Single-Variable-Calculus · Unit 17 · Video 1 · Interactive Practice

The Natural Log Defined as an Area Under 1/t

IKey Formulas

Formula	Name	What it gives
$L(x) = \displaystyle\int_1^x \frac{dt}{t}$ , for $x > 0$	Definition	The area under $1/t$ from $1$ to $x$ when $x > 1$ ; minus the area from $x$ to $1$ when $0 < x < 1$
$L'(x) = \dfrac{1}{x}$ and $L(1) = 0$ , for $x > 0$	FTC 2 + zero-width interval	Together these two facts determine $L$ uniquely
$L''(x) = -\dfrac{1}{x^2} < 0$ , for $x > 0$	Second derivative	With $L' > 0$ : $L$ is increasing and concave down on $(0, \infty)$
$L(ab) = L(a) + L(b)$ and $L\!\left(\tfrac{1}{x}\right) = -L(x)$ , for $a, b, x > 0$	Product rule for $L$	Products become sums; reciprocals change the sign

Key Insight: The logarithm comes first and $e$ is read off its graph: $e$ is the unique number with $L(e) = 1$ , that is $\int_1^e \frac{dt}{t} = 1$ .

IIVisualization 1 — Area Under 1/t and the Graph of L

The signed area under $1/t$ from $1$ to $x$ is the height $L(x)$ ; the height $1/x$ is its slope.

Endpoint x 2.50

IIIVisualization 2 — Why L(ab) = L(a) + L(b)

Stretching the region from $1$ to $b$ by a factor $k$ makes it $k$ times as wide and $1/k$ as tall, so its area never changes.

a 3 b 2 Stretch k 1

IVQuiz Questions

Problem 1 · A Product of Known Logarithms

Given: $L(x) = \int_1^x \frac{dt}{t}$ with $L(2) = p$ and $L(3) = q$ — find $L(18)$ in terms of $p$ and $q$ .

✅ Correct!

18 = 2 \cdot 3 \cdot 3

, and each factor contributes its own

L

L(18) = L(2) + L(3) + L(3) = p + 2q

❌ Not quite.

2p + q = L(2) + L(2) + L(3) = L(2 \cdot 2 \cdot 3) = L(12)

. The factorization of

18

2 \cdot 3 \cdot 3

❌ Not quite.

L

turns a product of inputs into a sum of outputs:

L(ab) = L(a) + L(b)

, never

L(a)\,L(b)

❌ Not quite.

L(9) = L(3 \cdot 3) = L(3) + L(3) = 2q

, not

q^2

— squaring the input doubles

L

; it does not square it.

Show solution

Factor $18 = 2 \cdot 9 = 2 \cdot 3 \cdot 3$ and apply $L(ab) = L(a) + L(b)$ twice:

L(18) = L(2) + L(9) = L(2) + L(3) + L(3) = p + 2q.

Check: $p + 2q \approx 0.69315 + 2(1.09861) = 2.89037$ , and $\int_1^{18} \frac{dt}{t} \approx 2.89037$ .

Problem 2 · A Quotient Below 1

Given: again $L(2) = p$ and $L(3) = q$ — find $L\!\left(\tfrac{2}{9}\right)$ , then identify the area under $y = 1/t$ that it equals.

What is $L\!\left(\tfrac{2}{9}\right)$ ?

Which statement describes $L\!\left(\tfrac{2}{9}\right)$ as an area?

✅ Correct!

L\!\left(\tfrac29\right) = p - 2q

, and since

\tfrac29 < 1

it is minus an ordinary area, so it is negative.

❌ Not quite.

L

turns division into subtraction, not division:

L\!\left(\tfrac{a}{b}\right) = L(a) + L\!\left(\tfrac1b\right) = L(a) - L(b)

❌ Not quite.

2q - p = L(9) - L(2) = L\!\left(\tfrac92\right)

— the reciprocal of

\tfrac29

, so the sign is reversed.

❌ Not quite.

-p - 2q = -L(18) = L\!\left(\tfrac{1}{18}\right)

. Only the denominator

9

picks up the minus sign; the numerator

2

contributes

+p

❌ Not quite. Right interval, wrong sign:

\int_1^{2/9} \frac{dt}{t} = -\int_{2/9}^1 \frac{dt}{t}

, because the limits run backwards.

❌ Not quite. The integral defining

L

starts at

t = 1

, not

t = 0

; the area under

1/t

near

0

is not even finite.

❌ Not quite. That area is

L\!\left(\tfrac92\right) = -L\!\left(\tfrac29\right)

: the right size, but positive, while

L

is negative on

(0, 1)

Show solution

Write $\tfrac{2}{9} = 2 \cdot \tfrac{1}{9}$ and use $L(ab) = L(a) + L(b)$ and $L\!\left(\tfrac1x\right) = -L(x)$ :

L\!\left(\tfrac{2}{9}\right) = L(2) + L\!\left(\tfrac{1}{9}\right) = L(2) - L(9) = p - 2q.

From the definition, reversing the limits introduces a minus sign:

L\!\left(\tfrac{2}{9}\right) = \int_1^{2/9} \frac{dt}{t} = -\int_{2/9}^{1} \frac{dt}{t},

and $\int_{2/9}^{1} \frac{dt}{t}$ runs from a smaller number to a larger one, so it is an ordinary positive area. Hence $L\!\left(\tfrac29\right) < 0$ ; numerically $p - 2q \approx 0.6931 - 2.1972 = -1.5041$ .

Problem 3 · Both Limits Move

Given: $G(x) = \displaystyle\int_x^{x^3} \frac{dt}{t}$ for $x > 0$ — find $G'(x)$ .

✅ Correct!

G(x) = L(x^3) - L(x)

, so

G'(x) = \frac{3x^2}{x^3} - \frac{1}{x} = \frac{2}{x}

❌ Not quite. The upper limit is

x^3

, not

x

\frac{d}{dx} L(x^3) = L'(x^3) \cdot 3x^2 = \frac{3x^2}{x^3}

. The chain-rule factor

3x^2

is missing.

❌ Not quite. The moving lower limit is subtracted:

G = L(x^3) - L(x)

, so its contribution is

-\frac{1}{x}

, not

+\frac1x

❌ Not quite.

\frac{3}{x}

is the derivative of

L(x^3)

alone; the lower limit

x

also moves and contributes

-\frac{1}{x}

Show solution

Split the integral at $1$ so both pieces are values of $L$ :

G(x) = \int_x^{1} \frac{dt}{t} + \int_1^{x^3} \frac{dt}{t} = -L(x) + L(x^3).

Differentiate with $L'(u) = \frac1u$ and the chain rule:

G'(x) = \frac{1}{x^3} \cdot 3x^2 - \frac{1}{x} = \frac{3}{x} - \frac{1}{x} = \frac{2}{x}.

Check: $L(x^3) - L(x) = L\!\left(\frac{x^3}{x}\right) = L(x^2) = 2L(x)$ , whose derivative is $\frac{2}{x}$ .

Problem 4 · Bounding L(2) to Locate e

Given: $[1, 2]$ split into two subintervals of width $\tfrac{1}{2}$ , with one rectangle on each, of height $1/t$ at an endpoint — find an upper bound for $L(2)$ and decide what it says about $e$ .

Which sum is an upper bound for $L(2)$ , and what is it?

What does this bound prove about $e$ ?

✅ Correct!

L(2) < \tfrac56 < 1 = L(e)

, and an increasing function only takes a smaller value at a smaller input, so

2 < e

❌ Not quite.

\tfrac{7}{12}

is the right-endpoint sum, but

1/t

is decreasing, so right endpoints give each rectangle the smallest height: it is a lower bound.

❌ Not quite. Left endpoints are right, but

1 + \tfrac23 = \tfrac53

adds heights only; each rectangle has width

\tfrac12

❌ Not quite.

\tfrac76 = \tfrac23 + \tfrac12

adds the right-endpoint heights without the widths

\tfrac12

— and the right-endpoint sum is a lower bound anyway, since

1/t

decreases.

❌ Not quite.

L(2) < 1 = L(e)

with

L

increasing means

2

comes before

e

2 < e

❌ Not quite. The conclusion is right but the reason is not: being concave down says nothing by itself about how large

L(2)

is. What this bound proves is

L(2) < \tfrac56 < 1 = L(e)

, and then

L

increasing gives

2 < e

❌ Not quite.

e

is defined by

L(e) = 1

, so comparing

L(2)

with

1

compares

2

with

e

, because

L

is increasing.

Show solution

The subintervals are $[1, \tfrac32]$ and $[\tfrac32, 2]$ . Since $1/t$ is decreasing, the left endpoint gives each rectangle its largest height, so the left-endpoint sum lies above the area:

L(2) < \tfrac{1}{2}\left(\frac{1}{1} + \frac{1}{3/2}\right) = \tfrac{1}{2}\left(1 + \tfrac{2}{3}\right) = \tfrac{5}{6}.

(The right-endpoint sum $\tfrac12\left(\tfrac23 + \tfrac12\right) = \tfrac{7}{12}$ is the matching lower bound, so $\tfrac{7}{12} < L(2) < \tfrac{5}{6}$ .)

By definition $L(e) = 1$ . Then $L(2) < \tfrac56 < 1 = L(e)$ , and because $L' = \tfrac1x > 0$ makes $L$ strictly increasing, $L(2) < L(e)$ forces $2 < e$ .

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