Definite Integral and Area

Introduction to the Definite Integral

Calculating the area under a curve is essential in calculus. This method enables us to determine the space beneath a curve, vital in fields such as physics, economics, and engineering.

Key Geometric Concepts

• Signed Areas: Measure the area between the curve and the x-axis, accounting for sign.

• Definite Integral Notation: Utilised to find the aggregate area under a curve.

  infoNote

  Definite Integral: $\int_{a}^{b} f(x) \, dx$: Used for determining the aggregate area beneath the curve of $f(x)$ from point $a$ to $b$.

Definition of Area Under a Curve

• Term: Area Under a Curve
  • Definition: The region between the curve and the x-axis, bounded by the intervals $[a, b]$.
  • Applications: Applicable to linear, quadratic, and sinusoidal functions.

Introduction to Approximation Methods

In numerous scenarios, precise area calculations for complex functions may be impractical. Approximation methods become indispensable:

• Engineering Applications: Simplify intricate calculations involving force and movement.
• Physical Simulations: Estimate trajectories in physics engines.
• Economics: Model uncertain financial trends.

Methods Overview

• Riemann Sums: Use rectangles to approximate the area under the curve.
• Trapezoidal Rule: Utilise trapezoids for enhanced accuracy.

Rectangular Approximation Method (RAM)

• Left Riemann Sum: Utilises the left endpoints of each interval.
• Right Riemann Sum: Employs the right endpoints for height calculation.
• Midpoint Riemann Sum: Bases rectangle's height on interval midpoints.

Trapezoidal Rule

• Definition: Technique where trapezoids, as opposed to rectangles, approximate the area under a curve.
• Formula: $\text{Area} = \frac{b-a}{2}(f(x_0) + 2 \sum_{i=1}^{n-1} f(x_i) + f(x_n))$

Introduction to Definite Integral Notation

A definite integral is used to find the total area under a curve between two designated points:

$\int_{a}^{b} f(x) \, dx$

Each part of this notation contributes to its meaning:

• Integral Sign $\int$: Denotes the integration process.
• Limits of Integration $a$ to $b$: Define the start and endpoint of the interval.
• Integrand $f(x)$: The function to be integrated.
• Differential $dx$: Specifies the variable of integration.

Function Examples with Geometric Shapes

Linear Function Example

• Determine the area under $f(x) = 2x$ over $[0, 3]$:
• Utilise integration: $\int_{0}^{3} 2x \, dx = 9$.

Worked Examples

Example 1: Polynomial Function

• Example: Calculate the area under $f(x) = x^2$ from $x = 1$ to $x = 3$.
  • Solution:
    1. Write the integral: $\int_{1}^{3} x^2 \, dx$
    2. Find the antiderivative: $\frac{x^3}{3}$
    3. Evaluate from 1 to 3: $\frac{3^3}{3} - \frac{1^3}{3} = 9 - \frac{1}{3} = \frac{26}{3}$

Detailed Calculation

• Calculate the area under a simple linear curve: $\int_0^2 (3x + 1) \, dx$.
  • Solution:
    • Integrate to find the antiderivative: $F(x) = \frac{3}{2}x^2 + x$
    • Evaluate at limits: $F(2) - F(0) = (\frac{3}{2}(2)^2 + 2) - (0) = 6 + 2 - 0 = 8$

Calculator Integration

• Calculator Steps:
  • Access and utilise the definite integral function.
  • Benefits: Expedite challenging integrals, particularly if functions are complex.

Practice Problems

• Polynomial: Integrate $f(x) = 3x^3$ from $1$ to $2$.

  • Solution: $\int_{1}^{2} 3x^3 \, dx = 3 \cdot \frac{x^4}{4} \big|_{1}^{2} = 3 \cdot (\frac{16}{4} - \frac{1}{4}) = 3 \cdot \frac{15}{4} = \frac{45}{4}$

• Trigonometric: Integrate $f(x) = \cos(x)$ from $0$ to $\pi/2$.

  • Solution: $\int_{0}^{\pi/2} \cos(x) \, dx = \sin(x) \big|_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$

• Exponential: Integrate $f(x) = 2e^x$ from $0$ to $\ln(2)$.

  • Solution: $\int_{0}^{\ln(2)} 2e^x \, dx = 2e^x \big|_{0}^{\ln(2)} = 2e^{\ln(2)} - 2e^0 = 2 \cdot 2 - 2 \cdot 1 = 4 - 2 = 2$

Techniques for Difficult Areas

• Substitution: Simplifies complex integrals.
• Example: $\int (2x+3)^2 \, dx$
  1. Let $u = 2x+3$.
  2. Then $du = 2dx$, so $dx = \frac{du}{2}$.
  3. Substituting: $\int (2x+3)^2 \, dx = \int u^2 \cdot \frac{du}{2} = \frac{1}{2} \int u^2 \, du = \frac{1}{2} \cdot \frac{u^3}{3} + C = \frac{(2x+3)^3}{6} + C$