Application of Integrals

8.1 Introduction

The **Application of Integrals** chapter extends the concept of definite integrals (studied in the previous chapter) to solve practical geometric problems. While elementary geometry provides formulae for simple figures like triangles, rectangles, and circles, these are inadequate for areas enclosed by curves. Integral calculus provides the mathematical tools to find:

Areas under simple curves

Areas between two curves

Areas bounded by lines and arcs of parabolas, ellipses, and circles

This chapter bridges the gap between abstract integration theory and real-world geometric applications, which is fundamental for engineering, physics, and economics.

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8.2 Area Under Simple Curves

Basic Concept and Formula

To find the area bounded by a curve y = f(x), the x-axis, and two ordinates x = a and x = b (where a < b):

**Approach**: Divide the region into thin vertical strips of width dx and height y = f(x). The area of each elementary strip is dA = y·dx. Summing all such strips from x = a to x = b using integration:

**Formula for area with vertical strips**:

A = ∫ₐᵇ y dx = ∫ₐᵇ f(x) dx

where y = f(x) is the equation of the curve.

Alternative: Area Using Horizontal Strips

If the curve is given as x = g(y), and we integrate between y = c and y = d:

**Formula for area with horizontal strips**:

A = ∫ₓ xdy = ∫ₐᵇ g(y) dy

Important Remarks

**Sign Convention**:

If the curve lies above the x-axis, f(x) > 0, and the integral gives positive area

If the curve lies below the x-axis, f(x) < 0, and the integral gives negative area (we take absolute value)

**If a curve crosses the x-axis** (partially above, partially below), calculate areas of each region separately and add their absolute values: A = |A₁| + |A₂|

**Step-by-Step Method**:

1. Identify the curve equation and limits

2. Check if the curve lies entirely above/below x-axis or crosses it

3. If it crosses, find intersection points with x-axis (set y = 0)

4. Split the integral at crossing points

5. Integrate each part and take absolute values

6. Add all areas

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Example 1: Area of a Circle

**Problem**: Find the area enclosed by the circle x² + y² = a².

**Solution**:

By symmetry, the total area = 4 × (area in first quadrant)

From x² + y² = a², we get: y = √(a² - x²) (positive in first quadrant)

Area of first quadrant = ∫₀ᵃ √(a² - x²) dx

Total Area = 4∫₀ᵃ √(a² - x²) dx

Using the standard formula: ∫√(a² - x²) dx = (x/2)√(a² - x²) + (a²/2)sin⁻¹(x/a) + C

A = 4[(x/2)√(a² - x²) + (a²/2)sin⁻¹(x/a)]₀ᵃ

A = 4[(a/2)·0 + (a²/2)sin⁻¹(1) - 0 - 0]

A = 4 × (a²/2) × (π/2) = **πa²**

**Verification**: This matches the standard formula for circle area.

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Example 2: Area of an Ellipse

**Problem**: Find the area enclosed by the ellipse (x²/a²) + (y²/b²) = 1.

**Solution**:

By symmetry, total area = 4 × (area in first quadrant)

From the ellipse equation: y = (b/a)√(a² - x²) (positive in first quadrant)

Area = 4∫₀ᵃ (b/a)√(a² - x²) dx

A = (4b/a) ∫₀ᵃ √(a² - x²) dx

Using the standard integration formula:

A = (4b/a) × [(x/2)√(a² - x²) + (a²/2)sin⁻¹(x/a)]₀ᵃ

A = (4b/a) × [(a²/2)sin⁻¹(1)]

A = (4b/a) × (a²/2) × (π/2) = **πab**

**Note**: When a = b = r, this reduces to πr², confirming the circle formula.

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Example 3: Area Bounded by a Line

**Problem**: Find the area of the region bounded by the line y = 3x + 2, the x-axis, and the ordinates x = -1 and x = 1.

**Solution**:

First, find where y = 0: 3x + 2 = 0 ⟹ x = -2/3

Since -1 < -2/3 < 1, the line crosses the x-axis within our limits.

From x = -1 to x = -2/3: the line is below x-axis (y < 0)

From x = -2/3 to x = 1: the line is above x-axis (y > 0)

Area = |∫₋₁⁻²/³ (3x + 2) dx| + |∫₋₂/₃¹ (3x + 2) dx|

First integral:

∫₋₁⁻²/³ (3x + 2) dx = [(3x²/2) + 2x]₋₁⁻²/³ = [3(4/9)/2 - 4/3] - [3/2 - 2] = [2/3 - 4/3] - [-1/2] = -2/3 + 1/2 = -1/6

Area₁ = |-1/6| = 1/6

Second integral:

∫₋₂/₃¹ (3x + 2) dx = [(3x²/2) + 2x]₋₂/₃¹ = [3/2 + 2] - [2/3 - 4/3] = 7/2 - (-2/3) = 7/2 + 2/3 = 25/6

Total Area = 1/6 + 25/6 = **26/6 = 13/3 square units**

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Example 4: Area Under Trigonometric Curves

**Problem**: Find the area bounded by y = cos x between x = 0 and x = 2π.

**Solution**:

The cosine function oscillates above and below the x-axis:

From 0 to π/2: cos x > 0

From π/2 to 3π/2: cos x < 0

From 3π/2 to 2π: cos x > 0

Area = |∫₀^(π/2) cos x dx| + |∫_(π/2)^(3π/2) cos x dx| + |∫_(3π/2)^(2π) cos x dx|

Each integral: ∫cos x dx = sin x + C

First: [sin x]₀^(π/2) = 1 - 0 = 1

Second: [sin x]_(π/2)^(3π/2) = -1 - 1 = -2, so |−2| = 2

Third: [sin x]_(3π/2)^(2π) = 0 - (-1) = 1

Total Area = 1 + 2 + 1 = **4 square units**

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Exam-Important Points

**Always check if the curve crosses the x-axis** within integration limits; if yes, split the integral

Use **absolute values** for regions below the x-axis

For **standard curves (circle, ellipse, parabola)**, memorize the standard integration formulas involving inverse trigonometric functions

When using **horizontal strips**, integrate with respect to y, and the curve should be expressed as x = g(y)

**Symmetry** reduces calculation: if a region is symmetric about x or y-axis, find area of one symmetric part and multiply

Always check **limits carefully** — which comes first (smaller value at lower limit)

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Common Mistakes to Avoid

Forgetting to take absolute values of negative areas

Not identifying where curves cross the axis before integrating

Confusing vertical strips (dx) with horizontal strips (dy)

Making errors in standard antiderivative formulas, especially for √(a² - x²) type integrals

Not using symmetry when it's available, leading to unnecessarily complex calculations