Definite Integral - Introduction

• The definite integral is a fundamental concept in calculus.
• It measures the area under a curve between two given points.
• It is denoted by ∫ (integral symbol) followed by a function f(x), and the limits of integration.
• It represents the net accumulation of a quantity over a given interval.

Definite Integral - Properties

• Linearity: ∫ (k * f(x) + g(x)) dx = k * ∫ f(x) dx + ∫ g(x) dx, where k is a constant.
• Interval Addition: ∫ f(x) dx from a to b = ∫ f(x) dx from a to c + ∫ f(x) dx from c to b, where a < c < b.
• Reverse Interval: ∫ f(x) dx from a to b = -∫ f(x) dx from b to a.
• Additive Property: ∫ f(x) dx from a to b + ∫ f(x) dx from b to c = ∫ f(x) dx from a to c.

Definite Integral - Geometric Interpretation

• The definite integral represents the area between the curve and the x-axis over a given interval.
• The curve can be above or below the x-axis.
• The area above the x-axis is positive, while the area below the x-axis is negative.
• The net area is the sum of these individual areas.

Definite Integral - Evaluation Techniques

1. Substitution Method:
   • Substitute a new variable to simplify the integral.
   • Differentiate the new variable to find dx.
   • Replace the integral with respect to the new variable, and solve.
   • Substitute back to obtain the final answer.

2. Integration by Parts:
   • Choose two parts of the integrand to differentiate and integrate.
   • Apply the formula ∫ u dv = u v - ∫ v du.
   • Simplify and integrate the remaining integrals.
   • Solve for the variable of interest.

Definite Integral - Examples

1. Calculate ∫ (3x^2 + 2x - 1) dx from 0 to 2.

2. Evaluate ∫ (sin(x) + cos(x)) dx from 0 to π/4.

3. Find ∫ (e^x / (1 + e^x))^2 dx from -∞ to ∞.

4. Determine ∫ (x^3 - 5x^2 + 4x - 3) dx from -2 to 3.

5. Solve ∫ (1/x^2 - x^2) dx from 1 to 2.

Definite Integral - Applications

• Area: Calculating the area enclosed by curves.
• Volume: Determining the volume of 3D shapes using cross-sectional areas.
• Work: Calculating work done by a force over a distance.
• Probability: Finding the probability density function from a continuous random variable.
• Moments: Calculating the moment of inertia for solid objects.
• Statistics: Calculating expected values and probabilities in statistics.

Definite Integral - Summary

• The definite integral measures the net accumulation of a quantity.
• It represents the area under a curve between two points.
• It has various properties and can be evaluated using different techniques.
• Applications include finding areas, volumes, work, probabilities, and moments.
• Practice solving examples to strengthen your understanding.

11. Definite Integral - Finite Sums Method

• The finite sums method is a technique used to approximate the value of a definite integral.
• It involves dividing the interval into smaller subintervals and approximating the area using rectangles.
• The width of each rectangle is determined by dividing the total interval width by the number of subintervals.
• The height of each rectangle is determined by evaluating the function at a specific point within the subinterval.
• The accuracy of the approximation increases as the number of subintervals increases.

12. Definite Integral - Finite Sums Method Example

• Example: Approximate the integral ∫ (2x + 1) dx from 0 to 4 using the finite sums method with 4 subintervals.
• Divide the interval [0, 4] into 4 equal subintervals: [0, 1], [1, 2], [2, 3], and [3, 4].
• Calculate the width of each subinterval: Δx = (4 - 0) / 4 = 1.
• Evaluate the function at a specific point within each subinterval, e.g., the left endpoint.
• Calculate the area of each rectangle: Area = base * height.

13. Definite Integral - Finite Sums Method Example (cont.)

• Subinterval [0, 1]:
  • Left endpoint: x = 0.
  • Height: f(x) = 2(0) + 1 = 1.
  • Area: Δx * f(x) = 1 * 1 = 1.
• Subinterval [1, 2]:
  • Left endpoint: x = 1.
  • Height: f(x) = 2(1) + 1 = 3.
  • Area: Δx * f(x) = 1 * 3 = 3.
• Subinterval [2, 3]:
  • Left endpoint: x = 2.
  • Height: f(x) = 2(2) + 1 = 5.
  • Area: Δx * f(x) = 1 * 5 = 5.
• Subinterval [3, 4]:
  • Left endpoint: x = 3.
  • Height: f(x) = 2(3) + 1 = 7.
  • Area: Δx * f(x) = 1 * 7 = 7.

14. Definite Integral - Finite Sums Method Example (cont.)

• Sum up the areas of all the rectangles:
  • Total Approximation = Area of 1st rectangle + Area of 2nd rectangle + …
  • Total Approximation = 1 + 3 + 5 + 7 = 16.
• Therefore, the approximate value of ∫ (2x + 1) dx from 0 to 4 using the finite sums method with 4 subintervals is 16.

15. Definite Integral - Integration by Substitution

• Integration by substitution is a powerful technique used to evaluate definite integrals.
• It involves making a substitution by introducing a new variable to simplify the integral.
• The choice of the new variable is typically based on the form of the function being integrated.
• After the substitution, the integral is transformed into a new integral with respect to the new variable.
• Once the new integral is solved, the substitution is reversed to obtain the final answer.

16. Definite Integral - Integration by Substitution Example

• Example: Evaluate the integral ∫ (2x + 1)^2 dx from 0 to 4 using integration by substitution.
• Let u = 2x + 1.
• Differentiate both sides of the equation with respect to x to find du/dx = 2.
• Solve for dx: dx = du/2.

17. Definite Integral - Integration by Substitution Example (cont.)

• Substitute the expression for x and dx in the integral:
  • New integral = ∫ u^2 * (du/2) from 0 to 4.
• Simplify the integral:
  • New integral = (1/2) * ∫ u^2 du from 0 to 4.
• Integrate with respect to the new variable:
  • New integral = (1/2) * (u^3/3) from 0 to 4.

18. Definite Integral - Integration by Substitution Example (cont.)

• Substitute the original variable back into the expression:
  • New integral = (1/2) * (2x + 1)^3/3 from 0 to 4.
• Evaluate the integral at the upper and lower limits:
  • New integral = [(1/2) * (2(4) + 1)^3/3] - [(1/2) * (2(0) + 1)^3/3].
• Simplify the expression:
  • New integral = [(1/2) * (9^3/3)] - [(1/2) * (1^3/3)].

19. Definite Integral - Integration by Substitution Example (cont.)

• Calculate the values:
  • New integral = [(1/2) * (729/3)] - [(1/2) * (1/3)].
• Simplify further:
  • New integral = (243/2) - (1/6).
• Therefore, the value of the integral ∫ (2x + 1)^2 dx from 0 to 4 using integration by substitution is (243/2) - (1/6).

20. Definite Integral - Review

• The finite sums method can be used to approximate definite integrals.
• Integration by substitution is a technique to evaluate definite integrals.
• Practice different types of examples to strengthen your understanding.
• Familiarize yourself with various techniques and properties of definite integrals.
• Understanding the geometric interpretation of definite integrals is crucial for solving real-life problems.

21. Definite Integral - Integration by Parts

• Integration by parts is another technique used to evaluate definite integrals.
• It involves choosing two parts of the integrand to differentiate and integrate.
• The formula for integration by parts is ∫ u dv = uv - ∫ v du.
• The choice of which part to differentiate and which part to integrate depends on simplification.
• This method is particularly useful for products of functions.

22. Definite Integral - Integration by Parts Example

• Example: Evaluate the integral ∫ x*e^x dx from 0 to 1 using integration by parts.
• Choose u = x and dv = e^x dx.
• Differentiate u to find du = dx and integrate dv to find v = e^x.

23. Definite Integral - Integration by Parts Example (cont.)

• Apply the integration by parts formula:
  • ∫ x*e^x dx = x * e^x - ∫ e^x dx.
• Simplify the new integral:
  • ∫ e^x dx = e^x.
• Substitute the new integral and apply the limits:
  • Integral from 0 to 1: x * e^x - e^x from 0 to 1.

24. Definite Integral - Integration by Parts Example (cont.)

• Evaluate the integral at the limits:
  • (1 * e^1 - e^1) - (0 * e^0 - e^0).
• Simplify the expression:
  • (e - e) - (0 - 1).
• Therefore, the value of the integral ∫ x*e^x dx from 0 to 1 using integration by parts is -1.

25. Definite Integral - Applications in Physics

• The definite integral has numerous applications in physics.
• Example 1: Calculate the work done by a force on an object over a given distance.
• Example 2: Determine the displacement of an object based on its velocity function.
• Example 3: Find the area under a velocity or acceleration graph to determine distance, speed, or change in speed.
• Example 4: Calculate the center of mass for an object with non-uniform density.
• Example 5: Determine the total mass or charge in a given region.

26. Definite Integral - Applications in Economics

• The definite integral also has applications in economics.
• Example 1: Calculate the total revenue of a business by integrating the demand function.
• Example 2: Find the consumer surplus by integrating the demand and price functions.
• Example 3: Determine the total cost of production by integrating the cost function.
• Example 4: Calculate the producer surplus by integrating the supply and price functions.
• Example 5: Find the equilibrium quantity and price in a market by equating the demand and supply functions.

27. Definite Integral - Applications in Probability

• The definite integral is used to calculate probabilities in continuous random variables.
• Example 1: Find the probability density function (PDF) from a cumulative distribution function (CDF).
• Example 2: Determine the mean or expected value by integrating the random variable multiplied by the PDF.
• Example 3: Calculate the variance by integrating the square of the random variable multiplied by the PDF.
• Example 4: Find the cumulative distribution function by integrating the PDF.
• Example 5: Determine quantiles or percentiles by finding the value where the CDF equals a given probability.

28. Definite Integral - Applications in Engineering

• The definite integral plays a vital role in various engineering disciplines.
• Example 1: Calculate the moment of inertia of an object about a given axis.
• Example 2: Determine the load distribution on a beam by integrating the distributed load function.
• Example 3: Find the deflection of a beam under a given load using the flexure formula.
• Example 4: Calculate the thermal energy transfer in a system using the heat transfer equation.
• Example 5: Determine the power output in an electrical circuit using the power formula.

29. Definite Integral - Importance of Practice

• To excel in integration and definite integrals, practice is crucial.
• Solve a wide variety of problems to develop fluency and speed.
• Understand the concepts, properties, and techniques thoroughly.
• Practice integrating different functions to gain expertise.
• Apply integration to real-world problems to develop problem-solving skills.

30. Definite Integral - Conclusion

• The definite integral is a powerful tool in calculus, with diverse applications.
• It measures the accumulation, area, and net effect of a quantity.
• Integration by substitution and integration by parts are two important techniques for solving definite integrals.
• The finite sums method provides an approximation of definite integrals.
• Understanding the geometric interpretation and the applications of definite integrals is vital for success.