Chapter 7 · Integrals

Exercise 7.10 · Q18: $\int_0^4|x-1|\,dx$

Exercise 7.10Properties of definite integrals (P2) · Section 7.10

The question

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

18. $\displaystyle\int_0^{4}|x - 1|\,dx$

Exam answer

Let $I = \displaystyle\int_0^{4}|x - 1|\,dx$ ⁠.

Here $|x - 1| = 1 - x$ for $0 \le x \le 1$ ⁠, and $|x - 1| = x - 1$ for $1 \le x \le 4$ ⁠.

Using $\displaystyle\int_a^b f(x)\,dx$ $= \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$ with $c = 1$ ⁠,

\begin{aligned} I &= \int_0^{1}(1 - x)\,dx + \int_1^{4}(x - 1)\,dx \\ &= \left[x - \frac{x^2}{2}\right]_0^1 + \left[\frac{x^2}{2} - x\right]_1^4 \\ &= \left(1 - \frac12\right) - 0 \\ &\quad + (8 - 4) - \left(\frac12 - 1\right) \\ &= \frac12 + \frac92 \end{aligned}

$\therefore\ I = 5$ ⁠.

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

The one idea this problem is built on

The modulus $|x - 1|$ is given by two different formulas: $1 - x$ to the left of $x = 1$ and $x - 1$ to the right. So we split the interval at its turning point $x = 1$ using P2, and integrate each piece with its own formula.

Given, to find, and the plan

Given: $f(x) = |x - 1|$ on $[0, 4]$ ⁠.
To find: $I = \displaystyle\int_0^{4}|x - 1|\,dx$ ⁠.
Plan: (1) Write $|x - 1|$ without the modulus on each side of $x = 1$ ⁠. (2) Split with P2 at $c = 1$ ⁠. (3) and (4) Integrate each piece. (5) Add, and check with triangle areas.

Basics you need

1Modulus

$|u| = u$ if $u \ge 0$ and $|u| = -u$ if $u < 0$ ⁠. Basics: symmetry and mirror images

2Property P2 (Section 7.10)

$\int_a^b f(x)\,dx$ $= \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$ ⁠. Lesson: properties of definite integrals

3Power rule and Theorem 2

$\int x\,dx = \frac{x^2}{2} + C$ ⁠; $\int_a^b f(x)\,dx = F(b) - F(a)$ ⁠. Basics: powers of x

4Area of a triangle

$\frac12 \times \text{base} \times \text{height}$ (used in the check). Basics: area as a sum of thin strips

Solution

Remove the modulus on each side of $x = 1$

The expression inside, $x - 1$ ⁠, changes sign at $x = 1$ ⁠:
$|x - 1| = \begin{cases} 1 - x, & 0 \le x \le 1,\\ x - 1, & 1 \le x \le 4. \end{cases}$
At $x = 1$ both formulas give 0, so they agree there. The function is continuous on $[0, 4]$ ⁠, with a corner at $x = 1$ ⁠.

No formula in the book's list integrates $|x - 1|$ directly, but on each side of $x = 1$ it is a polynomial, which we can integrate. The turning point $x = 1$ lies inside $[0, 4]$ ⁠, so both formulas are needed.

For $x < 1$ ⁠, $x - 1 < 0$ ⁠, so $|x - 1| = -(x - 1) = 1 - x$ ⁠.
Split the interval with P2 at $c = 1$
$I = \int_0^{1}(1 - x)\,dx + \int_1^{4}(x - 1)\,dx .$
P2 holds for any $c$ (and any continuous $f$ ⁠); we choose $c = 1$ because that is where the formula changes. On each piece the integrand is a polynomial, continuous on the closed sub-interval, so Theorem 2 applies to each.

P2, Section 7.10. Lesson: properties
The left piece, $[0, 1]$
$\begin{aligned}\int_0^{1}(1 - x)\,dx &= \left[x - \frac{x^2}{2}\right]_0^1 \\ &= \left(1 - \frac12\right) - 0 = \frac12 .\end{aligned}$
An antiderivative of $1 - x$ is $x - \frac{x^2}{2}$ (check: its derivative is $1 - x$ ⁠). Then Theorem 2: value at the upper limit minus value at the lower limit.

Power rule: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ ⁠. Basics: powers of x
The right piece, $[1, 4]$
$\begin{aligned}\int_1^{4}(x - 1)\,dx &= \left[\frac{x^2}{2} - x\right]_1^4 \\ &= (8 - 4) - \left(\frac12 - 1\right) \\ &= 4 + \frac12 = \frac92 .\end{aligned}$
An antiderivative of $x - 1$ is $\frac{x^2}{2} - x$ (its derivative is $x - 1$ ⁠). At $x = 4$ ⁠: $\frac{16}{2} - 4 = 4$ ⁠. At $x = 1$ ⁠: $\frac12 - 1 = -\frac12$ ⁠. Subtracting a negative number adds it.

Theorem 2, Section 7.8.3. Lesson: definite integrals
Add, and check with triangles
$I = \frac12 + \frac92 = 5 .$
Geometrically, the region under $y = |x - 1|$ is two right triangles: on $[0, 1]$ ⁠, base 1 and height $|0 - 1| = 1$ ⁠, area $\frac12$ ⁠; on $[1, 4]$ ⁠, base 3 and height $|4 - 1| = 3$ ⁠, area $\frac92$ ⁠. Total 5.

$y = |x - 1|$ on $[0, 4]$ ⁠: a V with its corner at $(1, 0)$ ⁠. The region under it consists of two triangles of areas $\frac12$ and $\frac92$ ⁠.

Since $|x - 1| \ge 0$ ⁠, the integral is a genuine area, and the triangle areas confirm the answer without any antiderivative.

Area of a triangle $= \frac12\times\text{base}\times\text{height}$ ⁠. Basics: area as a sum of thin strips

Result

$\int_0^{4}|x - 1|\,dx = 5$ (Textbook answer: 5.)

Check it with numbers

Numerical integration over $[0, 1]$ and $[1, 4]$ gives $0.500000000000$ and $4.500000000000$ ⁠, and over $[0, 4]$ it gives $5.000000000000$ ⁠.

A wrong route for comparison: $\int_0^4 (x - 1)\,dx = 8 - 4 = 4$ ⁠, not 5. Dropping the modulus counts the left triangle as $-\frac12$ instead of $+\frac12$ ⁠, and $5 - 2 \times \frac12 = 4$ ⁠.

Common mistakes

Ignoring the modulus and integrating $x - 1$ over $[0, 4]$ ⁠, which gives 4.
Splitting at the wrong point, such as the midpoint 2. The split must be where $x - 1$ changes sign.
Sign slip at the lower limit in Step 4: $-\left(\frac12 - 1\right) = +\frac12$ ⁠.

What to take away

For $\int |g(x)|\,dx$ ⁠: find where $g$ changes sign inside the interval, split there with P2, and use $g$ or $-g$ on each piece.
Compare Q5, Q6 and Example 28, which split at the zeros of the expression inside the modulus in the same way.

Every fact used, and where it comes from

Fact	Used in	Source
Definition of $\|u\|$	Step 1	Class XI, modulus function
P2: splitting the interval at $c$	Step 2	Section 7.10, property P2 (Lesson: properties)
Power rule	Steps 3, 4	Section 7.2, list of standard integrals
Theorem 2 (second fundamental theorem)	Steps 3, 4	Section 7.8.3
Area of a triangle	Step 5	Elementary geometry

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Exercise 7.10 · Q18: ∫04∣x−1∣ dx\int_0^4|x-1|\,dx∫04​∣x−1∣dx

Basics you need

Solution

Remove the modulus on each side of x=1x = 1x=1

Split the interval with P2 at c=1c = 1c=1

The left piece, [0,1][0, 1][0,1]

The right piece, [1,4][1, 4][1,4]

Add, and check with triangles

Every fact used, and where it comes from

Exercise 7.10 · Q18: $\int_0^4|x-1|\,dx$

Remove the modulus on each side of $x = 1$

Split the interval with P2 at $c = 1$

The left piece, $[0, 1]$

The right piece, $[1, 4]$