Definite Integral - Application Of Definite Integration With Examples Motivating The Definite Integral

Definite Integral - Application of definite integration with examples motivating the definite integral

Introduction to definite integral
Motivation behind using definite integral
Applications of definite integral in real-life scenarios
Examples illustrating the concept of definite integral
Techniques to evaluate definite integrals

Introduction to Definite Integral

Definition: The definite integral of a function f between two points a and b represents the area under the curve of the function over the interval [a, b].
Denoted as ∫_a^b f(x) dx
It is a fundamental concept in calculus that deals with the accumulation of infinitesimal quantities.

Motivation behind using Definite Integral

Calculating displacement from velocity function
Finding total cost from cost function
Determining total work done from the force function
Measuring total change in population over time

Applications of Definite Integral in Real-Life Scenarios

Area Calculation:
- Finding the area enclosed by a curve
- Determining the area of irregular shapes

Physics:
- Calculating work done by a variable force
- Determining the center of mass of an object

Economics:
- Calculating total revenue or profit
- Determining consumer surplus

Probability:
- Calculating probabilities using density functions
- Finding expected values of random variables

Examples Illustrating the Concept of Definite Integral

Example 1: For the function f(x) = x² over the interval [0, 2], find the definite integral ∫₀² x² dx. Example 2: A car’s velocity function is given by v(t) = 3t² + 2t. Find the displacement of the car between t=1 and t=3. Example 3: The population of a city at time t is given by the function P(t) = 100e^0.05t, where t is measured in years. Find the total change in population from t=0 to t=10.

Techniques to Evaluate Definite Integrals

Use the fundamental theorem of calculus:
- If F(x) is the antiderivative of f(x), then ∫_a^b f(x) dx = F(b) - F(a)

Substitution method:
- Substitute a variable to simplify the integrand

Integration by parts:
- Use the product rule for differentiation in reverse

Partial fraction decomposition:
- Break down a fraction into simpler fractions

Trigonometric identities:
- Use trigonometric identities to simplify the integrand

Summary

The definite integral calculates the area under a curve over a given interval.
It has numerous applications in various fields such as physics, economics, and probability.
Techniques like substitution, integration by parts, and trigonometric identities are used to evaluate definite integrals.

Techniques to Evaluate Definite Integrals (Continued)

Integration by substitution:
- Substitute a part of the integrand using a new variable
Integration using trigonometric identities:
- Use trigonometric identities to simplify the integrand
Integration of rational functions:
- Divide the numerator by the denominator and express it as a sum of partial fractions
Integration of trigonometric functions:
- Use trigonometric identities and trigonometric substitutions to integrate trigonometric functions
Integration of exponential and logarithmic functions:
- Apply the rules of differentiation in reverse to integrate exponential and logarithmic functions

Example: Evaluate ∫₁³ x³ dx using the substitution method.

Example: Evaluate ∫₀^π/2 (sin x + cos x) dx using trigonometric identities.

Example: Evaluate ∫(x² + 3x + 2)/(x+1)(x+2) dx using partial fraction decomposition.

Example: Evaluate ∫(2x + 1) √(4x + 2) dx using integration by parts.

Applications of Definite Integrals in Physics

Calculating work done by a varying force:
- The integral of force with respect to displacement gives the work done over a certain distance.
Finding the center of mass:
- The definite integral can be used to determine the center of mass of an object with varying density.
Determining the moment of inertia:
- By integrating the square of the distance from the axis of rotation, the moment of inertia can be found.

Examples: Determining Work Done and Center of Mass

Example 1: A variable force F(x) = 3x acts on an object moving from x=0 to x=5. Find the work done.
Example 2: A rod has density function ρ(x) = 2x, where x is the distance from the end of the rod. Find the center of mass of the rod from x=0 to x=4.

Applications of Definite Integrals in Economics

Calculating total revenue or profit:
- The definite integral of the revenue or profit function gives the total revenue or profit over a certain time period.
Determining consumer surplus:
- The area under the demand curve and above the price line represents consumer surplus.

Example: Calculating Total Profit and Consumer Surplus

Example 1: The profit function for a company is given by P(x) = 4x - x², where x is the quantity sold. Find the total profit over the range 0 to 5.
Example 2: The demand for a product is given by Q(x) = 100 - 2x, where x is the price per unit. Find the consumer surplus when the price is set at $20.

Applications of Definite Integrals in Probability

Calculating probabilities using density functions:
- The definite integral of a probability density function gives the probability of an event occurring within a certain range.
Finding expected values of random variables:
- The definite integral can be used to calculate the expected value of a random variable. Examples:
Example 1: The probability density function of a random variable X is given by f(x) = 3x² for 0 ≤ x ≤ 1. Find the probability that X lies between 0.2 and 0.8.
Example 2: The probability density function of a continuous random variable Y is given by f(y) = 2e^(-2y) for y ≥ 0. Find the expected value of Y.

Techniques to Evaluate Definite Integrals (Continued)

Integration by substitution:
- Substitute a part of the integrand using a new variable
- Example: Evaluate ∫₀¹ 2x e^x² dx using the substitution u = x²
Integration using trigonometric identities:
- Use trigonometric identities to simplify the integrand
- Example: Evaluate ∫₀^π/2 sin²x dx using the identity sin²x = (1 - cos(2x))/2
Integration of rational functions:
- Divide the numerator by the denominator and express it as a sum of partial fractions
- Example: Evaluate ∫(x² + 3x + 2)/(x+1)(x+2) dx using partial fraction decomposition
Integration of trigonometric functions:
- Use trigonometric identities and trigonometric substitutions to integrate trigonometric functions
- Example: Evaluate ∫sin³x cos²x dx using the substitution u = sin x
Integration of exponential and logarithmic functions:
- Apply the rules of differentiation in reverse to integrate exponential and logarithmic functions
- Example: Evaluate ∫x e^2x dx using integration by parts

Example: Evaluate ∫₁³ x³ dx using the substitution method.

Let’s substitute u = x², then du = 2x dx
The integral becomes ∫(1/2)u² du
Integrate u² to get (1/6)u³
Substitute back x² for u and evaluate the integral from 1 to 3: ∫₁³ (1/2)x² dx = (1/6)x³ |₁⁄³ = (1/6)(3³) - (1/6)(1³)

= (1/6)(27) - (1/6)(1)

= 27/6 - 1/6

= 26/6

= 13/3

Example: Evaluate ∫₀^π/2 (sin x + cos x) dx using trigonometric identities.

Use the identity sin²x + cos²x = 1
Rearrange the integral to ∫(1 + sin x cos x) dx
Apply the identity to get ∫1 dx + ∫(sin x cos x) dx
The first integral becomes x + C
Evaluate the second integral: ∫sin x cos x dx = (1/2)∫sin(2x) dx

= (1/2)(-1/2)cos(2x) + C

= (-1/4)cos(2x) + C
Add the two integrals together: x + (-1/4)cos(2x) + C

Example: Evaluate ∫(2x + 1) √(4x + 2) dx using integration by parts.

Use the formula for integration by parts: ∫u dv = uv - ∫v du
Let u = (2x + 1), dv = √(4x + 2) dx
Calculate du and v: du = 2 dx

v = (2/3)(4x + 2)^(3/2)
Apply the formula: ∫(2x + 1) √(4x + 2) dx = (2x + 1)(2/3)(4x + 2)^(3/2) - ∫(2/3)(4x + 2)^(3/2) dx
Simplify the integrals and evaluate: = (4/3)(2x + 1)(4x + 2)^(3/2) - (2/9)(4x + 2)^(5/2) + C

Applications of Definite Integrals in Physics

Calculating work done by a varying force:
- The integral of force with respect to displacement gives the work done over a certain distance.
- Example: Find the work done by a force F(x) = 3x when an object moves from x=0 to x=5.
Finding the center of mass:
- The definite integral can be used to determine the center of mass of an object with varying density.
- Example: Find the center of mass of a rod with density function ρ(x) = 2x from x=0 to x=4.
Determining the moment of inertia:
- By integrating the square of the distance from the axis of rotation, the moment of inertia can be found.
- Example: Find the moment of inertia of a thin rod of length L rotating about its center.

Example: Determining Work Done and Center of Mass

Example 1: A variable force F(x) = 3x acts on an object moving from x=0 to x=5. Find the work done.
- The work done is given by the definite integral ∫₀⁵ F(x) dx.
- Substitute the given force function and evaluate the integral:
  
  ∫₀⁵ (3x) dx = (3/2)x² |₀⁄⁵
  
  = (3/2)(5)² - (3/2)(0)²
  
  = (3/2)(25) - (3/2)(0)
  
  = 75/2
Example 2: A rod has density function ρ(x) = 2x, where x is the distance from the end of the rod. Find the center of mass of the rod from x=0 to x=4.
- The center of mass is given by the definite integral ∫x ρ(x) dx / ∫ρ(x) dx.
- Substitute the given density function and evaluate the integrals:
  
  ∫₀⁴ x(2x) dx / ∫₀⁴ (2x) dx
  
  = (∫₀⁴ 2x³ dx) / (∫₀⁴ 2x dx)
  
  = [(1/2)x⁴] |₀⁄⁴ / [x²] |₀⁄⁴
  
  = (1/2)(4)⁴/(4)²
  
  = 16/4
  
  = 4

Applications of Definite Integrals in Economics

Calculating total revenue or profit:
- The definite integral of the revenue or profit function gives the total revenue or profit over a certain time period.
- Example: Find the total profit over the range 0 to 5 given the profit function P(x) = 4x - x².
Determining consumer surplus:
- The area under the demand curve and above the price line represents consumer surplus.
- Example: Find the consumer surplus when the price is set at $20 for the demand function Q(x) = 100 - 2x.

Example: Calculating Total Profit and Consumer Surplus

Example 1: The profit function for a company is given by P(x) = 4x - x², where x is the quantity sold. Find the total profit over the range 0 to 5.
- The total profit is given by the definite integral ∫₀⁵ P(x) dx.
- Substitute the given profit function and evaluate the integral:
  
  ∫₀⁵ (4x - x²) dx = (2x² - (1/3)x³) |₀⁄⁵
  
  = (2(5)² - (1/3)(5)³) - (2(0)² - (1/3)(0)³)
  
  = (2(25) - (1/3)(125)) - (0 - 0)
  
  = 50 - 125/3
  
  = 150/3 - 125/3
  
  = 25/3
Example 2: The demand for a product is given by Q(x) = 100 - 2x, where x is the price per unit. Find the consumer surplus when the price is set at $20.
- The consumer surplus is given by the definite integral ∫₀²⁰ (100 - 2x) dx.
- Substitute the given demand function and evaluate the integral:
  
  ∫₀²⁰ (100 - 2x) dx = (100x - x²) |₀⁄²⁰
  
  = (100(20) - (20)²) - (100(0) - (0)²)
  
  = (2000 - 400) - (0 - 0)
  
  = 1600

Applications of Definite Integrals in Probability

Calculating probabilities using density functions:
- The definite integral of a probability density function gives the probability of an event occurring within a certain range.
- Example: Find the probability that a random variable X lies between 0.2 and 0.8, given its probability density function f(x) = 3x² for 0 ≤ x ≤ 1.
Finding expected values of random variables:
- The definite integral can be used to calculate the expected value of a random variable.
- Example: Find the expected value of a continuous random variable Y with probability density function f(y) = 2e^(-2y) for y ≥ 0.

Example: Calculating Probabilities and Expected Value

Example 1: The probability density function of a random variable X is given by f(x) = 3x² for 0 ≤ x ≤ 1. Find the probability that X lies between 0.2 and 0.8.
- The probability is given by the definite integral ∫_0.2^0.8 3x² dx.
- Substitute the given density function and evaluate the integral:
  
  ∫_0.2^0.8 3x² dx = x³ |_0.2⁄^0.8
  
  = (0.8)³ - (0.2)³
  
  = 0.512 - 0.008
  
  = 0.504
Example 2: The probability density function of a continuous random variable Y is given by f(y) = 2e