Chapter 8 · Application of Integrals

Miscellaneous · Q2: Graph of $y = |x + 3|$ and $\int_{-6}^{0}|x + 3|\,dx$

Miscellaneous ExerciseA modulus inside the integral: split at the corner · Miscellaneous Exercise

The question

2. Sketch the graph of $y = |x + 3|$ and evaluate $\displaystyle\int_{-6}^{0}|x + 3|\,dx$ ⁠.

Exam answer

(figure) The graph of $y = |x + 3|$ is a V with vertex $(-3, 0)$ ⁠, through $(-6, 3)$ and $(0, 3)$ ⁠.

Here $|x + 3| = x + 3$ for $x \ge -3$ ⁠, and $|x + 3| = -(x + 3)$ for $x < -3$ ⁠.

Let $I = \displaystyle\int_{-6}^{0}|x + 3|\,dx$ ⁠. Using $\displaystyle\int_a^b f(x)\,dx$ $\displaystyle = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$ with $c = -3$ ⁠,

\begin{aligned} I &= -\int_{-6}^{-3}(x + 3)\,dx \\ &\quad + \int_{-3}^{0}(x + 3)\,dx \\ &= -\left[\frac{(x + 3)^2}{2}\right]_{-6}^{-3} \\ &\quad + \left[\frac{(x + 3)^2}{2}\right]_{-3}^{0} \\ &= -\left(0 - \frac{9}{2}\right) + \left(\frac{9}{2} - 0\right) \\ &= \frac{9}{2} + \frac{9}{2} \end{aligned}

$\therefore\ I = 9$ ⁠.

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

The one idea this problem is built on

The bars change their formula where the inside is zero: $x + 3 = 0$ at $x = -3$ ⁠. On the left of $-3$ ⁠, $|x + 3| = -(x + 3)$ ⁠; on the right, $|x + 3| = x + 3$ ⁠. So the graph is a V with its corner at $(-3, 0)$ ⁠, and the integral is split at $x = -3$ into two plain integrals, each worth $\frac92$ ⁠. Since $|x + 3| \ge 0$ ⁠, the value 9 is the area of the region between the V and the $x$ -axis from $x = -6$ to $x = 0$ ⁠: two triangles.

Given, to find, and the plan

Given: The function $y = |x + 3|$ and the integral $I = \int_{-6}^{0}|x + 3|\,dx$ ⁠.
To find: A sketch of the graph, and the exact value of $I$ ⁠.
Plan: (1) Write $|x + 3|$ without bars on each side of $x = -3$ ⁠. (2) Plot a few points and sketch the V. (3) Split the integral at $-3$ ⁠. (4) Integrate each piece. (5) Add, and check with the areas of the two triangles.

Basics you need

1Absolute value

$|u| = u$ when $u \ge 0$ ⁠, and $|u| = -u$ when $u < 0$ ⁠. So $|u|$ is never negative. Test: $\lvert -2\rvert = -(-2) = 2$ ⁠.

2Splitting an integral at a point

$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$ for any $c$ (property P2, Chapter 7, Section 7.10). Choosing $c$ at the corner gives one formula on each piece. Lesson: splitting an integral

3A definite integral is a number

$\int_a^b f(x)\,dx = F(b) - F(a)$ where $F' = f$ ⁠. Basics: the definite integral

4Integral of a non-negative function as an area

If $y \ge 0$ on $[a, b]$ ⁠, then $\int_a^b y\,dx$ is the area between the graph and the $x$ -axis there. Basics: a strip has area y dx

Solution

Write $|x + 3|$ without the bars

The inside, $x + 3$ ⁠, is $\ge 0$ exactly when $x \ge -3$ ⁠. So
$|x + 3| = \begin{cases} -(x + 3), & x < -3,\\ x + 3, & x \ge -3.\end{cases}$
Both formulas give 0 at $x = -3$ ⁠, so the two pieces join there.

On each side of $x = -3$ the function is a plain straight line, which is easy to draw and to integrate. The point where the inside is zero is the only place where the formula changes.

Test on each side: at $x = -5$ ⁠, $\lvert -5 + 3\rvert =\lvert -2\rvert = 2$ and $-(-5 + 3) = 2$ ✓; at $x = 0$ ⁠, $|3| = 3 = 0 + 3$ ✓.
Plot points and sketch the graph

A few values (each from the formula for its side):

$x$ $-7$ $-6$ $-5$ $-4$ $-3$ $-2$ $-1$ 0 1

$y$ 4 3 2 1 0 1 2 3 4

To the left of $-3$ the graph is the line $y = -x - 3$ (slope $-1$ ⁠); to the right it is $y = x + 3$ (slope 1). Together they form a V with its corner (vertex) at $(-3, 0)$ ⁠. It meets the $y$ -axis at $(0, 3)$ ⁠, and above $x = -6$ it is at height 3.

The graph of $y = |x + 3|$ ⁠, drawn with equal scales. Between $x = -6$ and $x = 0$ the region under the V is two triangles: on $[-6, -3]$ (gold) and on $[-3, 0]$ (blue). Each has base 3 and height 3. No part is below the $x$ -axis.

The question asks for the sketch, and the sketch also shows what the integral means: since $|x + 3| \ge 0$ ⁠, the integral from $-6$ to 0 is the area of the shaded region. The two triangles are mirror images in the line $x = -3$ ⁠, which already suggests two equal pieces.

Exercises: this question, step by step
Split the integral at $x = -3$
$\begin{aligned} I &= \int_{-6}^{-3}|x + 3|\,dx + \int_{-3}^{0}|x + 3|\,dx \\ &= \int_{-6}^{-3}-(x + 3)\,dx + \int_{-3}^{0}(x + 3)\,dx . \end{aligned}$
Property P2 lets us cut $[-6, 0]$ at $c = -3$ ⁠, which lies inside it. On $[-6, -3]$ we use the formula for $x < -3$ ⁠; at the single point $x = -3$ both formulas agree (value 0), so it does not matter which piece includes it.

P2, Chapter 7, Section 7.10. Lesson: splitting an integral
The left piece, on $[-6, -3]$
$\begin{aligned} \int_{-6}^{-3}-(x + 3)\,dx &= -\left[\frac{(x + 3)^2}{2}\right]_{-6}^{-3} \\ &= -\left(\frac{0^2}{2} - \frac{(-3)^2}{2}\right) \\ &= -\left(0 - \frac92\right) = \frac92 . \end{aligned}$
An antiderivative of $x + 3$ is $\frac{(x + 3)^2}{2}$ ⁠: by the chain rule its derivative is $\frac{2(x + 3)}{2} = x + 3$ ⁠. At $x = -3$ the bracket is $\frac{0^2}{2} = 0$ ⁠; at $x = -6$ it is $\frac{(-3)^2}{2} = \frac92$ ⁠. The minus sign in front turns the negative difference into a positive one, as it must for a region above the axis. (With the antiderivative $\frac{x^2}{2} + 3x$ instead: $-\big[(\frac92 - 9) - (18 - 18)\big] = \frac92$ ⁠, the same.)

Basics: the definite integral is F(b) − F(a)
The right piece, on $[-3, 0]$
$\begin{aligned} \int_{-3}^{0}(x + 3)\,dx &= \left[\frac{(x + 3)^2}{2}\right]_{-3}^{0} \\ &= \frac{3^2}{2} - \frac{0^2}{2} = \frac92 . \end{aligned}$
Same antiderivative; at $x = 0$ the bracket is $\frac{9}{2}$ ⁠, at $x = -3$ it is 0.

Basics: the definite integral is F(b) − F(a)
Add the pieces, and check with the triangles
$I = \frac92 + \frac92 = 9 .$
Check by geometry: each triangle in the figure has base 3 and height 3, so area $\frac12 \cdot 3 \cdot 3 = \frac92$ ⁠. Two of them make 9 ✓.

Because $|x + 3|$ is never negative, the integral is exactly the shaded area, and for straight-line pieces the triangle formula gives it directly. The two pieces are equal because the V is symmetric about $x = -3$ ⁠, the midpoint of $[-6, 0]$ ⁠.

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

$x$	$-7$	$-6$	$-5$	$-4$	$-3$	$-2$	$-1$	0	1
$y$	4	3	2	1	0	1	2	3	4

Result

The graph of $y = |x + 3|$ is a V with its vertex at $(-3, 0)$ ⁠, through $(-6, 3)$ and $(0, 3)$ ⁠, and

\int_{-6}^{0}|x + 3|\,dx = 9 .

This is the area between the graph and the $x$ -axis from $x = -6$ to $x = 0$ ⁠. (Textbook answer: 9.)

Check it with numbers

A computer's numerical integration of $|x + 3|$ over $[-6, 0]$ ⁠, with a break at $-3$ ⁠, gives 9.0000000000; over $[-6, -3]$ and over $[-3, 0]$ it gives 4.5000000000 each, which is $\frac92$ ✓.

A rough check with six strips of width 1, using the heights at the middles $-5.5, -4.5, \ldots, -0.5$ ⁠: $2.5 + 1.5 + 0.5 + 0.5 + 1.5 + 2.5 = 9$ ✓ (exact here, because each piece is a straight line and no strip crosses the corner).

Common mistakes

Dropping the bars: $\int_{-6}^{0}(x + 3)\,dx = \frac92 - \frac92 = 0$ ⁠. Without the bars the left triangle counts as negative and cancels the right one.
Using $x + 3$ on $[-6, -3]$ : there $x + 3 < 0$ ⁠, so $|x + 3| = -(x + 3)$ ⁠.
Putting the corner at $x = 3$ : the inside $x + 3$ is zero at $x = -3$ ⁠, so the V sits to the left of the $y$ -axis.
Losing the minus sign in front of the left bracket, which gives $-\frac92 + \frac92 = 0$ again.

What to take away

For $\int|g(x)|\,dx$ ⁠, find where $g(x) = 0$ ⁠, split there, and use $g$ or $-g$ on each piece.
The graph of $y = |x - h|$ or $|x + k|$ is a V with its corner on the $x$ -axis where the inside is zero.
An integral of a function that is never negative is an area; for straight-line graphs, triangles give a quick check. The same split appears in Miscellaneous Q5 for $y = x|x|$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\|u\| = u$ for $u \ge 0$ ⁠, $-u$ for $u < 0$	Step 1	Definition of absolute value
P2: $\int_a^b f = \int_a^c f + \int_c^b f$	Step 3	Chapter 7, Section 7.10 (Lesson)
$\frac{d}{dx}\frac{(x + 3)^2}{2} = x + 3$	Steps 4 and 5	Chain rule (Chapter 5)
$\int_a^b f(x)\,dx = F(b) - F(a)$	Steps 4 and 5	Chapter 7, Theorem 2, Section 7.8.3 (Basics)
Integral of $y \ge 0$ is the area under the graph	Steps 2 and 6	Section 8.2 (Basics)
Area of a triangle $= \frac12 bh$	Step 6	Mensuration

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Miscellaneous · Q2: Graph of y=∣x+3∣y = |x + 3|y=∣x+3∣ and ∫−60∣x+3∣ dx\int_{-6}^{0}|x + 3|\,dx∫−60​∣x+3∣dx

Basics you need

Solution

Write ∣x+3∣|x + 3|∣x+3∣ without the bars

Plot points and sketch the graph

Split the integral at x=−3x = -3x=−3

The left piece, on [−6,−3][-6, -3][−6,−3]

The right piece, on [−3,0][-3, 0][−3,0]

Add the pieces, and check with the triangles

Every fact used, and where it comes from

Miscellaneous · Q2: Graph of $y = |x + 3|$ and $\int_{-6}^{0}|x + 3|\,dx$

Write $|x + 3|$ without the bars

Split the integral at $x = -3$

The left piece, on $[-6, -3]$

The right piece, on $[-3, 0]$