High School Mathematics · Grades 10–12

25Integration

Integration answers two questions at once: what is the area under a curve, and how can one recover a function from its rate of change? The fundamental theorem of calculus states that these are the same question — the deepest and most useful discovery of seventeenth-century mathematics.

25.1 The integral of a continuous function

Definition 25.1 (Integral of a positive function)

Let $f$ be continuous and positive on $\intcc{a}{b}$ . The integral

\int_a^b f(x)\,\dd x

is the area, in units of area, of the region bounded by the curve of $f$ , the $x$ -axis, and the vertical lines $x = a$ and $x = b$ .

For a function of arbitrary sign, areas below the $x$ -axis are counted negatively; and one sets $\int_b^a f(x)\,\dd x = -\int_a^b f(x)\,\dd x$ .

Signed areas: ∈t_02π x\, x = 0, the region below the axis (red) cancelling the region above (blue). — Signed areas: $\int_0^{2\pi} \sin x\,\dd x = 0$ , the region below the axis (red) cancelling the region above (blue).

Proposition 25.2 (Properties of the integral)

Let $f, g$ be continuous on an interval containing $a, b, c$ , and $\lambda \in \R$ .

Linearity: $\displaystyle\int_a^b (f + \lambda g) = \int_a^b f + \lambda \int_a^b g$ .
Chasles relation: $\displaystyle\int_a^c f = \int_a^b f + \int_b^c f$ .
Positivity: if $a \leq b$ and $f \geq 0$ on $\intcc{a}{b}$ , then $\displaystyle\int_a^b f \geq 0$ ; if $f \leq g$ , then $\displaystyle\int_a^b f \leq \int_a^b g$ .

Proof. Chasles and positivity are immediate from the area interpretation (areas add up when regions are juxtaposed; a region of positive height has positive area). The comparison follows by applying positivity to $g - f$ . Linearity is intuitively clear for the sum of positive functions (stack the areas) and is proved rigorously at university; it is admitted here. ∎

25.2 The fundamental theorem of calculus

Definition 25.3 (Primitive)

A primitive (or antiderivative) of $f$ on an interval $I$ is a differentiable function $F$ on $I$ such that $F' = f$ .

Proposition 25.4

If $F$ is a primitive of $f$ on an interval $I$ , the primitives of $f$ on $I$ are exactly the functions $F + c$ , $c \in \R$ . Given $x_0 \in I$ and $y_0 \in \R$ , there is a unique primitive with $F(x_0) = y_0$ .

Proof. If $G' = F' = f$ , then $(G - F)' = 0$ on the interval $I$ , so $G - F$ is constant.¹ Conversely every $F + c$ is a primitive. The condition $F(x_0) = y_0$ pins down the constant. ∎

Theorem 25.5 (Fundamental theorem of calculus)

Let $f$ be continuous on an interval $I$ and $a \in I$ . The function

F \colon x \longmapsto \int_a^x f(t)\,\dd t

is the primitive of $f$ on $I$ vanishing at $a$ . Consequently, for any primitive $G$ of $f$ and $a, b \in I$ :

\int_a^b f(t)\,\dd t = \bigl[G(t)\bigr]_a^b = G(b) - G(a).

The idea of the proof: the increment F(x+h) - F(x) is the area of the narrow strip (orange), squeezed between the rectangles of heights f(x) and f(x+h) over a base of length h. — The idea of the proof: the increment $F(x+h) - F(x)$ is the area of the narrow strip (orange), squeezed between the rectangles of heights $f(x)$ and $f(x+h)$ over a base of length $h$ .

Proof when $f$ is increasing. Fix $x \in I$ and $h > 0$ with $x + h \in I$ . By Chasles,

F(x+h) - F(x) = \int_x^{x+h} f(t)\,\dd t .

Since $f$ is increasing, $f(x) \leq f(t) \leq f(x+h)$ for $t \in \intcc{x}{x+h}$ , and by comparison of integrals (the integral of a constant $c$ over an interval of length $h$ is $ch$ ):

h\,f(x) \leq F(x+h) - F(x) \leq h\,f(x+h),

f(x) \leq \frac{F(x+h) - F(x)}{h} \leq f(x+h).

As $h \to 0^+$ , $f(x+h) \to f(x)$ by continuity, and the squeeze theorem gives that the difference quotient tends to $f(x)$ ; the case $h < 0$ is symmetric. Hence $F' = f$ , and $F(a) = 0$ . The general (non-monotonic) continuous case is proved at university.

Finally, if $G$ is any primitive, $G = F + c$ (Proposition 25.4), so $G(b) - G(a) = F(b) - F(a) = \int_a^b f$ . ∎

Example 25.6

$\displaystyle\int_0^1 x^2\,\dd x = \left[\frac{x^3}{3}\right]_0^1 = \frac13$ : the area under the parabola is one third of the unit square, as Archimedes knew.

The table of primitives is read from the table of derivatives ( $u$ denotes a differentiable function, $c$ an arbitrary constant):

$f(x)$	primitive	$f$	primitive
$x^n \ (n \neq -1)$	$\dfrac{x^{n+1}}{n+1} + c$	$u'u^n \ (n \neq -1)$	$\dfrac{u^{n+1}}{n+1} + c$
$\dfrac1x \ (x > 0)$	$\ln x + c$	$\dfrac{u'}{u} \ (u > 0)$	$\ln u + c$
$\eu^x$	$\eu^x + c$	$u'\eu^u$	$\eu^u + c$
$\cos x$	$\sin x + c$	$\dfrac{u'}{\sqrt u} \ (u>0)$	$2\sqrt u + c$
$\sin x$	$-\cos x + c$

Method 25.7 (Recognizing the form $u' \times (\text{something in } u)$ )

To integrate a product, look for a factor which is the derivative of an inner function $u$ , up to a multiplicative constant. For instance in $\int_0^1 x\,\eu^{x^2}\dd x$ , the factor $x$ is $\frac12 (x^2)'$ :

\int_0^1 x\,\eu^{x^2}\dd x = \frac12\left[\eu^{x^2}\right]_0^1 = \frac{\eu - 1}{2}.

Theorem 25.8 (Integration by parts)

Let $u, v$ be differentiable on $\intcc{a}{b}$ with continuous derivatives. Then

\int_a^b u'(t)\,v(t)\,\dd t = \bigl[u(t)\,v(t)\bigr]_a^b - \int_a^b u(t)\,v'(t)\,\dd t .

Proof. The product rule gives $(uv)' = u'v + uv'$ ; integrating both sides over $\intcc{a}{b}$ and using the fundamental theorem for the left-hand side yields $\bigl[uv\bigr]_a^b = \int_a^b u'v + \int_a^b uv'$ . ∎

Example 25.9

$\displaystyle\int_0^1 t\,\eu^{t}\,\dd t$ : take $u' = \eu^t$ , $v = t$ , so $u = \eu^t$ , $v' = 1$ :

\int_0^1 t\,\eu^t\,\dd t = \bigl[t\,\eu^t\bigr]_0^1 - \int_0^1 \eu^t\,\dd t = \eu - (\eu - 1) = 1 .

25.3 Applications

Definition 25.10 (Mean value)

The mean value of a continuous function $f$ on $\intcc{a}{b}$ ( $a<b$ ) is

\mu = \frac{1}{b-a}\int_a^b f(t)\,\dd t .

Proposition 25.11

If $m \leq f \leq M$ on $\intcc{a}{b}$ , then $m \leq \mu \leq M$ .

Proof. Integrate the inequalities $m \leq f(t) \leq M$ over $\intcc{a}{b}$ and divide by $b - a > 0$ . ∎

Method 25.12 (Area between two curves)

If $f \geq g$ on $\intcc{a}{b}$ , the area between the two curves is $\int_a^b \bigl(f(x) - g(x)\bigr)\dd x$ . If the curves cross, split the interval at the crossing points and integrate $\abs{f - g}$ piece by piece.

The area between the line y = x + 1 (red) and the parabola y = x2 - 1 (blue) is ∈t_-12 ((x+1) - (x2-1) ) x = 9/2. — The area between the line $y = x + 1$ (red) and the parabola $y = x^2 - 1$ (blue) is $\int_{-1}^{2} \bigl((x+1) - (x^2-1)\bigr)\dd x = \frac92$ .

25.4 Exercises

Exercise 25.1 ★

Compute

\int_1^2 \left(3x^2 - \frac{1}{x^2}\right)\dd x, \qquad \int_0^{\pi/2} \cos t \,\dd t, \qquad \int_0^{1} \frac{\dd t}{2t+1} .

Solution

Solution of Exercise 25.1.

\int_1^2 \left(3x^2 - \frac{1}{x^2}\right)\dd x = \left[x^3 + \frac1x\right]_1^2 = \left(8 + \tfrac12\right) - 2 = \frac{13}{2}.

\int_0^{\pi/2} \cos t\,\dd t = \bigl[\sin t\bigr]_0^{\pi/2} = 1 .

With $u = 2t+1$ , $\frac{1}{2t+1} = \frac12\,\frac{u'}{u}$ :

\int_0^1 \frac{\dd t}{2t+1} = \frac12\bigl[\ln(2t+1)\bigr]_0^1 = \frac{\ln 3}{2}.

Exercise 25.2 ★

Find the primitive $F$ of $f(x) = x\eu^{x^2}$ on $\R$ such that $F(0) = 1$ .

Solution

Solution of Exercise 25.2.

$x\,\eu^{x^2} = \frac12\,(x^2)'\,\eu^{x^2}$ has primitives $\frac12\eu^{x^2} + c$ . The condition $F(0) = 1$ gives $\frac12 + c = 1$ , so

F(x) = \frac{\eu^{x^2} + 1}{2}.

Exercise 25.3 ★

Compute the mean value of $f(t) = \sin t$ on $\intcc{0}{\pi}$ , and interpret the result on a graph.

Solution

Solution of Exercise 25.3.

\mu = \frac{1}{\pi}\int_0^\pi \sin t\,\dd t = \frac{1}{\pi}\bigl[-\cos t\bigr]_0^\pi = \frac{2}{\pi} \approx 0.64 .

Graphically, the rectangle of base $\intcc{0}{\pi}$ and height $\frac{2}{\pi}$ has the same area as the arch of the sine curve.

Exercise 25.4 ★★

Using integration by parts, compute

\int_1^{\eu} \ln t\,\dd t \qquad\text{and}\qquad \int_0^{\pi} t \sin t\,\dd t .

Solution

Solution of Exercise 25.4.

Take $u' = 1$ , $v = \ln t$ , so $u = t$ , $v' = \frac1t$ :

\int_1^{\eu} \ln t\,\dd t = \bigl[t\ln t\bigr]_1^{\eu} - \int_1^{\eu} 1\,\dd t = \eu - (\eu - 1) = 1 .

Take $u' = \sin t$ , $v = t$ , so $u = -\cos t$ , $v' = 1$ :

\int_0^\pi t\sin t\,\dd t = \bigl[-t\cos t\bigr]_0^\pi + \int_0^\pi \cos t\,\dd t = \pi + 0 = \pi .

Exercise 25.5 ★★

Compute the area of the region between the parabola $y = x^2$ and the line $y = x + 2$ .

Solution

Solution of Exercise 25.5.

The curves intersect where $x^2 = x + 2$ , i.e. $x \in \{-1, 2\}$ , and on $\intcc{-1}{2}$ the line is above the parabola. The area is

\int_{-1}^{2} \bigl(x + 2 - x^2\bigr)\dd x = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2} = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{9}{2}.

Exercise 25.6 ★★

Let $I = \displaystyle\int_0^1 \frac{\dd t}{1 + t}$ .

Compute $I$ .
For $n \in \N$ , let $I_n = \displaystyle\int_0^1 \frac{t^n}{1+t}\dd t$ . Show that $I_n + I_{n+1} = \dfrac{1}{n+1}$ , and that $0 \leq I_n \leq \dfrac{1}{n+1}$ .
Deduce that $1 - \frac12 + \frac13 - \dots + \frac{(-1)^{n-1}}{n} \xrightarrow[n \to +\infty]{} \ln 2$ .

Solution

Solution of Exercise 25.6.

1. $I = \bigl[\ln(1+t)\bigr]_0^1 = \ln 2$ .

I_n + I_{n+1} = \int_0^1 \frac{t^n(1 + t)}{1+t}\,\dd t = \int_0^1 t^n \,\dd t = \frac{1}{n+1}.

On $\intcc{0}{1}$ , $0 \leq \dfrac{t^n}{1+t} \leq t^n$ , so by comparison $0 \leq I_n \leq \frac{1}{n+1}$ .

3. From the recurrence, by a telescoping induction,

I_n = (-1)^n\left(I_0 - \left(1 - \frac12 + \dots + \frac{(-1)^{n-1}}{n}\right)\right),

i.e. $\displaystyle\sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} = \ln 2 - (-1)^n I_n$ (recall $I_0 = \ln 2$ ). Since $\abs{I_n} \leq \frac{1}{n+1} \to 0$ , the alternating harmonic sum tends to $\ln 2$ .

Exercise 25.7 ★★

The speed of a train (in m/s) during the first $100$ seconds after departure is modeled by $v(t) = 30\bigl(1 - \eu^{-t/50}\bigr)$ . Compute the distance traveled during these $100$ seconds, and the train’s average speed over the interval.

Solution

Solution of Exercise 25.7.

Distance:

d = \int_0^{100} 30\left(1 - \eu^{-t/50}\right)\dd t = 30\left[t + 50\,\eu^{-t/50}\right]_0^{100} = 30\left(100 + 50\eu^{-2} - 50\right) = 1500 + 1500\,\eu^{-2}.

Numerically $d \approx 1500 + 203 = 1703$ m. Average speed: $\frac{d}{100} = 15\bigl(1 + \eu^{-2}\bigr) \approx 17.0$ m/s.

Exercise 25.8 ★★★

For $n \geq 1$ , let $S_n = \dfrac1n \displaystyle\sum_{k=1}^{n} \frac{1}{1 + k/n}$ .

Interpret $S_n$ as an area of rectangles approximating a region under the curve of $t \mapsto \frac{1}{1+t}$ on $\intcc{0}{1}$ .
Using the monotonicity of $t \mapsto \frac{1}{1+t}$ , show that

$S_n \leq \int_0^1 \frac{\dd t}{1+t} \leq S_n + \frac1n\left(1 - \frac12\right),$

and deduce $\lim\limits_{n\to+\infty} S_n = \ln 2$ .

Solution

Solution of Exercise 25.8.

1. $S_n$ is the total area of $n$ rectangles of width $\frac1n$ and heights $f\left(\frac kn\right)$ , $k = 1, \dots, n$ , where $f(t) = \frac{1}{1+t}$ : it is the “right endpoint” rectangle approximation of the area under $f$ on $\intcc{0}{1}$ .

2. $f$ is decreasing, so on each interval $\intcc{\frac{k-1}{n}}{\frac kn}$ , $f\left(\frac kn\right) \leq f(t) \leq f\left(\frac{k-1}{n}\right)$ ; integrating and summing over $k$ :

S_n \leq \int_0^1 f(t)\,\dd t \leq S_n + \frac1n\bigl(f(0) - f(1)\bigr) = S_n + \frac{1}{2n}.

(The middle term of the exercise’s display, with $f(0) - f(1) = 1 - \frac12$ , is exactly this bound.) Hence $0 \leq \ln 2 - S_n \leq \frac{1}{2n} \to 0$ , so $S_n \to \ln 2$ .

Exercise 25.9 ★★★

(Wallis integrals.) For $n \in \N$ , let $W_n = \displaystyle\int_0^{\pi/2} \sin^n t\,\dd t$ .

Compute $W_0$ and $W_1$ .
Writing $\sin^{n+2}t = \sin t \cdot \sin^{n+1} t$ and integrating by parts, show that $W_{n+2} = \dfrac{n+1}{n+2}\,W_n$ .
Deduce $W_2$ , $W_3$ , $W_4$ and show that $(W_n)$ is decreasing and positive.

Solution

Solution of Exercise 25.9.

1. $W_0 = \int_0^{\pi/2} \dd t = \frac{\pi}{2}$ and $W_1 = \bigl[-\cos t\bigr]_0^{\pi/2} = 1$ .

2. Integrate by parts with $u' = \sin t$ , $v = \sin^{n+1} t$ , so $u = -\cos t$ , $v' = (n+1)\sin^n t \cos t$ :

W_{n+2} = \bigl[-\cos t \sin^{n+1} t\bigr]_0^{\pi/2} + (n+1)\int_0^{\pi/2} \cos^2 t\,\sin^n t\,\dd t = (n+1)\int_0^{\pi/2} (1 - \sin^2 t)\sin^n t\,\dd t,

the bracket vanishing at both ends. Hence $W_{n+2} = (n+1)(W_n - W_{n+2})$ , i.e. $W_{n+2} = \frac{n+1}{n+2} W_n$ .

3. $W_2 = \frac12 W_0 = \frac{\pi}{4}$ , $W_3 = \frac23 W_1 = \frac23$ , $W_4 = \frac34 W_2 = \frac{3\pi}{16}$ . On $\intoo{0}{\frac\pi2}$ , $0 < \sin t < 1$ , so $\sin^{n+1} t \leq \sin^n t$ with strict inequality on the interior; integrating, $0 < W_{n+1} \leq W_n$ (in fact $<$ ): $(W_n)$ is decreasing and positive.

25.5 Problem: Archimedes against the machine

Problem 25.1

Weekend problem — the area under the parabola, computed three ways across twenty-two centuries, and the logarithm’s secret identity as an area

Around 240 BC, Archimedes computed the exact area of a parabolic segment — without coordinates, without limits, without algebra. Nineteen centuries later, Riemann’s rectangles redid it by brute squeeze; and the fundamental theorem of calculus (Theorem 25.5) now does it in one line. This problem plays all three matches, then uses the same machine to reveal what the logarithm really is: an area with a scaling symmetry.

Part I — Fluency.

Compute $\displaystyle\int_0^1 (3x^2 - 2x + 1)\,\dd x$ , $\displaystyle\int_1^{\eu} \frac{\dd x}{x}$ , and $\displaystyle\int_0^{\pi/2} \cos x\,\dd x$ .
Spot the $u'u$ -forms (Method 25.7): $\displaystyle\int_0^1 x\,\eu^{x^2}\dd x$ and $\displaystyle\int_0^1 \frac{2x}{x^2 + 1}\,\dd x$ .
By parts (Theorem 25.8): $\displaystyle\int_0^1 x\,\eu^x \dd x$ and $\displaystyle\int_1^{\eu} \ln x\,\dd x$ .
Compute the mean value (Definition 25.10) of $\sin$ over $\intcc{0}{\pi}$ — and note it is not $\frac12$ .
Compute the area between the line $y = x$ and the parabola $y = x^2$ over $\intcc{0}{1}$ (Method 25.12).

Part II — The parabola, three ways.

Riemann’s setup for the area under $y = x^2$ on $\intcc{0}{1}$ : write the lower sum $L_n$ and the upper sum $U_n$ over $n$ equal rectangles (as in Exercise 25.8).
Prove by induction (Problem 20.1’s machine) the sum-of-squares formula

$1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6} .$
Deduce closed forms for $U_n$ and $L_n$ , compute their common limit, and conclude: the area is $\frac13$ .
Now the machine: compute $\int_0^1 x^2\,\dd x$ by the fundamental theorem, in one line. Compare the efforts.
Archimedes stated it otherwise: a parabolic segment is $\frac43$ of its inscribed triangle. For the segment cut from $y = x^2$ by the chord joining $(-1, 1)$ to $(1, 1)$ : compute the segment’s area with an integral, the inscribed triangle’s area (apex at the vertex $(0, 0)$ ), and verify the master’s ratio.
Archimedes’ own method: fill the segment with the big triangle $T$ , then two triangles totalling $\frac T4$ , then four totalling $\frac{T}{16}$ , and so on. Sum the geometric series and recover $\frac43 T$ — the endlessly bitten chocolate bar of the Middle School volume, eaten by a Greek geometer.
One sentence each: exhaustion (Archimedes), squeeze (Riemann), anti-differentiation (Newton–Leibniz) — what does each need, and what does each give?

Part III — The logarithm is an area. For $x > 0$ , set $A(x) = \displaystyle\int_1^x \frac{\dd t}{t}$ .

Give $A(1)$ and $A'(x)$ (Theorem 25.5), and conclude that $A$ is exactly the natural logarithm of Definition 23.5.
The scaling miracle: fix $a > 0$ and study $g(x) = A(ax) - A(x)$ . Compute $g'$ (chain rule), deduce that $g$ is constant, evaluate the constant — and conclude the functional equation

$\ln(ab) = \ln a + \ln b ,$

proved by pure calculus: the area from $1$ to $ab$ splits into scaled copies.
Deduce from question 14: $\ln(a^n) = n\ln a$ and $\ln\frac1a = -\ln a$ .
Exercise 25.8 reads $\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n}$ as rectangles under $\frac{1}{1 + t}$ : compute this sum for $n = 10$ (three decimals) and compare with $\ln 2$ . Which harmonic-flavored sums, divergent term by term, here converge to an area?

Part IV — Accumulation.

A car accelerates with speed $v(t) = 3t^2$ m/s for $t \in \intcc{0}{10}$ . Compute the distance covered, the mean speed, and the instant at which the instantaneous speed equals the mean speed.
A $4$ -meter rod has linear density $\rho(x) = 2 + x$ kg/m. Compute its total mass and its center of mass $\dfrac{\int_0^4 x\,\rho(x)\,\dd x}{\int_0^4 \rho(x)\,\dd x}$ — the balance point of the Middle School volume’s cardboard triangle, finally computed with weights that vary.
Wallis’s integrals (Exercise 25.9): compute $W_0$ , $W_1$ , and $W_2 = \int_0^{\pi/2} \sin^2 t\,\dd t$ using the linearization of Problem 24.1. (Their infinite ladder climbs, in the university volumes, to a product formula for $\pi$ and to the normalization of the bell curve.)
Finale — integration’s three faces: an area by definition, an accumulation in the world (distance, mass), an anti-derivative by the fundamental theorem; and its history in three names. Close with the chapter’s pearl: which everyday button on the calculator is secretly the area under $\frac1t$ — and which identity did that area prove?

Solution

Solution of Problem 25.1.

1. $\left[x^3 - x^2 + x\right]_0^1 = 1$ ; $\left[\ln x\right]_1^{\eu} = 1$ ; $\left[\sin x\right]_0^{\pi/2} = 1$ .

2. $\left[\frac12 \eu^{x^2}\right]_0^1 = \frac{\eu - 1}{2}$ ; $\left[\ln(x^2 + 1)\right]_0^1 = \ln 2$ .

3. $\int_0^1 x\eu^x = \left[x\eu^x\right]_0^1 - \int_0^1 \eu^x = \eu - (\eu - 1) = 1$ . And $\int_1^{\eu} \ln x = \left[x\ln x\right]_1^{\eu} - \int_1^{\eu} 1 = \eu - (\eu - 1) = 1$ .

4. $\frac{1}{\pi}\int_0^\pi \sin x\,\dd x = \frac{2}{\pi} \approx 0.64$ : the average height of the sine arch is $\frac2\pi$ , noticeably more than one half — the arch is fat near its top.

5. $\int_0^1 (x - x^2)\,\dd x = \frac12 - \frac13 = \frac16$ .

6. $L_n = \frac1n \sum_{k=0}^{n-1} \left(\frac kn\right)^2$ and $U_n = \frac1n \sum_{k=1}^{n} \left(\frac kn\right)^2$ : rectangles of width $\frac1n$ with heights read at the left (below the curve) or right (above) endpoints.

7. True for $n = 1$ ( $1 = \frac{1 \cdot 2 \cdot 3}{6}$ ). If it holds for $n$ , then adding $(n+1)^2$ :

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

the formula at $n + 1$ : heredity, done.

8. $U_n = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(1 + \frac1n)(2 + \frac1n)}{6} \to \frac26 = \frac13$ , and $L_n = U_n - \frac1n \to \frac13$ : squeezed between its staircases, the area is exactly $\frac13$ .

9. $\int_0^1 x^2\,\dd x = \left[\frac{x^3}{3}\right]_0^1 = \frac13$ : one line. The fundamental theorem converts an infinite squeeze into a single anti-derivative evaluation — that is why it is called fundamental.

10. Segment: $\int_{-1}^{1}(1 - x^2)\,\dd x = 2 - \frac23 = \frac43$ . Triangle with base the chord (length $2$ , at height $1$ ) and apex $(0,0)$ : area $\frac12 \times 2 \times 1 = 1$ . Ratio: $\frac43$ — exactly Archimedes’ theorem, verified by the machine he lacked.

11. $T\left(1 + \frac14 + \frac{1}{16} + \dots\right) = T \cdot \frac{1}{1 - \frac14} = \frac43 T$ (geometric series of ratio $\frac14$ ): with $T = 1$ , the segment’s area $\frac43$ again. Archimedes summed the series by a purely geometric argument — the quarter-bites of the chocolate-bar kind, two millennia before limits were words.

12. Exhaustion needs a clever geometric self-similarity and gives exact areas case by case; Riemann needs only monotonicity and gives a universal definition at the price of computing sums; the fundamental theorem needs the whole edifice of derivatives and repays it by making areas one-line computations.

13. $A(1) = 0$ and $A'(x) = \frac1x$ (fundamental theorem): $A$ is the primitive of $\frac1x$ on $\intoo{0}{+\infty}$ vanishing at $1$ — which is precisely the natural logarithm.

14. $g'(x) = \frac{a}{ax} - \frac1x = 0$ : $g$ is constant; at $x = 1$ : $g(1) = A(a) - A(1) = A(a)$ . Hence $A(ax) = A(a) + A(x)$ for all $x$ , and with $x = b$ : $\ln(ab) = \ln a + \ln b$ . The hyperbola’s area does not care about horizontal scale: stretching $t$ by $a$ squashes $\frac1t$ by $a$ , and the area survives — addition of logarithms is a scaling symmetry.

15. Iterating: $\ln(a^n) = n\ln a$ (induction on $n$ ). And $0 = \ln 1 = \ln\left(a \cdot \frac1a\right) = \ln a + \ln\frac1a$ : $\ln\frac1a = -\ln a$ .

16. $\frac{1}{11} + \frac{1}{12} + \dots + \frac{1}{20} \approx 0.669$ , against $\ln 2 \approx 0.693$ : ten rectangles under $\frac{1}{1+t}$ , already close. The harmonic series diverges, but its slices from $n$ to $2n$ settle on $\ln 2$ : divergence at a crawl, measured by an area.

17. Distance: $\int_0^{10} 3t^2\,\dd t = 1000$ m. Mean speed: $100$ m/s. Instant with $v = 100$ : $3t^2 = 100$ : $t = \sqrt{\frac{100}{3}} \approx 5.77$ s.

18. Mass: $\int_0^4 (2 + x)\,\dd x = 8 + 8 = 16$ kg. Moment: $\int_0^4 (2x + x^2)\,\dd x = 16 + \frac{64}{3} = \frac{112}{3}$ . Center of mass: $\frac{112/3}{16} = \frac73 \approx 2.33$ m — pushed beyond the middle by the heavier right end, as intuition demands.

19. $W_0 = \frac\pi2$ ; $W_1 = \left[-\cos t\right]_0^{\pi/2} = 1$ ; and $W_2 = \int_0^{\pi/2} \frac{1 - \cos 2t}{2}\,\dd t = \frac\pi4$ .

20. Area: the definition, squeezed by staircases. Accumulation: distances from speeds, masses from densities — the integral as a running total. Anti-derivative: the fundamental theorem, Newton’s and Leibniz’s shortcut through Archimedes’ patience. The pearl: the $\ln$ button is the area under $\frac1t$ , and that area’s scaling symmetry is the identity $\ln(ab) = \ln a + \ln b$ .

That a function with zero derivative on an interval is constant follows from Theorem 22.7: it is both increasing and decreasing. ↩