MATH

INTEGRAL

DAFTAR ISI MATERI INTEGRAL

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): \( |x - 1| \)

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

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

Apply absolute value definition:

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

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

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

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

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

Recognize perfect square:

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

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

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

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

Factor the quadratic:

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

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

Positive condition (using quadratic inequality):

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

Negative condition:

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

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

Use definition from Example 1a:

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

Split integral at x = 1:

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

Apply definitions in each interval:

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

Compute first integral:

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

Compute second integral:

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

Add the results:

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

Final Answer:

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

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

Use definition from Example 1b:

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

Check integration interval:

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

Simplify:

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

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

Recognize as absolute value:

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

Use definition:

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

Split integral at x = 2:

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

Compute first integral:

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

Compute second integral:

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

Add results:

\[ 2 + 4.5 = 6.5 \]

Final Answer:

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

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

Use definition from Example 1d:

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

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

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

Compute first integral:

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

Compute second integral:

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

Compute third integral:

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

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:

Definition of \( |x| \):

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

Rewrite the function:

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

Split integral at x = 0:

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

Compute first integral:

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

Compute second integral:

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

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:

Simplify the expression:

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

Factor:

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

Find critical points:

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

Positive condition (quadratic inequality):

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

Negative condition:

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

Apply absolute value definition:

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

Split integral:

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

Compute first integral:

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

Compute second integral:

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

Compute third integral:

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

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

Absolute Value Definition: \( |f(x)| = \begin{cases} f(x), & f(x) \geq 0 \\ -f(x), & f(x) < 0 \end{cases} \)
Property: \( |f(x)| = \sqrt{(f(x))^2} \)
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
For quadratic expressions: Factor and use quadratic inequalities to determine sign
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.

Determining Arc Length Using Integrals

Introduction

Besides calculating area and volume of solids of revolution, integrals have another important application: determining arc length of a curve. This application further demonstrates the importance and versatility of integral calculus.

Prerequisites

To understand this material, you should master:

Distance between two points
Riemann sums
Derivatives of algebraic functions
Definite integrals of algebraic and trigonometric functions

Note: The theory is relatively simple, but the integration calculations can be challenging.

Conceptual Foundation

Idea Behind Arc Length Formula

To calculate arc length, we approximate the curve with many small straight line segments (red lines in the diagram).

The length of each small segment from point \(C(x_{k-1}, y_{k-1})\) to \(D(x_k, y_k)\) is:

\[ \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2} \]

The total approximate length is the sum of all segments:

\[ \text{Approximate length} = \sum_{k=1}^n \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2} \]

As we take more and smaller segments (\(n \to \infty\)), this becomes an integral:

\[ \text{Arc length} = \lim_{n \to \infty} \sum_{k=1}^n \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2} \]

Arc Length Formulas

Standard Arc Length Formulas

For curve \(y = f(x)\) from point \(A(a, c)\) to \(B(b, d)\):

Using X-boundaries (integrate with respect to x):

\[ \text{Arc length} = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]

Using Y-boundaries (integrate with respect to y):

\[ \text{Arc length} = \int_c^d \sqrt{\left(\frac{dx}{dy}\right)^2 + 1} \, dy \]

Note: Choose the formula that gives the easier integration.

Example Problems

Example 1: Curve \(9y^2 = 4x^3\)

Find the arc length of curve \(9y^2 = 4x^3\) from \(A(0, 0)\) to \(B(3, 2\sqrt{3})\).

Solution (using X-boundaries):

1. Rewrite function: \(9y^2 = 4x^3 \Rightarrow y = \frac{2}{3}x^{3/2}\)

2. Find derivative: \(\frac{dy}{dx} = \frac{2}{3} \cdot \frac{3}{2}x^{1/2} = x^{1/2}\)

3. Set up integral (x from 0 to 3):

Arc length = \(\int_0^3 \sqrt{1 + (x^{1/2})^2} \, dx = \int_0^3 \sqrt{1 + x} \, dx\)

4. Calculate: \(= \int_0^3 (1 + x)^{1/2} \, dx = \left[ \frac{2}{3}(1 + x)^{3/2} \right]_0^3\)

5. Evaluate: \(= \frac{2}{3}(4^{3/2}) - \frac{2}{3}(1^{3/2}) = \frac{2}{3}(8) - \frac{2}{3}(1) = \frac{16}{3} - \frac{2}{3} = \frac{14}{3}\)

Answer: Arc length = \(\frac{14}{3}\) units.

Note: Using Y-boundaries would be more difficult for this problem.

Example 2: Straight Line \(y = 3x\)

Find the arc length of line \(y = 3x\) from \(A(0, 0)\) to \(B(2, 6)\) using both methods.

Solution Method 1 (X-boundaries):

1. Derivative: \(\frac{dy}{dx} = 3\)

2. Arc length: \(\int_0^2 \sqrt{1 + 3^2} \, dx = \int_0^2 \sqrt{10} \, dx\)

3. Calculate: \(= \left[ \sqrt{10}x \right]_0^2 = 2\sqrt{10}\)

Solution Method 2 (Y-boundaries):

1. Rewrite: \(y = 3x \Rightarrow x = \frac{1}{3}y\)

2. Derivative: \(\frac{dx}{dy} = \frac{1}{3}\)

3. Arc length: \(\int_0^6 \sqrt{\left(\frac{1}{3}\right)^2 + 1} \, dy = \int_0^6 \sqrt{\frac{1}{9} + 1} \, dy = \int_0^6 \sqrt{\frac{10}{9}} \, dy\)

4. Calculate: \(= \int_0^6 \frac{\sqrt{10}}{3} \, dy = \left[ \frac{\sqrt{10}}{3}y \right]_0^6 = 2\sqrt{10}\)

Answer: Arc length = \(2\sqrt{10}\) units (both methods give same result).

Parametric Form Arc Length

Arc Length for Parametric Equations

For parametric equations \(x = f(t)\), \(y = g(t)\) with \(a \leq t \leq b\):

\[ \text{Arc length} = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

Derivation: Starting from \(\int_a^b \sqrt{(dx)^2 + (dy)^2}\), multiply by \(\frac{dt}{dt}\):

\[ = \int_a^b \frac{\sqrt{(dx)^2 + (dy)^2}}{dt} \, dt = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

Example 3: Parametric Path of a Particle

A particle moves along a path described by \(x = 3t\) and \(y = \frac{8}{3}t^{3/2}\), where \(t\) is in minutes. Find the distance traveled in the first minute.

Solution:

1. Find derivatives:

\(\frac{dx}{dt} = 3\)

\(\frac{dy}{dt} = \frac{8}{3} \cdot \frac{3}{2}t^{1/2} = 4t^{1/2}\)

2. Set up integral (t from 0 to 1):

Distance = \(\int_0^1 \sqrt{3^2 + (4t^{1/2})^2} \, dt = \int_0^1 \sqrt{9 + 16t} \, dt\)

3. Calculate using substitution \(u = 9 + 16t\), \(du = 16 dt\):

\(= \frac{1}{16} \int \sqrt{u} \, du = \frac{1}{16} \cdot \frac{2}{3} u^{3/2} = \frac{1}{24} u^{3/2}\)

4. Evaluate from t=0 to t=1:

\(= \left[ \frac{1}{24}(9 + 16t)^{3/2} \right]_0^1 = \frac{1}{24}(25^{3/2}) - \frac{1}{24}(9^{3/2})\)

5. Simplify: \(= \frac{1}{24}(125) - \frac{1}{24}(27) = \frac{1}{24}(98) = \frac{49}{12} = 4\frac{1}{12}\)

Answer: Distance traveled = \(4\frac{1}{12}\) units.

Key Points and Tips

Choosing the Right Formula

1. Standard form \(y = f(x)\): Use \(\int \sqrt{1 + (dy/dx)^2} \, dx\)

2. Standard form \(x = f(y)\): Use \(\int \sqrt{(dx/dy)^2 + 1} \, dy\)

3. Parametric form: Use \(\int \sqrt{(dx/dt)^2 + (dy/dt)^2} \, dt\)

Tip: Choose the form that gives the simplest derivative and integral.

Common Integration Techniques

Arc length problems often require:

Substitution (u-substitution)
Trigonometric substitution
Integration of square roots
Recognizing perfect squares under the radical

Verification Strategy

For straight lines, you can verify your answer using the distance formula:

\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

For Example 2: \(\sqrt{(2-0)^2 + (6-0)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}\) ✓

Summary of Formulas

Form	Arc Length Formula
\(y = f(x)\)	\(\displaystyle L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\)
\(x = f(y)\)	\(\displaystyle L = \int_c^d \sqrt{\left(\frac{dx}{dy}\right)^2 + 1} \, dy\)
Parametric	\(\displaystyle L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\)

Volume of Solids of Revolution Using Integrals

Introduction

Besides calculating area, integrals are also used to calculate volumes of solids of revolution. A solid of revolution is formed when a region bounded by curves is rotated around a specific line (usually the X-axis or Y-axis) through a full rotation of 360°.

Visualization: Rotating a triangular region around the X-axis forms a cone. Rotating a semicircular region around the X-axis forms a sphere.

Two Main Methods

Disk Method

Characteristic: Rotation direction matches integration boundaries.

Example: Rotate around X-axis, integrate with X-boundaries.

Forms solid disks/washers perpendicular to axis of rotation.

Shell Method

Characteristic: Rotation direction differs from integration boundaries.

Example: Rotate around Y-axis, integrate with X-boundaries.

Forms cylindrical shells parallel to axis of rotation.

Disk Method

Rotation Around X-Axis

Region bounded by \(y = f(x)\), X-axis, \(x = a\), \(x = b\), rotated 360° around X-axis:

\[ V = \pi \int_a^b [f(x)]^2 dx \]

Example 1: Single Curve Around X-Axis

Find volume when region bounded by \(y = x\), X-axis, and \(x = 3\) is rotated 360° around X-axis.

Solution:

\(V = \pi \int_0^3 [x]^2 dx = \pi \int_0^3 x^2 dx\)

\(= \pi \left[ \frac{1}{3}x^3 \right]_0^3\)

\(= \pi \left( \frac{1}{3}(27) - 0 \right)\)

\(= \pi (9) = 9\pi\)

Answer: Volume = \(9\pi\) cubic units.

Rotation Around Y-Axis

Region bounded by \(x = f(y)\), Y-axis, \(y = a\), \(y = b\), rotated 360° around Y-axis:

\[ V = \pi \int_a^b [f(y)]^2 dy \]

Example 2: Single Curve Around Y-Axis

Find volume when region bounded by \(y = x^2\), Y-axis, \(y = 2\), and \(y = 5\) is rotated 360° around Y-axis.

Solution:

First, rewrite: \(y = x^2 \Rightarrow x = \sqrt{y}\)

\(V = \pi \int_2^5 [\sqrt{y}]^2 dy = \pi \int_2^5 y dy\)

\(= \pi \left[ \frac{1}{2}y^2 \right]_2^5\)

\(= \pi \left( \frac{25}{2} - \frac{4}{2} \right) = \pi \left( \frac{21}{2} \right)\)

Answer: Volume = \(\frac{21}{2}\pi\) cubic units.

Two Curves - Washer Method

Around X-axis: Region between \(y_1 = f(x)\) and \(y_2 = g(x)\) where \(|f(x)| \geq |g(x)|\) on \([a, b]\):

\[ V = \pi \int_a^b \left( [f(x)]^2 - [g(x)]^2 \right) dx \]

Around Y-axis: Region between \(x_1 = f(y)\) and \(x_2 = g(y)\) where \(|f(y)| \geq |g(y)|\) on \([a, b]\):

\[ V = \pi \int_a^b \left( [f(y)]^2 - [g(y)]^2 \right) dy \]

Important Rule

Always subtract: curve farther from rotation axis minus curve closer to rotation axis.

Example 3: Two Curves Around X-Axis

Find volume when region bounded by \(y = 6x - x^2\) and \(y = x\) is rotated 360° around X-axis.

Solution:

1. Find intersections: \(6x - x^2 = x \Rightarrow x^2 - 5x = 0 \Rightarrow x(x-5) = 0\)

Intersections at \(x = 0\) and \(x = 5\)

2. On [0,5], \(6x - x^2 \geq x\) (check at \(x=2\): \(8 > 2\))

So \(f(x) = 6x - x^2\), \(g(x) = x\)

3. \(V = \pi \int_0^5 \left( (6x - x^2)^2 - x^2 \right) dx\)

\(= \pi \int_0^5 (36x^2 - 12x^3 + x^4 - x^2) dx\)

\(= \pi \int_0^5 (x^4 - 12x^3 + 35x^2) dx\)

\(= \pi \left[ \frac{1}{5}x^5 - 3x^4 + \frac{35}{3}x^3 \right]_0^5\)

\(= \pi \left( \frac{3125}{5} - 1875 + \frac{4375}{3} \right)\)

\(= \pi \left( 625 - 1875 + \frac{4375}{3} \right) = \pi \left( \frac{625}{3} \right)\)

Answer: Volume = \(208\frac{1}{3}\pi\) cubic units.

Shell Method

Shell Method Formulas

Rotate around Y-axis (X-boundaries):

\[ V = 2\pi \int_a^b x \cdot f(x) dx \]

Rotate around X-axis (Y-boundaries):

\[ V = 2\pi \int_a^b y \cdot f(y) dy \]

When to use: When it's difficult to rewrite \(y = f(x)\) as \(x = f(y)\) or vice versa.

Example 4: Shell Method - Single Curve

Find volume when region bounded by \(y = -x^3 + 4x\), \(x = 0\), \(x = 1\), and X-axis is rotated 360° around Y-axis.

Solution:

Why use shell method? \(y = -x^3 + 4x\) is difficult to rewrite as \(x = f(y)\), so shell method is easier.

\(V = 2\pi \int_0^1 x \cdot (-x^3 + 4x) dx\)

\(= 2\pi \int_0^1 (-x^4 + 4x^2) dx\)

\(= 2\pi \left[ -\frac{1}{5}x^5 + \frac{4}{3}x^3 \right]_0^1\)

\(= 2\pi \left( -\frac{1}{5} + \frac{4}{3} \right) = 2\pi \left( \frac{-3 + 20}{15} \right)\)

\(= 2\pi \left( \frac{17}{15} \right) = \frac{34}{15}\pi = 2\frac{4}{15}\pi\)

Answer: Volume = \(2\frac{4}{15}\pi\) cubic units.

Shell Method for Two Curves

Rotate around Y-axis (X-boundaries):

\[ V = 2\pi \int_a^b x \cdot [f(x) - g(x)] dx \]

Where \(|f(x)| \geq |g(x)|\) on \([a, b]\)

Rotate around X-axis (Y-boundaries):

\[ V = 2\pi \int_a^b y \cdot [f(y) - g(y)] dy \]

Example 5: Shell Method - Two Curves

Find volume when region bounded by \(y = \frac{1}{3}x^2\), \(y = x\), \(x = 0\), \(x = 2\), and X-axis is rotated 360° around Y-axis.

Solution:

On [0,2], \(x \geq \frac{1}{3}x^2\) (check at \(x=1\): \(1 > \frac{1}{3}\))

So \(f(x) = x\), \(g(x) = \frac{1}{3}x^2\)

\(V = 2\pi \int_0^2 x \cdot \left( x - \frac{1}{3}x^2 \right) dx\)

\(= 2\pi \int_0^2 \left( x^2 - \frac{1}{3}x^3 \right) dx\)

\(= 2\pi \left[ \frac{1}{3}x^3 - \frac{1}{12}x^4 \right]_0^2\)

\(= 2\pi \left( \frac{8}{3} - \frac{16}{12} \right) = 2\pi \left( \frac{8}{3} - \frac{4}{3} \right)\)

\(= 2\pi \left( \frac{4}{3} \right) = \frac{8}{3}\pi = 2\frac{2}{3}\pi\)

Answer: Volume = \(2\frac{2}{3}\pi\) cubic units.

Method Selection Guide

When to Use Disk/Washer Method

• Rotation axis matches integration variable (rotate around X, integrate dx)

• Easy to express functions in terms of rotation axis variable

• Forms solid perpendicular slices to rotation axis

When to Use Shell Method

• Rotation axis differs from integration variable (rotate around Y, integrate dx)

• Difficult to rewrite functions in terms of rotation axis variable

• Forms cylindrical shells parallel to rotation axis

Fundamental Skills Required

To master volume calculations using integrals, you need:

Graphing skills: Sketch curves and identify regions
Integration skills: Calculate definite integrals of algebraic functions
Analytical thinking: Determine which method applies
Visualization: Imagine the 3D solid formed by rotation

Even for difficult problems, the process remains the same as in these examples.

Quick Reference - Volume Formulas

Method	Rotation	Formula
Disk	Around X-axis	\(V = \pi \int_a^b [f(x)]^2 dx\)
Disk	Around Y-axis	\(V = \pi \int_a^b [f(y)]^2 dy\)
Washer	Around X-axis	\(V = \pi \int_a^b ([f(x)]^2 - [g(x)]^2) dx\)
Washer	Around Y-axis	\(V = \pi \int_a^b ([f(y)]^2 - [g(y)]^2) dy\)
Shell	Around Y-axis	\(V = 2\pi \int_a^b x f(x) dx\)
Shell	Around X-axis	\(V = 2\pi \int_a^b y f(y) dy\)

Proof of Quick Formulas for Calculating Area Using Integrals

Introduction

Previously, we learned quick methods for calculating area using integrals. These methods are helpful for solving certain types of problems quickly, but they are limited in scope and don't apply to all problems. As critical thinkers, we shouldn't just accept these formulas without understanding where they come from, even though they give the same results as the fundamental method when applied correctly.

Prerequisites

To understand these proofs, you should be familiar with:

Indefinite and definite integrals of algebraic functions
Calculating area using integrals
Operations with roots of quadratic equations

Proof 1: Discriminant Formula

Discriminant Formula

For a quadratic equation \(ax^2 + bx + c = 0\), the discriminant is \(D = b^2 - 4ac\).

\[ \text{Area} = \frac{D\sqrt{D}}{6a^2} \]

Conditions: This formula applies only to regions bounded by:

Two parabolas, or
A parabola and a straight line

Proof

Case 1: Two parabolas

Let the parabolas be: \(y_1 = a_1x^2 + b_1x + c_1\) and \(y_2 = a_2x^2 + b_2x + c_2\)

Set them equal to find intersection points:

\(a_1x^2 + b_1x + c_1 = a_2x^2 + b_2x + c_2\)

\((a_1 - a_2)x^2 + (b_1 - b_2)x + (c_1 - c_2) = 0\)

Let \(a = a_1 - a_2\), \(b = b_1 - b_2\), \(c = c_1 - c_2\)

Case 2: Parabola and line

Let: \(y_1 = a_1x^2 + b_1x + c_1\) and \(y_2 = b_2x + c_2\)

Set them equal:

\(a_1x^2 + b_1x + c_1 = b_2x + c_2\)

\(a_1x^2 + (b_1 - b_2)x + (c_1 - c_2) = 0\)

Let \(a = a_1\), \(b = b_1 - b_2\), \(c = c_1 - c_2\)

In both cases, we get: \(ax^2 + bx + c = 0\) with roots \(x_1\) and \(x_2\) (intersection points).

Properties of roots:

\(x_1 + x_2 = -\frac{b}{a}\), \(x_1x_2 = \frac{c}{a}\), \(x_2 - x_1 = \frac{\sqrt{D}}{a}\)

\(x_2^2 - x_1^2 = (x_2 - x_1)(x_2 + x_1) = \frac{\sqrt{D}}{a} \cdot \frac{(-b)}{a}\)

\(x_2^3 - x_1^3 = (x_2 - x_1)^3 + 3x_1x_2(x_2 - x_1) = \left(\frac{\sqrt{D}}{a}\right)^3 + 3 \cdot \frac{c}{a} \cdot \frac{\sqrt{D}}{a}\)

Area calculation using integration:

\(\text{Area} = \int_{x_1}^{x_2} (y_1 - y_2) dx = \int_{x_1}^{x_2} (ax^2 + bx + c) dx\)

\(\text{Area} = \left[ \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx \right]_{x_1}^{x_2}\)

\(= \frac{a}{3}(x_2^3 - x_1^3) + \frac{b}{2}(x_2^2 - x_1^2) + c(x_2 - x_1)\)

Substitute root properties:

\(= \frac{a}{3}\left( \frac{D\sqrt{D}}{a^3} + \frac{3c\sqrt{D}}{a^2} \right) + \frac{b}{2}\left( -\frac{b\sqrt{D}}{a^2} \right) + c\left( \frac{\sqrt{D}}{a} \right)\)

\(= \frac{\sqrt{D}}{a^2}\left( \frac{D}{3} + c - \frac{b^2}{2a} + \frac{ac}{a} \right)\)

\(= \frac{\sqrt{D}}{a^2}\left( \frac{2D + 6ac - 3b^2 + 6ac}{6a} \right)\)

\(= \frac{\sqrt{D}}{a^2}\left( \frac{2D - 3(b^2 - 4ac)}{6a} \right)\)

Since \(D = b^2 - 4ac\):

\(= \frac{\sqrt{D}}{a^2}\left( \frac{2D - 3D}{6a} \right) = \frac{\sqrt{D}}{a^2}\left( -\frac{D}{6a} \right)\)

Since area is always positive:

\(\text{Area} = \frac{D\sqrt{D}}{6a^2}\)

Proof 2: Intersection Points Formula

Intersection Points Formula

If two curves intersect at \(x_1\) and \(x_2\), and the combined equation is \(ax^2 + bx + c = 0\):

\[ \text{Area} = \frac{a}{6}|x_1 - x_2|^3 \]

Same conditions as the discriminant formula.

Proof

We'll prove this using the discriminant formula and properties of roots.

From root properties: \(x_2 - x_1 = \frac{\sqrt{D}}{a}\)

Therefore: \(\sqrt{D} = a(x_2 - x_1)\) and \(D = a^2(x_2 - x_1)^2\)

Using the discriminant formula:

\(\text{Area} = \frac{D\sqrt{D}}{6a^2} = \frac{a^2(x_2 - x_1)^2 \cdot a(x_2 - x_1)}{6a^2}\)

Simplify: \(= \frac{a(x_2 - x_1)^3}{6}\)

Since area is always positive: \(= \frac{a}{6}|x_1 - x_2|^3\)

Proof complete. This shows the connection between the two formulas.

Proof 3: Rectangle Ratio Formula

Rectangle Ratio Formula

For quadratic functions (parabolas) where the region's sides pass through the vertex:

\[ \text{Larger part} = \frac{2}{3} \times \text{rectangle area} \] \[ \text{Smaller part} = \frac{1}{3} \times \text{rectangle area} \]

Condition: The "fatter" part inside the curve has 2:1 area ratio with the "thinner" part outside.

Proof

We'll use the simplest quadratic function \(y = ax^2\). Other quadratics \(y = ax^2 + bx + c\) are just translations of this basic form and don't change area ratios.

Consider the parabola \(y = ax^2\) and the horizontal line \(y = ak^2\) (where \(k > 0\)):

Area of larger part (A):

Bounded by \(y = ak^2\) (top) and \(y = ax^2\) (bottom) from 0 to \(k\):

\(\text{Area A} = \int_0^k (ak^2 - ax^2) dx\)

\(= \left[ ak^2x - \frac{a}{3}x^3 \right]_0^k\)

\(= \left( ak^3 - \frac{a}{3}k^3 \right) - 0\)

\(= \frac{2}{3}ak^3\)

Area of smaller part (B):

Bounded by \(y = ax^2\) and x-axis from 0 to \(k\):

\(\text{Area B} = \int_0^k ax^2 dx\)

\(= \left[ \frac{a}{3}x^3 \right]_0^k\)

\(= \frac{a}{3}k^3\)

Area ratio:

\(\frac{\text{Area A}}{\text{Area B}} = \frac{\frac{2}{3}ak^3}{\frac{1}{3}ak^3} = \frac{2}{1}\)

So Area A : Area B = 2 : 1

Total rectangle area:

The rectangle formed has width \(k\) and height \(ak^2\):

Rectangle area = \(k \times ak^2 = ak^3\)

Relation to rectangle area:

Area A = \(\frac{2}{3}ak^3 = \frac{2}{3} \times \text{rectangle area}\)

Area B = \(\frac{1}{3}ak^3 = \frac{1}{3} \times \text{rectangle area}\)

Key insight: The ratio 2:1 between the "inside" and "outside" areas of a parabola relative to a horizontal line through a point on the parabola is constant and independent of the specific quadratic coefficients (as long as we're measuring from the vertex).

Summary of Proofs

Formula 1: Discriminant Method

Proof Strategy: Start with general quadratic forms, find intersection points, use root properties and integration.

Key Step: \(x_2 - x_1 = \frac{\sqrt{D}}{a}\) connects geometry to algebra.

Formula 2: Intersection Points Method

Proof Strategy: Derived directly from Formula 1 using root distance.

Key Insight: Area depends on cube of distance between intersections.

Formula 3: Rectangle Ratio Method

Proof Strategy: Use simplest parabola \(y = ax^2\), calculate areas with integration.

Key Result: Constant 2:1 ratio for areas inside/outside parabola relative to horizontal line.

Important Notes

1. These proofs show why the quick formulas work, but remember they have limited applicability.

2. The fundamental integration method works for all cases, while these formulas work only for specific situations.

3. Understanding these proofs helps develop mathematical intuition and connects algebra, geometry, and calculus.

Mathematical Connections Demonstrated

These proofs beautifully connect:

Algebra: Discriminant and root properties of quadratics
Geometry: Area measurements and spatial relationships
Calculus: Integration as a tool for area calculation
Analytic Geometry: Equations of curves and their intersections

This demonstrates the power of mathematics as an interconnected discipline.