(x - 1).

f(x) = x²
secant through P and Q
tangent line (numerical slope)

Point x₀x₀ = 1Shrink h toward 0h = 1

Slope of the secant PQ

(f(x₀ + h) − f(x₀)) / h = (4 − 1) / 1 = 3

Slope of the tangent at P

f′(1) ≈ 2

Numerical estimate (central difference with a tiny step).

Difference quotients as h shrinks

h	(f(x₀ + h) − f(x₀)) / h
1	3
0.1	2.1
0.01	2.01
0.001	2.001

As h shrinks, Q slides towards P, the secant turns into the tangent, and its slope settles on the derivative f′(x₀).

Now the area under $y = x^2$ from $0$ to $1$ , which we will prove is exactly $\tfrac13$ . Compare left, right and midpoint sums, and watch the errors shrink as the strips multiply.

Riemann sums for the parabola

Right sums overestimate and left sums underestimate an increasing function. Which rule is most accurate?

f(x) = x²

Number of strips nn = 8

Right sum with 8 strips

R₈ = 0.398438

Exact value of the integral

∫₀¹ f(x) dx = 0.333333

Error (sum − integral)

0.065104

All four rules with n = 8

Rule	Sum	Error
Left (L₈)	0.273438	−0.059896
Right (R₈)	0.398438	0.065104
Midpoint (M₈)	0.332031	−0.001302
Trapezoid (T₈)	0.335938	0.002604

Double n and watch the error: left and right sums roughly halve it, midpoint and trapezoid sums cut it to about a quarter.

Finally, $x^3$ at $0$ : the tangent is horizontal, yet the graph keeps climbing through it.

Horizontal tangents

Move the sliders and trace along the curves to read off values.

y = x³
tangent to x³ at 0
y = x³ − 3x

Trace along xx = 1

y = x³: y = 1
tangent to x³ at 0: y = 0
y = x³ − 3x: y = −2

ProofH

The rules of differentiation

Theorem— Theorem 1 (differentiable implies continuous)

If $f$ is differentiable at $a$ , then $\lim_{h \to 0} f(a + h) = f(a)$ .

Proof

For $h \ne 0$ , $\;f(a + h) - f(a) = \dfrac{f(a + h) - f(a)}{h}\cdot h$ . As $h \to 0$ the first factor tends to $f'(a)$ and the second to $0$ , so by the product law the difference tends to $f'(a)\cdot 0 = 0$ .

∎

The converse is false: $|x|$ is continuous at $0$ , but its difference quotient $\tfrac{|h|}{h}$ is $1$ for $h > 0$ and $-1$ for $h < 0$ , so it has no limit. A graph with a corner has no tangent there.

Theorem— Theorem 2 (sums and constant multiples)

If $f$ and $g$ are differentiable at $a$ and $c$ is a constant, then $(f + g)'(a) = f'(a) + g'(a)$ and $(cf)'(a) = c f'(a)$ .

Proof

The difference quotient of $f + g$ is the sum of the difference quotients of $f$ and $g$ , and that of $cf$ is $c$ times that of $f$ ; apply the limit laws.

∎

Theorem— Theorem 3 (product rule)

If $f$ and $g$ are differentiable at $a$ , then $(fg)'(a) = f'(a)g(a) + f(a)g'(a)$ .

Proof

Add and subtract $f(a + h)g(a)$ :

\frac{f(a + h)g(a + h) - f(a)g(a)}{h} = f(a + h)\,\frac{g(a + h) - g(a)}{h} + g(a)\,\frac{f(a + h) - f(a)}{h}.

As $h \to 0$ : $f(a + h) \to f(a)$ by Theorem 1, and the two quotients tend to $g'(a)$ and $f'(a)$ . By the limit laws the whole expression tends to $f(a)g'(a) + g(a)f'(a)$ .

∎

Theorem— Theorem 4 (power rule)

For every integer $n \ge 1$ : $\;\dfrac{d}{dx} x^n = n x^{n-1}$ .

Proof

Induction on $n$ . For $n = 1$ , the difference quotient of $x$ is $\tfrac{(a + h) - a}{h} = 1$ , so the derivative is $1 = 1 \cdot x^0$ . Suppose $\tfrac{d}{dx}x^n = nx^{n-1}$ . Writing $x^{n+1} = x^n \cdot x$ , the product rule gives

\frac{d}{dx}x^{n+1} = nx^{n-1}\cdot x + x^n \cdot 1 = (n + 1)x^n.

∎

With Theorem 2, every polynomial can now be differentiated term by term: $\tfrac{d}{dx}(4x^3 - 5x + 2) = 12x^2 - 5$ . (The constant $2$ has derivative $0$ : its difference quotient is $\tfrac{2 - 2}{h} = 0$ .)

Theorem— Theorem 5 (Fermat's theorem on interior extrema)

Suppose $f$ is defined on an open interval containing $a$ , has a local maximum or minimum at $a$ , and is differentiable at $a$ . Then $f'(a) = 0$ .

Proof

Take a local maximum: $f(a + h) \le f(a)$ for all small $h$ . For small $h > 0$ the difference quotient $\tfrac{f(a + h) - f(a)}{h}$ has numerator $\le 0$ and denominator $> 0$ , so it is $\le 0$ ; since limits respect inequalities, $f'(a) \le 0$ . For small $h < 0$ the numerator is $\le 0$ and the denominator $< 0$ , so the quotient is $\ge 0$ , giving $f'(a) \ge 0$ . Hence $f'(a) = 0$ . For a minimum, apply this to $-f$ .

∎

The converse is false. For $f(x) = x^3$ , $f'(0) = 3 \cdot 0^2 = 0$ , but $0$ is neither a maximum nor a minimum: $t^3 > 0$ for every $t > 0$ and $(-t)^3 < 0$ , however small $t$ is. A zero derivative only marks a candidate; to decide, look at the sign of $f'$ on each side (as in the worked example) or compare values.

ProofH

Area under a parabola, and the fundamental theorem

Archimedes found the area of a parabolic segment more than 2,200 years ago. Here is the modern version of that result, by Riemann sums.

Theorem— Lemma 6 (sum of squares)

For every $n \ge 0$ : $\;0^2 + 1^2 + 2^2 + \dots + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}$ .

Proof

Induction on $n$ . For $n = 0$ both sides are $0$ . If the formula holds for $n$ , then

\frac{n(n + 1)(2n + 1)}{6} + (n + 1)^2 = \frac{(n + 1)\big(2n^2 + n + 6n + 6\big)}{6} = \frac{(n + 1)(n + 2)(2n + 3)}{6},

which is the formula for $n + 1$ , because $2n^2 + 7n + 6 = (n + 2)(2n + 3)$ .

∎

Theorem— Theorem 7 (the area under y = x² on [0, 1] is 1/3)

The right and left Riemann sums of $f(x) = x^2$ on $[0, 1]$ with $n$ strips are

R_n = \frac{(n + 1)(2n + 1)}{6n^2}, \qquad L_n = \frac{(n - 1)(2n - 1)}{6n^2},

and both tend to $\tfrac13$ as $n \to \infty$ . Hence $\displaystyle\int_0^1 x^2\,dx = \frac13$ .

Proof

The $k$ -th strip is $\big[\tfrac{k - 1}{n}, \tfrac kn\big]$ , of width $\tfrac1n$ . Using right endpoints and Lemma 6,

R_n = \sum_{k=1}^{n} \Big(\frac kn\Big)^2 \frac1n = \frac{1^2 + \dots + n^2}{n^3} = \frac{n(n + 1)(2n + 1)}{6n^3} = \frac{(n + 1)(2n + 1)}{6n^2}.

With left endpoints we get $\tfrac{0^2 + \dots + (n - 1)^2}{n^3} = \tfrac{(n - 1)(2n - 1)}{6n^2}$ in the same way. Dividing top and bottom by $n^2$ ,

R_n = \frac{\big(1 + \frac1n\big)\big(2 + \frac1n\big)}{6} \to \frac{1 \cdot 2}{6} = \frac13, \qquad L_n = \frac{\big(1 - \frac1n\big)\big(2 - \frac1n\big)}{6} \to \frac13.

Since $x^2$ is increasing on $[0, 1]$ , every Riemann sum with $n$ strips lies between $L_n$ and $R_n$ (each strip's height lies between the heights at its two ends), so every Riemann sum tends to $\tfrac13$ .

∎

Computing every area this way would be exhausting. The fundamental theorem gives a shortcut.

Theorem— Fundamental theorem of calculus

If $f$ is continuous on $[a, b]$ and $F$ is an antiderivative of $f$ there, then

\int_a^b f(x)\,dx = F(b) - F(a).

We give the idea rather than a complete proof (which belongs to real analysis). Let $A(x) = \int_a^x f(t)\,dt$ be the area accumulated from $a$ up to $x$ . Increasing $x$ by a small $h$ adds a thin strip of width $h$ and height about $f(x)$ , so $A(x + h) - A(x) \approx f(x)\,h$ , and dividing by $h$ suggests $A'(x) = f(x)$ — making this precise uses the continuity of $f$ . So $A$ and $F$ have the same derivative, and two functions with the same derivative on an interval differ by a constant (a consequence of the mean value theorem). Hence $A(x) = F(x) + C$ ; putting $x = a$ gives $0 = F(a) + C$ , and putting $x = b$ gives $\int_a^b f = F(b) - F(a)$ .

Example— Integrals by antiderivatives

$\displaystyle\int_0^1 x^2\,dx = \Big[\tfrac{x^3}{3}\Big]_0^1 = \tfrac13$ , matching Theorem 7 — two very different routes to the same number.
$\displaystyle\int_0^2 x^2\,dx = \tfrac83 - 0 = \tfrac83$ .
$\displaystyle\int_0^{\pi} \sin x\,dx = \big[-\cos x\big]_0^{\pi} = 1 + 1 = 2$ , but $\displaystyle\int_0^{2\pi} \sin x\,dx = -\cos 2\pi + \cos 0 = 0$ : the area below the axis on $[\pi, 2\pi]$ counts negatively and cancels the area above.

Formal mathematicsF

Calculus in Lean

Mathlib states that $f$ has derivative $f'$ at $a$ as HasDerivAt f f' a, and writes limits with filters: Tendsto g (𝓝[≠] 0) (𝓝 L) says " $g(h) \to L$ as $h \to 0$ with $h \ne 0$ ". First the derivative of $x^2$ straight from the definition, in three steps that mirror the human computation:

diff_quotient_sqLean 4Proved here from basic factsFormally verified

/-- **The difference quotient of `x²`.** For `h ≠ 0`,
`((a + h)² − a²)/h = (2ah + h²)/h = 2a + h`. -/
theorem diff_quotient_sq (a h : ℝ) (hh : h ≠ 0) : ((a + h) ^ 2 - a ^ 2) / h = 2 * a + h := by
  field_simp
  ring