Chapter 7 · Integrals

Exercise 7.8 · Q2: $\int_2^3\frac{1}{x}\,dx$

Exercise 7.8Fundamental theorem of calculus · Section 7.8

The question

Evaluate the definite integral

\int_2^3\frac{1}{x}\,dx.

Exam answer

Let $I = \displaystyle\int_2^3\frac{dx}{x}$ ⁠.

Using $\displaystyle\int\frac{dx}{x} = \log|x| + C,$

\begin{aligned} I &= \Big[\log x\Big]_2^3 \\ &= \log 3 - \log 2 \end{aligned}

$\therefore\ I = \log\dfrac32$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The second fundamental theorem (Theorem 2, Section 7.8.3): for $f$ continuous on $[a, b]$ with antiderivative $F$ ⁠, $\int_a^b f(x)\,dx = F(b) - F(a)$ ⁠. The antiderivative of $\frac1x$ is $\log|x|$ ⁠, and on $[2, 3]$ the number $x$ is positive, so the bars can be dropped. The answer $\log 3 - \log 2$ then combines into one logarithm.

Given, to find, and the plan

Given: The integrand $f(x) = \frac1x$ on the interval $[2, 3]$ ⁠.
To find: The exact value of $\int_2^3\frac{dx}{x}$ ⁠.
Plan: (1) Check $f$ is continuous on $[2, 3]$ ⁠. (2) Antiderivative $\log|x|$ ⁠, checked by differentiating. (3) Compute $F(3) - F(2)$ and combine the logs. (4) Check the size of the answer against the graph and a numerical integration.

Basics you need

1Second fundamental theorem

$\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a)$ when $f$ is continuous on $[a, b]$ and $F' = f$ ⁠. Lesson: definite integrals and the fundamental theorem

2The integral of $\frac1x$

$\int\frac{dx}{x} = \log|x| + C$ for $x \ne 0$ (standard formula (xii), Section 7.2). In this book "log" means the natural logarithm, base e. Basics: logarithms and 1/x

3Laws of logarithms

$\log a - \log b = \log\frac{a}{b}$ for $a, b > 0$ ⁠. Basics: logarithms

4Area meaning

For $f \ge 0$ on $[a, b]$ ⁠, $\int_a^b f$ is the area under the graph (Section 7.8.1). Basics: area as a sum of thin strips

Solution

Check continuity on $[2, 3]$

$f(x) = \frac1x$ is defined and continuous at every $x \ne 0$ ⁠. The only bad point, $x = 0$ ⁠, is not in $[2, 3]$ ⁠. So $f$ is continuous on the whole closed interval $[2, 3]$ ⁠.

A quotient of continuous functions is continuous wherever its denominator is not 0. If the interval had contained 0 (say $\int_{-1}^{1}\frac{dx}{x}$ ⁠), Theorem 2 could not be used at all (Remark (iv) after Theorem 2).

Theorem 2, Section 7.8.3. Lesson: the fundamental theorem
Find an antiderivative

By standard formula (xii),
$\int\frac{dx}{x} = \log|x| + C.$
For $x$ in $[2, 3]$ ⁠, $x > 0$ ⁠, so $|x| = x$ ⁠. Take
$F(x) = \log x \quad (x > 0).$
The absolute value is there so that the formula also works for negative $x$ ⁠. On our interval every $x$ is positive, so $\log|x|$ and $\log x$ are the same function there.

Standard formula (xii), Section 7.2. Basics: logarithms and 1/x
Check: differentiate $F$ back
$F'(x) = \frac{d}{dx}\log x = \frac1x = f(x) \quad (x > 0).$
The derivative of the natural logarithm is $\frac1x$ (Chapter 5). So $F$ really is an antiderivative of $f$ on $[2, 3]$ ⁠.

$\frac{d}{dx}\log x = \frac1x$ for $x > 0$ ⁠. Basics: integration undoes differentiation
Evaluate $F(3) - F(2)$
$\begin{aligned}\int_2^3\frac{dx}{x} &= \big[\log x\big]_2^3 = \log 3 - \log 2 \\ &= \log\frac32 \approx 0.4054651.\end{aligned}$
Upper limit 3 first, then subtract the value at the lower limit 2. The law $\log a - \log b = \log\frac{a}{b}$ turns the difference into the single logarithm in the book's answer.

$\log 3 \approx 1.0986123$ and $\log 2 \approx 0.6931472$ ⁠; their difference is $0.4054651$ ⁠. Basics: logarithms
Does the size make sense?

On $[2, 3]$ the curve $y = \frac1x$ falls from $\frac12$ to $\frac13$ ⁠. So the region under it, on a base of width 1, has area between $\frac13\times 1$ and $\frac12\times 1$ ⁠:
$0.333\ldots < 0.405 < 0.5.$

The shaded region under $y = \frac1x$ from 2 to 3 has area $\log\frac32 \approx 0.405$ ⁠. The curve's height falls from $\frac12$ to $\frac13$ across a base of width 1, so the area must lie between $\frac13$ and $\frac12$ ⁠.

A short "between the smallest and the largest rectangle" estimate catches mistakes such as $\log 3 + \log 2 = \log 6 \approx 1.79$ ⁠, which is far too big.

Definite integral as area, Section 7.8.1. Basics: area as a sum of thin strips

Result

\int_2^3\frac{1}{x}\,dx = \log\frac32 \approx 0.4054651.

(Textbook answer: $\log\frac32$ ⁠, the same.)

Check it with numbers

Numerical integration of $\frac1x$ from 2 to 3 (thin strips, no antiderivative used) gives $0.4054651$ ⁠, and $\log 1.5 \approx 0.4054651$ ⁠: they agree.

Common mistakes

Using the power rule with $n = -1$ ⁠. $\frac{x^{0}}{0}$ is meaningless; $n = -1$ is exactly the case the power rule excludes, and $\log|x|$ replaces it.
Writing $\log 3 - \log 2 = \log 1$ ⁠. Logs of a difference do not work like that: $\log 3 - \log 2 = \log\frac32$ ⁠.
Ignoring where $\frac1x$ is defined. On an interval containing 0 the fundamental theorem cannot be used.

What to take away

$\int_a^b\frac{dx}{x} = \log\frac{b}{a}$ whenever $0 < a < b$ ⁠.
Check that the interval avoids every point where the integrand is undefined before using Theorem 2.
Bounding the area between two rectangles is a quick sanity check.

Every fact used, and where it comes from

Fact	Used in	Source
$\int_a^b f = F(b) - F(a)$ for continuous $f$	Steps 1, 4	Theorem 2, Section 7.8.3 (Lesson)
$\int\frac{dx}{x} = \log\|x\| + C$	Step 2	Standard formula (xii), Section 7.2 (Basics)
$\frac{d}{dx}\log x = \frac1x$	Step 3	Chapter 5, Section 5.4
$\log a - \log b = \log\frac{a}{b}$	Step 4	Laws of logarithms (Basics)
Definite integral as area	Step 5	Section 7.8.1 (Basics)

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Exercise 7.8 · Q2: ∫231x dx\int_2^3\frac{1}{x}\,dx∫23​x1​dx

Basics you need

Solution

Check continuity on [2,3][2, 3][2,3]

Find an antiderivative

Check: differentiate FFF back

Evaluate F(3)−F(2)F(3) - F(2)F(3)−F(2)

Does the size make sense?

Every fact used, and where it comes from

Exercise 7.8 · Q2: $\int_2^3\frac{1}{x}\,dx$

Check continuity on $[2, 3]$

Check: differentiate $F$ back

Evaluate $F(3) - F(2)$