Determining Integrals of Absolute Value Functions

Introduction

After learning integration techniques for algebraic and trigonometric functions, we now focus on a special type of integral: integrals of absolute value functions. As the title suggests, this involves integrating functions containing absolute values, whether algebraic or trigonometric functions. The absolute value of a function \( f(x) \) is denoted by \( |f(x)| \) and is always non-negative for all \( x \).

To make learning Determining Integrals of Absolute Value Functions easier, it's best to master the material on integrating algebraic functions and trigonometric functions as well as existing integration techniques. Besides that, we must also relearn the definition of absolute value (or modulus).

Prerequisites

To understand this material thoroughly, you should master:

• Integration of algebraic functions
• Integration of trigonometric functions
• Basic integration techniques (substitution, integration by parts, etc.)
• Definition and properties of absolute values
• Quadratic inequalities and solving inequalities

Note: Understanding absolute value definitions is crucial for solving these problems.

Definition of Absolute Value of a Function

Definition of Absolute Value Function

The absolute value of a function \( f(x) \) is denoted by \( |f(x)| \).

The definition of absolute value is:

\[ |f(x)| = \begin{cases} f(x), & f(x) \geq 0 \\ -f(x), & f(x) < 0 \end{cases} \]

This means \( |f(x)| = f(x) \) or \( |f(x)| = -f(x) \) depending on the value of \( f(x) \).

Property of Absolute Value: \( |f(x)| = \sqrt{(f(x))^2} \) where the square and square root cannot be eliminated.

With the definition of absolute value, the absolute value of any number or function is always non-negative.

Determining Integrals of Absolute Value Functions

In general, the steps in determining integrals of absolute value functions are: first change the absolute value function based on its definition to determine when the function is positive and when it is negative. This means the absolute value function will be divided into several integral boundaries depending on how many absolute value functions we want to integrate.

\[ \int_a^b |f(x)| dx = \int_a^c |f(x)| dx + \int_c^b |f(x)| dx \]

where \( c \) is the point where \( f(x) = 0 \).

Example 1: Removing Absolute Values from Expressions

Problem: Break down the following absolute value functions based on the definition of absolute value (remove the absolute value form):

a) \( |x - 1| \)

b) \( |2x + 5| \)

c) \( \sqrt{x^2 - 4x + 4} \)

d) \( |x^2 - x - 6| \)

Solution for (a): \( |x - 1| \)

1

Positive condition: \( x - 1 \geq 0 \Rightarrow x \geq 1 \)

2

Negative condition: \( x - 1 < 0 \Rightarrow x < 1 \)

3

Apply absolute value definition:

\[ |x - 1| = \begin{cases} x - 1, & x \geq 1 \\ -(x - 1), & x < 1 \end{cases} \]

4

Simplify:

\[ |x - 1| = \begin{cases} x - 1, & x \geq 1 \\ 1 - x, & x < 1 \end{cases} \]

Final Answer:

\[ |x - 1| = \begin{cases} x - 1, & x \geq 1 \\ 1 - x, & x < 1 \end{cases} \]

Solution for (b): \( |2x + 5| \)

1

Positive condition: \( 2x + 5 \geq 0 \Rightarrow x \geq -\frac{5}{2} \)

2

Negative condition: \( 2x + 5 < 0 \Rightarrow x < -\frac{5}{2} \)

3

Apply absolute value definition:

\[ |2x + 5| = \begin{cases} 2x + 5, & x \geq -\frac{5}{2} \\ -(2x + 5), & x < -\frac{5}{2} \end{cases} \]

Final Answer:

\[ |2x + 5| = \begin{cases} 2x + 5, & x \geq -\frac{5}{2} \\ -2x - 5, & x < -\frac{5}{2} \end{cases} \]

Solution for (c): \( \sqrt{x^2 - 4x + 4} \)

1

Recognize perfect square:

\[ \sqrt{x^2 - 4x + 4} = \sqrt{(x - 2)^2} = |x - 2| \]

2

Positive condition: \( x - 2 \geq 0 \Rightarrow x \geq 2 \)

3

Negative condition: \( x - 2 < 0 \Rightarrow x < 2 \)

4

Apply absolute value definition:

\[ \sqrt{x^2 - 4x + 4} = |x - 2| = \begin{cases} x - 2, & x \geq 2 \\ -(x - 2), & x < 2 \end{cases} \]

Final Answer:

\[ \sqrt{x^2 - 4x + 4} = \begin{cases} x - 2, & x \geq 2 \\ 2 - x, & x < 2 \end{cases} \]

Solution for (d): \( |x^2 - x - 6| \)

1

Factor the quadratic:

\[ x^2 - x - 6 = (x + 2)(x - 3) \]

2

Find roots: \( (x + 2)(x - 3) = 0 \Rightarrow x = -2 \text{ or } x = 3 \)

3

Positive condition (using quadratic inequality):

\( (x + 2)(x - 3) \geq 0 \Rightarrow x \leq -2 \text{ or } x \geq 3 \)

4

Negative condition:

\( (x + 2)(x - 3) < 0 \Rightarrow -2 < x < 3 \)

5

Apply absolute value definition:

\[ |x^2 - x - 6| = \begin{cases} x^2 - x - 6, & x \leq -2 \text{ or } x \geq 3 \\ -(x^2 - x - 6), & -2 < x < 3 \end{cases} \]

Final Answer:

\[ |x^2 - x - 6| = \begin{cases} x^2 - x - 6, & x \leq -2 \text{ or } x \geq 3 \\ 6 + x - x^2, & -2 < x < 3 \end{cases} \]

Determining Integrals of Absolute Value Functions

General Strategy

Suppose we want to determine the integral of the absolute value function \( |f(x)| \) from \( a \) to \( b \), with the absolute value function broken down into:

\[ |f(x)| = \begin{cases} f(x), & x \geq c \\ -f(x), & x < c \end{cases} \]

Then the integral can be calculated by:

\[ \int_a^b |f(x)| dx = \int_a^c |f(x)| dx + \int_c^b |f(x)| dx \]

where \( c \) is the point where \( f(x) = 0 \).

Example 2: Integral of Absolute Value Functions

Problem: Determine the results of the following absolute value function integrals:

a) \( \int_{-1}^3 |x - 1| dx \)

b) \( \int_0^2 |2x + 5| dx \)

c) \( \int_0^5 \sqrt{x^2 - 4x + 4} dx \)

d) \( \int_{-3}^5 |x^2 - x - 6| dx \)

Solution for (a): \( \int_{-1}^3 |x - 1| dx \)

1

Use definition from Example 1a:

\[ |x - 1| = \begin{cases} x - 1, & x \geq 1 \\ 1 - x, & x < 1 \end{cases} \]

2

Split integral at x = 1:

\[ \int_{-1}^3 |x - 1| dx = \int_{-1}^1 |x - 1| dx + \int_1^3 |x - 1| dx \]

3

Apply definitions in each interval:

\[ = \int_{-1}^1 (1 - x) dx + \int_1^3 (x - 1) dx \]

4

Compute first integral:

\[ \int_{-1}^1 (1 - x) dx = \left[ x - \frac{x^2}{2} \right]_{-1}^1 \] \[ = \left(1 - \frac{1}{2}\right) - \left(-1 - \frac{1}{2}\right) = \frac{1}{2} - \left(-\frac{3}{2}\right) = \frac{1}{2} + \frac{3}{2} = 2 \]

5

Compute second integral:

\[ \int_1^3 (x - 1) dx = \left[ \frac{x^2}{2} - x \right]_1^3 \] \[ = \left(\frac{9}{2} - 3\right) - \left(\frac{1}{2} - 1\right) = \frac{3}{2} - \left(-\frac{1}{2}\right) = \frac{3}{2} + \frac{1}{2} = 2 \]

6

Add the results:

\[ \text{Total} = 2 + 2 = 4 \]

Final Answer:

\[ \int_{-1}^3 |x - 1| dx = 4 \]

Solution for (b): \( \int_0^2 |2x + 5| dx \)

1

Use definition from Example 1b:

\[ |2x + 5| = \begin{cases} 2x + 5, & x \geq -\frac{5}{2} \\ -2x - 5, & x < -\frac{5}{2} \end{cases} \]

2

Check integration interval:

The interval [0, 2] is entirely in the region where \( x \geq -\frac{5}{2} \)

3

Simplify:

For all \( x \) in [0, 2], \( |2x + 5| = 2x + 5 \)

4

Compute integral:

\[ \int_0^2 (2x + 5) dx = \left[ x^2 + 5x \right]_0^2 \] \[ = (2^2 + 5 \times 2) - (0^2 + 5 \times 0) = (4 + 10) - 0 = 14 \]

Final Answer:

\[ \int_0^2 |2x + 5| dx = 14 \]

Solution for (c): \( \int_0^5 \sqrt{x^2 - 4x + 4} dx \)

1

Recognize as absolute value:

\[ \sqrt{x^2 - 4x + 4} = \sqrt{(x - 2)^2} = |x - 2| \]

2

Use definition:

\[ |x - 2| = \begin{cases} x - 2, & x \geq 2 \\ 2 - x, & x < 2 \end{cases} \]

3

Split integral at x = 2:

\[ \int_0^5 |x - 2| dx = \int_0^2 (2 - x) dx + \int_2^5 (x - 2) dx \]

4

Compute first integral:

\[ \int_0^2 (2 - x) dx = \left[ 2x - \frac{x^2}{2} \right]_0^2 \] \[ = (4 - 2) - 0 = 2 \]

5

Compute second integral:

\[ \int_2^5 (x - 2) dx = \left[ \frac{x^2}{2} - 2x \right]_2^5 \] \[ = \left(\frac{25}{2} - 10\right) - \left(2 - 4\right) = \frac{5}{2} - (-2) = \frac{5}{2} + 2 = \frac{9}{2} = 4.5 \]

6

Add results:

\[ 2 + 4.5 = 6.5 \]

Final Answer:

\[ \int_0^5 \sqrt{x^2 - 4x + 4} dx = 6.5 \]

Solution for (d): \( \int_{-3}^5 |x^2 - x - 6| dx \)

1

Use definition from Example 1d:

\[ |x^2 - x - 6| = \begin{cases} x^2 - x - 6, & x \leq -2 \text{ or } x \geq 3 \\ 6 + x - x^2, & -2 < x < 3 \end{cases} \]

2

Split integral at x = -2 and x = 3:

\[ \int_{-3}^5 |x^2 - x - 6| dx = \int_{-3}^{-2} (x^2 - x - 6) dx + \int_{-2}^3 (6 + x - x^2) dx + \int_3^5 (x^2 - x - 6) dx \]

3

Compute first integral:

\[ \int_{-3}^{-2} (x^2 - x - 6) dx = \left[ \frac{x^3}{3} - \frac{x^2}{2} - 6x \right]_{-3}^{-2} = \frac{17}{6} \]

4

Compute second integral:

\[ \int_{-2}^3 (6 + x - x^2) dx = \left[ 6x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^3 = \frac{125}{6} \]

5

Compute third integral:

\[ \int_3^5 (x^2 - x - 6) dx = \left[ \frac{x^3}{3} - \frac{x^2}{2} - 6x \right]_3^5 = \frac{38}{3} \]

6

Add all results:

\[ \frac{17}{6} + \frac{125}{6} + \frac{38}{3} = \frac{17}{6} + \frac{125}{6} + \frac{76}{6} = \frac{218}{6} = \frac{109}{3} \]

Final Answer:

\[ \int_{-3}^5 |x^2 - x - 6| dx = \frac{109}{3} \]

Example 3: Integral with Mixed Absolute Value

Problem: \( \int_{-1}^2 (x^2 - 2|x| + 5) dx \)

Only \( |x| \) has absolute value, so we remove the absolute value from \( |x| \) only using the absolute value definition.

Complete Solution:

1

Definition of \( |x| \):

\[ |x| = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases} \]

2

Rewrite the function:

\[ x^2 - 2|x| + 5 = \begin{cases} x^2 - 2x + 5, & x \geq 0 \\ x^2 + 2x + 5, & x < 0 \end{cases} \]

3

Split integral at x = 0:

\[ \int_{-1}^2 (x^2 - 2|x| + 5) dx = \int_{-1}^0 (x^2 + 2x + 5) dx + \int_0^2 (x^2 - 2x + 5) dx \]

4

Compute first integral:

\[ \int_{-1}^0 (x^2 + 2x + 5) dx = \left[ \frac{x^3}{3} + x^2 + 5x \right]_{-1}^0 \] \[ = 0 - \left(-\frac{1}{3} + 1 - 5\right) = 0 - \left(-\frac{13}{3}\right) = \frac{13}{3} \]

5

Compute second integral:

\[ \int_0^2 (x^2 - 2x + 5) dx = \left[ \frac{x^3}{3} - x^2 + 5x \right]_0^2 \] \[ = \left(\frac{8}{3} - 4 + 10\right) - 0 = \frac{8}{3} + 6 = \frac{26}{3} \]

6

Add results:

\[ \frac{13}{3} + \frac{26}{3} = \frac{39}{3} = 13 \]

Final Answer:

\[ \int_{-1}^2 (x^2 - 2|x| + 5) dx = 13 \]

Example 4: Complex Square Root Integral

Problem: \( \int_{-2}^1 \sqrt{9x^4 - 12x^3 + 4x^2} dx \)

Complete Solution:

1

Simplify the expression:

\[ \sqrt{9x^4 - 12x^3 + 4x^2} = \sqrt{(3x^2 - 2x)^2} = |3x^2 - 2x| \]

2

Factor:

\[ 3x^2 - 2x = x(3x - 2) \]

3

Find critical points:

\( x(3x - 2) = 0 \Rightarrow x = 0 \text{ or } x = \frac{2}{3} \)

4

Positive condition (quadratic inequality):

\( x(3x - 2) \geq 0 \Rightarrow x \leq 0 \text{ or } x \geq \frac{2}{3} \)

5

Negative condition:

\( x(3x - 2) < 0 \Rightarrow 0 < x < \frac{2}{3} \)

6

Apply absolute value definition:

\[ |3x^2 - 2x| = \begin{cases} 3x^2 - 2x, & x \leq 0 \text{ or } x \geq \frac{2}{3} \\ -(3x^2 - 2x), & 0 < x < \frac{2}{3} \end{cases} \]

7

Split integral:

\[ \int_{-2}^1 |3x^2 - 2x| dx = \int_{-2}^0 (3x^2 - 2x) dx + \int_0^{\frac{2}{3}} (-3x^2 + 2x) dx + \int_{\frac{2}{3}}^1 (3x^2 - 2x) dx \]

8

Compute first integral:

\[ \int_{-2}^0 (3x^2 - 2x) dx = \left[ x^3 - x^2 \right]_{-2}^0 = 0 - (-8 - 4) = 12 \]

9

Compute second integral:

\[ \int_0^{\frac{2}{3}} (-3x^2 + 2x) dx = \left[ -x^3 + x^2 \right]_0^{\frac{2}{3}} = -\frac{8}{27} + \frac{4}{9} = \frac{4}{27} \]

10

Compute third integral:

\[ \int_{\frac{2}{3}}^1 (3x^2 - 2x) dx = \left[ x^3 - x^2 \right]_{\frac{2}{3}}^1 = (1 - 1) - \left(\frac{8}{27} - \frac{4}{9}\right) = \frac{4}{27} \]

11

Add results:

\[ 12 + \frac{4}{27} + \frac{4}{27} = 12 + \frac{8}{27} = \frac{332}{27} \]

Final Answer:

\[ \int_{-2}^1 \sqrt{9x^4 - 12x^3 + 4x^2} dx = \frac{332}{27} \]

Summary of Key Concepts

1. Absolute Value Definition: \( |f(x)| = \begin{cases} f(x), & f(x) \geq 0 \\ -f(x), & f(x) < 0 \end{cases} \)
2. Property: \( |f(x)| = \sqrt{(f(x))^2} \)
3. General Strategy for Integration:
   • Find where \( f(x) = 0 \) (critical points)
   • Remove absolute values using piecewise definition
   • Split integral at critical points within integration bounds
   • Integrate each piece separately
   • Sum the results
4. For quadratic expressions: Factor and use quadratic inequalities to determine sign
5. For square roots: Remember that \( \sqrt{f(x)^2} = |f(x)| \)

Conclusion

This completes our discussion on Determining Integrals of Absolute Value Functions and the examples. Please also read other materials related to integrals. Hopefully this material can help those who need it.