Chapter 3: Integration

Overview

Integration is the reverse process of differentiation and is fundamental to calculus. This chapter explores indefinite and definite integrals, their properties, and various applications including area under curves and volumes of revolution. Mastery of integration is essential for solving problems in physics, engineering, economics, and other fields where accumulation and total quantities need to be calculated.

Learning Objectives

After completing this chapter, you will be able to:

• Understand the concept of integration as the reverse of differentiation
• Compute indefinite and definite integrals
• Apply integration formulas and techniques
• Find areas under curves and between curves
• Calculate volumes of revolution
• Use integration in real-world applications

Key Concepts

3.1 Integration as the Inverse of Differentiation

Fundamental Theorem of Calculus

Integration is the inverse operation of differentiation. If d/dx[F(x)] = f(x), then:

$$\int f(x) \, dx = F(x) + c$$

Where:

• ∫ is the integral sign
• f(x) is the integrand
• F(x) is the antiderivative
• c is the constant of integration

Geometric Interpretation

Integration can be interpreted as:

• Finding the area under a curve y = f(x)
• Accumulating quantities over an interval
• Reversing the differentiation process

3.2 Indefinite and Definite Integrals

Indefinite Integrals

An indefinite integral (antiderivative) represents a family of functions:

$$\int f(x) \, dx = F(x) + c$$

Where c is an arbitrary constant determined by initial conditions.

Definite Integrals

A definite integral evaluates to a specific numerical value:

$$\int_a^b f(x) \, dx = [F(x)]_a^b = F(b) - F(a)$$

Where:

• a and b are the limits of integration
• F(x) is the antiderivative of f(x)
• [F(x)]_a^b means evaluate F at b and subtract F evaluated at a

Properties of Definite Integrals

1. Order of Limits: ∫ₐᵇ f(x) dx = -∫ᵇₐ f(x) dx
2. Addition Property: ∫ₐᵇ f(x) dx + ∫ᵇᶜ f(x) dx = ∫ₐᶜ f(x) dx
3. Constant Multiple: ∫ₐᵇ kf(x) dx = k ∫ₐᵇ f(x) dx
4. Sum/Difference: ∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f(x) dx ± ∫ₐᵇ g(x) dx

3.3 Application of Integration

Area Under a Curve

The area under y = f(x) from x = a to x = b is:

$$A = \int_a^b f(x) \, dx$$

Conditions:

• If f(x) ≥ 0, area = ∫ₐᵇ f(x) dx
• If f(x) ≤ 0, area = -∫ₐᵇ f(x) dx
• If f(x) changes sign, split the integral at zeros

Area Between Two Curves

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

$$A = \int_a^b |f(x) - g(x)| \, dx$$

Where the upper function minus the lower function is integrated.

Volume of Revolution

1. Rotation about x-axis:

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

2. Rotation about y-axis:

$$V = \pi \int_c^d x^2 \, dy = \pi \int_c^d [g(y)]^2 \, dy$$

Definite Integration by Parts

For definite integrals:

$$\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du$$

Important Formulas and Methods

Basic Integration Formulas

Integration Techniques

1. Power Rule: ∫axⁿ dx = axⁿ⁺¹/(n+1) + c, n ≠ -1
2. Substitution: ∫f(g(x))g'(x) dx = ∫f(u) du where u = g(x)
3. Integration by Parts: ∫u dv = uv - ∫v du
4. Partial Fractions: For rational functions

Area Calculation Methods

Area Under Single Curve:

1. Identify the function and interval
2. Find the antiderivative
3. Apply limits: F(b) - F(a)
4. Take absolute value if necessary

Area Between Two Curves:

1. Find intersection points (limits)
2. Determine upper and lower functions
3. Integrate (upper - lower)
4. Take absolute value of result

Volume Calculation Methods

Volume of Revolution:

1. Identify axis of rotation
2. Use appropriate formula (π∫$y^2$ dx or π∫$x^2$ dy)
3. Find antiderivative of [f(x)]² or [g(y)]²
4. Apply limits and multiply by π

Solved Examples

Example 1: Basic Integration

Find the following integrals: a) ∫(3$x^2$ + 2x - 5) dx b) ∫(4eˣ - sin x) dx

Solutions:

a) ∫(3$x^2$ + 2x - 5) dx = 3($x^3$/3) + 2($x^2$/2) - 5x + c = $x^3$ + $x^2$ - 5x + c

b) ∫(4eˣ - sin x) dx = 4eˣ - (-cos x) + c = 4eˣ + cos x + c

Example 2: Definite Integrals

Evaluate the following definite integrals: a) ∫₀² ($x^3$ + 2x) dx b) ∫₁³ (3$e^2$ˣ) dx

Solutions:

a) ∫₀² ($x^3$ + 2x) dx = [$x^4$/4 + $x^2$]₀² = [(16/4 + 4) - (0 + 0)] = [4 + 4] = 8

b) ∫₁³ (3$e^2$ˣ) dx = 3 ∫₁³ $e^2$ˣ dx = 3 [$e^2$ˣ/2]₁³ = (3/2)[$e^6$ - $e^2$] ≈ (3/2)[403.429 - 7.389] ≈ (3/2)(396.04) ≈ 594.06

Example 3: Area Under Curve

Find the area under the curve y = 4 - $x^2$ from x = -2 to x = 2.

Solution:

First, check if function is above x-axis in interval: y = 4 - $x^2$, which is positive when |x| < 2

Area = ∫₋₂² (4 - $x^2$) dx = [4x - $x^3$/3]₋₂² = [(4(2) - (8)/3) - (4(-2) - (-8)/3)] = [(8 - 8/3) - (-8 + 8/3)] = [(24/3 - 8/3) - (-24/3 + 8/3)] = [16/3 - (-16/3)] = 16/3 + 16/3 = 32/3 ≈ 10.67 square units

Example 4: Area Between Curves

Find the area between the curves y = $x^2$ and y = x + 2.

Solution:

First, find intersection points: $x^2$ = x + 2 ⇒ $x^2$ - x - 2 = 0 ⇒ (x - 2)(x + 1) = 0 ⇒ x = -1, x = 2

In interval [-1, 2], y = x + 2 is above y = $x^2$

Area = ∫₋₁² [(x + 2) - $x^2$] dx = [$x^2$/2 + 2x - $x^3$/3]₋₁² = [(4/2 + 4 - 8/3) - (1/2 - 2 - (-1)/3)] = [(2 + 4 - 8/3) - (1/2 - 2 + 1/3)] = [(6 - 8/3) - (-3/2 + 1/3)] = [(18/3 - 8/3) - (-9/6 + 2/6)] = [10/3 - (-7/6)] = 10/3 + 7/6 = 20/6 + 7/6 = 27/6 = 4.5 square units

Example 5: Volume of Revolution

Find the volume generated when the area bounded by y = √x, x = 0, x = 4, and y = 0 is rotated about the x-axis.

Solution:

Area bounded by y = √x, x-axis, from x = 0 to x = 4 Rotate about x-axis using V = π∫$y^2$ dx

V = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = π [$x^2$/2]₀⁴ = π [16/2 - 0] = 8π cubic units

Example 6: Volume About y-axis

Find the volume generated when the area bounded by y = $x^2$, y = 0, x = 1, x = 2 is rotated about the y-axis.

Solution:

Area bounded by y = $x^2$, x-axis, from x = 1 to x = 2 Rotate about y-axis using V = π∫$x^2$ dy

But we need to express x in terms of y: x = √y When x = 1, y = 1; x = 2, y = 4

V = π∫₁⁴ (√y)² dy = π∫₁⁴ y dy = π [$y^2$/2]₁⁴ = π [16/2 - 1/2] = π [8 - 0.5] = 7.5π cubic units

Mathematical Derivations

Fundamental Theorem of Calculus

Part 1: If F(x) = ∫ₐˣ f(t) dt, then F'(x) = f(x)

Part 2: ∫ₐᵇ f(x) dx = F(b) - F(a) where F'(x) = f(x)

Proof Sketch: By definition of the derivative and properties of limits, we can show that the derivative of the integral function is the original function.

Integration by Parts Derivation

Start with the product rule for differentiation: d(uv)/dx = u dv/dx + v du/dx

Integrate both sides with respect to x: uv = ∫u dv/dx dx + ∫v du/dx dx

Since dv/dx dx = dv and du/dx dx = du: uv = ∫u dv + ∫v du

Therefore: ∫u dv = uv - ∫v du

Volume of Revolution Formula

For rotation about x-axis, consider thin disks of radius y and thickness dx:

• Volume of disk = π$r^2$h = π$y^2$ dx
• Total volume = ∫π$y^2$ dx = π∫$y^2$ dx

Real-World Applications

1. Physics - Work and Energy

Work Done:

• Work = ∫Force × distance
• For variable force: W = ∫ₐᵇ F(x) dx

Example: Find work done to stretch a spring from x=0 to x=1 if force F = kx.

2. Engineering - Center of Mass

Center of Mass:

• x̄ = (1/A)∫ₐᵇ x f(x) dx
• ȳ = (1/A)∫ₐᵇ ½[f(x)]² dx

Example: Find center of mass of a uniform wire bent into curve y = $x^2$ from x=0 to x=1.

3. Economics - Consumer and Producer Surplus

Consumer Surplus: CS = ∫₀^($Q_0$) [Demand function] dQ - $P_0Q_0$

Producer Surplus: PS = $P_0Q_0$ - ∫₀^($Q_0$) [Supply function] dQ

Example: Calculate consumer surplus if demand function is P = 100 - 2Q and equilibrium price is $P_0$ = 60.

4. Biology - Population Growth

Total Population:

• If growth rate is given, total population over time is ∫rate dt

Example: If population grows at rate dP/dt = 100e^(0.1t), find total population growth from t=0 to t=10.

Complex Problem-Solving Techniques

Problem: Find the area between the curves y = sin x and y = cos x from x = 0 to x = π/2.

Solution:

Find intersection points in [0, π/2]: sin x = cos x ⇒ tan x = 1 ⇒ x = π/4

In [0, π/4]: cos x ≥ sin x In [π/4, π/2]: sin x ≥ cos x

Area = ∫₀^(π/4) (cos x - sin x) dx + ∫_(π/4)^(π/2) (sin x - cos x) dx

Calculate first integral: ∫(cos x - sin x) dx = sin x + cos x From 0 to π/4: (sin π/4 + cos π/4) - (sin 0 + cos 0) = (√2/2 + √2/2) - (0 + 1) = √2 - 1

Calculate second integral: ∫(sin x - cos x) dx = -cos x - sin x From π/4 to π/2: (-cos π/2 - sin π/2) - (-cos π/4 - sin π/4) = (0 - 1) - (-√2/2 - √2/2) = -1 + √2

Total area = (√2 - 1) + (-1 + √2) = 2√2 - 2 ≈ 2(1.414) - 2 ≈ 0.828 square units

Problem: Find the volume generated when the region bounded by y = $x^3$, x = 1, x = 2, and y = 0 is rotated about:

a) The x-axis b) The y-axis

Solution:

a) Rotation about x-axis: V = π∫₁² ($x^3$)² dx = π∫₁² $x^6$ dx = π [$x^7$/7]₁² = π [(128/7) - (1/7)] = 127π/7 ≈ 56.94 cubic units

b) Rotation about y-axis: Express x in terms of y: x = y^(1/3) When x = 1, y = 1; x = 2, y = 8

V = π∫₁⁸ (y^(1/3))² dy = π∫₁⁸ y^(2/3) dy = π [y^(5/3)/(5/3)]₁⁸ = (3π/5)[y^(5/3)]₁⁸ = (3π/5)[8^(5/3) - 1^(5/3)] = (3π/5)[32 - 1] = (3π/5)(31) = 93π/5 ≈ 58.48 cubic units

Problem: Find the area bounded by y = $x^2$ and y = 2 - $x^2$.

Solution:

Find intersection points: $x^2$ = 2 - $x^2$ ⇒ 2$x^2$ = 2 ⇒ $x^2$ = 1 ⇒ x = ±1

In interval [-1, 1], y = 2 - $x^2$ is above y = $x^2$

Area = ∫₋₁¹ [(2 - $x^2$) - $x^2$] dx = ∫₋₁¹ (2 - 2$x^2$) dx = [2x - 2$x^3$/3]₋₁¹ = [(2 - 2/3) - (-2 + 2/3)] = [(6/3 - 2/3) - (-6/3 + 2/3)] = [4/3 - (-4/3)] = 4/3 + 4/3 = 8/3 ≈ 2.667 square units

Summary Points

• Integration is the reverse of differentiation
• Indefinite integral: ∫f(x) dx = F(x) + c (family of functions)
• Definite integral: ∫ₐᵇ f(x) dx = F(b) - F(a) (numerical value)
• Area under curve: ∫ₐᵇ f(x) dx (for f(x) ≥ 0)
• Area between curves: ∫ₐᵇ |f(x) - g(x)| dx
• Volume of revolution: π∫$y^2$ dx (x-axis) or π∫$x^2$ dy (y-axis)
• Applications span physics, engineering, economics, and biology

Common Mistakes to Avoid

1. Constant of integration - Don't forget +c for indefinite integrals
2. Limit errors - Apply upper limit minus lower limit correctly
3. Area sign errors - Take absolute value for area calculations
4. Volume formula errors - Use correct formula for rotation axis
5. Integration errors - Check antiderivatives by differentiating back

SPM Exam Tips

Exam Strategies

1. Memorize formulas - Basic integral formulas and area/volume formulas
2. Practice substitution - Be comfortable with u-substitution
3. Sketch regions - Draw graphs to understand area/volume bounds
4. Show working - Step-by-step integration for partial marks
5. Check answers - Differentiate antiderivatives to verify

Key Exam Topics

• Basic integration (25% of questions)
• Definite integrals (20% of questions)
• Area under curves (25% of questions)
• Area between curves (15% of questions)
• Volumes of revolution (15% of questions)

Time Management Tips

• Basic integration: 3-4 minutes
• Definite integrals: 4-5 minutes
• Area calculations: 6-7 minutes
• Volume problems: 7-9 minutes

Practice Problems

Level 1: Basic Integration

1. Find the integrals: a) ∫(2$x^3$ - 3$x^2$ + 4x - 5) dx b) ∫($e^2$ˣ + cos 2x) dx c) ∫(1/x + x⁻²) dx

2. Evaluate definite integrals: a) ∫₀³ ($x^2$ + 2x) dx b) ∫₁² (3x - 4) dx c) ∫₀^(π/2) sin x dx

Level 2: Area Under Curves

3. Find the area under the curves: a) y = $x^2$ + 1 from x = 0 to x = 2 b) y = 4 - $x^2$ from x = -2 to x = 2 c) y = eˣ from x = 0 to x = 1

Level 3: Area Between Curves

4. Find areas between curves: a) y = x and y = $x^2$ b) y = 2x and y = $x^2$ + 1 c) y = sin x and y = cos x from 0 to π/2

Level 4: Volumes of Revolution

5. Find volumes generated by rotating about x-axis: a) y = $x^2$ from x = 0 to x = 2 b) y = √x from x = 0 to x = 4 c) y = 2x from x = 1 to x = 3

6. Find volumes generated by rotating about y-axis: a) y = $x^2$ from y = 0 to y = 4 b) y = √x from x = 1 to x = 4 c) x = $y^2$ from y = 0 to y = 2

Level 5: Applications

7. Physics: A force F(x) = 3$x^2$ + 2x acts on an object. Find work done moving from x=0 to x=2.

8. Economics: Demand function P = 100 - Q, supply function P = 20 + Q. Find consumer surplus.

9. Engineering: Find the center of mass of a lamina bounded by y = $x^2$, x=0, x=1, y=0 with uniform density.

Did You Know? 📚

The concept of integration was developed independently by Isaac Newton and Gottfried Wilhelm Leibniz in the late 17th century. The integral symbol ∫ was introduced by Leibniz in 1675, resembling an elongated S for "summa" (sum). The fundamental theorem of calculus, connecting differentiation and integration, is considered one of the most important theorems in mathematics, bridging the seemingly unrelated concepts of slopes and areas.

Quick Reference Guide

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Integration provides powerful tools for calculating accumulated quantities and is essential for understanding the relationship between rates of change and total amounts. Master these concepts as they form the foundation for advanced calculus and its applications in science, engineering, and economics.