Unit 8: Applications of Integration

📋 Unit Overview

🎯 Learning Objectives: This unit focuses on applying integration to solve real-world problems including area, volume, arc length, and work calculations.

Unit 8 builds upon the integration techniques from Unit 6 to solve practical problems. You'll learn to calculate areas between curves, volumes of revolution, arc lengths, and work done by variable forces.

Key Topics:

Area between curves
Volumes of revolution (disk/washer methods)
Volumes by cross-sections
Arc length calculations
Work and force applications
Average value of functions

Skills You'll Master:

Setting up integral expressions
Choosing appropriate methods
Interpreting results in context
Solving optimization problems
Working with parametric equations
Applying to physics problems

📐 Area Between Curves

Basic Concept

The area between two curves y = f(x) and y = g(x) from x = a to x = b is:

A = ∫_a^b |f(x) - g(x)| dx

💡 Key Steps:

Find points of intersection
Determine which function is "on top"
Set up the integral with correct bounds
Evaluate the integral

Example 1: Basic Area Between Curves

Find the area between y = x² and y = 2x.

Solution:

1) Find intersection points: x² = 2x → x = 0, 2

2) On [0,2], 2x ≥ x²

3) A = ∫₀² (2x - x²) dx = [x² - x³/3]₀² = 4 - 8/3 = 4/3

🔄 Volumes of Revolution

Disk Method

When rotating around a horizontal axis, the volume is:

V = π ∫_a^b [f(x)]² dx

Example 2: Disk Method

Find the volume when y = √x from x = 0 to x = 4 is rotated about the x-axis.

Solution:

V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π[x²/2]₀⁴ = π(8) = 8π

Washer Method

When there's a "hole" in the solid, use the washer method:

V = π ∫_a^b [R(x)² - r(x)²] dx

Where R(x) is the outer radius and r(x) is the inner radius.

Example 3: Washer Method

Find the volume when the region between y = x² and y = 2x is rotated about the x-axis.

Solution:

1) Intersection points: x = 0, 2

2) R(x) = 2x, r(x) = x²

3) V = π ∫₀² [(2x)² - (x²)²] dx = π ∫₀² (4x² - x⁴) dx

4) = π[4x³/3 - x⁵/5]₀² = π(32/3 - 32/5) = 64π/15

📏 Volumes by Cross-Sections

General Formula

For any cross-sectional area A(x):

V = ∫_a^b A(x) dx

🔍 Common Cross-Sections:

Squares: A(x) = [f(x) - g(x)]²
Rectangles: A(x) = h[f(x) - g(x)]
Semicircles: A(x) = π[f(x) - g(x)]²/8
Equilateral triangles: A(x) = √3[f(x) - g(x)]²/4

Example 4: Square Cross-Sections

Find the volume of a solid with square cross-sections perpendicular to the x-axis, where the base is the region between y = x² and y = 4.

Solution:

1) Intersection: x² = 4 → x = ±2

2) A(x) = (4 - x²)² = 16 - 8x² + x⁴

3) V = ∫_-2² (16 - 8x² + x⁴) dx = 2∫₀² (16 - 8x² + x⁴) dx

4) = 2[16x - 8x³/3 + x⁵/5]₀² = 2(32 - 64/3 + 32/5) = 512/15

📏 Arc Length

Arc Length Formula

For a function y = f(x) from x = a to x = b:

L = ∫_a^b √(1 + [f'(x)]²) dx

Example 5: Arc Length

Find the arc length of y = x³/2 from x = 0 to x = 2.

Solution:

1) f'(x) = 3x²/2

2) L = ∫₀² √(1 + (3x²/2)²) dx = ∫₀² √(1 + 9x⁴/4) dx

3) This requires numerical methods or substitution techniques

⚡ Work and Force Applications

Work Formula

When force varies with position:

W = ∫_a^b F(x) dx

🔧 Common Applications:

Springs: F(x) = kx (Hooke's Law)
Pumping liquids: F(x) = ρgA(x)h(x)
Variable forces: Any force that depends on position

Example 6: Spring Work

A spring has natural length 10 cm and requires 5 N to stretch it 2 cm. How much work is required to stretch it from 12 cm to 15 cm?

Solution:

1) Find spring constant: F = kx → 5 = k(0.02) → k = 250 N/m

2) W = ∫_0.02^0.05 250x dx = 250[x²/2]_0.02^0.05

3) = 125[(0.05)² - (0.02)²] = 125(0.0025 - 0.0004) = 0.2625 J

📊 Average Value of Functions

Average Value Formula

The average value of f(x) on [a,b] is:

f_avg = (1/(b-a)) ∫_a^b f(x) dx

Example 7: Average Value

Find the average value of f(x) = x² on [0,3].

Solution:

f_avg = (1/3) ∫₀³ x² dx = (1/3)[x³/3]₀³ = (1/3)(9) = 3

🎯 Practice Problems

Problem 1: Area Between Curves

Find the area of the region bounded by y = x³ and y = x.

Solution:

1) Find intersections: x³ = x → x(x² - 1) = 0 → x = -1, 0, 1

2) On [-1,0]: x³ ≥ x, On [0,1]: x ≥ x³

3) A = ∫_-1⁰ (x³ - x) dx + ∫₀¹ (x - x³) dx

4) = [x⁴/4 - x²/2]_-1⁰ + [x²/2 - x⁴/4]₀¹ = 1/4 + 1/4 = 1/2

Problem 2: Volume by Washer Method

Find the volume when the region between y = x² and y = x is rotated about y = 2.

Solution:

1) Intersections: x = 0, 1

2) R(x) = 2 - x², r(x) = 2 - x

3) V = π ∫₀¹ [(2-x²)² - (2-x)²] dx

4) = π ∫₀¹ (4-4x²+x⁴-4+4x-x²) dx = π ∫₀¹ (4x-5x²+x⁴) dx

5) = π[2x²-5x³/3+x⁵/5]₀¹ = π(2-5/3+1/5) = 8π/15

Problem 3: Work Application

A cable weighing 2 lb/ft is used to lift a 200 lb load 50 ft. How much work is required?

Solution:

1) At height x, the cable length is 50-x

2) Total weight at height x: 200 + 2(50-x) = 300 - 2x

3) W = ∫₀⁵⁰ (300 - 2x) dx = [300x - x²]₀⁵⁰

4) = 15000 - 2500 = 12500 ft-lb

💡 Exam Tips & Strategies

🎯 Problem-Solving Strategy:

Sketch the region - Always draw a diagram first
Find intersection points - These determine your bounds
Identify the method - Disk, washer, or cross-sections
Set up the integral - Be careful with radius expressions
Evaluate carefully - Check your arithmetic

⚠️ Common Mistakes:

Forgetting to square the radius in volume formulas
Using wrong bounds of integration
Mixing up inner and outer radius in washer method
Not accounting for units in work problems
Forgetting the π in volume calculations

🔑 Key Insight: Most application problems follow the same pattern: identify the quantity you want to find, express it as a sum of infinitesimal pieces, and convert that sum to an integral.

🧠 Ready to Test Your Knowledge?

📝 Unit 8 Practice Test

Now that you've mastered the applications of integration, test your understanding with our comprehensive practice test. The test includes:

15 Multiple Choice Questions - Covering all major topics
6 Free Response Questions - With detailed solutions
45-minute timer - Simulating real exam conditions
Instant feedback - Check your answers as you go
Score calculation - See how you perform

Take Practice Test

Recommended after completing all unit content

💡 Pro Tip: Take the practice test under timed conditions to simulate the real AP exam experience. Review any topics you miss and retake the test to track your improvement!

➡️ Next Unit Preview

Ready for the next challenge? Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions introduces new coordinate systems and ways to describe curves.

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