Differential Equations - Definite Integrals

• Introduction to definite integrals
• Understanding the concept of a limit
• Definition of a definite integral
• Notation for definite integrals

Properties of Definite Integrals

• Linearity property
• Constant multiple rule
• Splitting an integral into two parts
• Integrating over a symmetric interval

Fundamental Theorem of Calculus

• Statement of the fundamental theorem
• Relationship between derivatives and integrals
• Evaluating definite integrals using antiderivatives

Techniques for Evaluating Definite Integrals

• Substitution method
• Integration by parts
• Partial fractions
• Trigonometric substitutions

Properties of Definite Integrals (cont.)

• Changing the limits of integration
• Using the mean value theorem for integrals
• Relation between the value of a definite integral and the area under a curve

Applications of Definite Integrals

• Finding the area between two curves
• Calculating volumes of solids of revolution
• Computing the average value of a function
• Solving motion problems with definite integrals

Definite Integrals of Continuous Functions

• Definition of continuity
• Theorem for existence of definite integrals
• Riemann sums and the limit as the partition becomes finer

Definite Integrals of Discontinuous Functions

• Types of discontinuities
• Jump discontinuities
• Infinite discontinuities
• Discontinuous functions with integrable singularities

Advanced Techniques for Evaluating Definite Integrals

• Integration by parts with parametric equations
• Trigonometric substitutions with nested radicals
• Trigonometric substitutions with multiple trigonometric functions
• Integration involving hyperbolic functions

Summary

• Definition and notation of definite integrals
• Properties of definite integrals
• Fundamental theorem of calculus
• Techniques for evaluating definite integrals
• Applications of definite integrals

Definite Integrals - Techniques for Evaluating (cont.)

• Integration by trigonometric substitution
  • Example: Evaluating the integral of √(9 - x^2) dx using trigonometric substitution
• Integration involving rational functions
  • Example: Evaluating the integral of (3x^2 + 2x - 1) / (x^3 + x^2) dx using partial fractions
• Integration involving exponential and logarithmic functions
  • Example: Evaluating the integral of e^x / (e^x + 1) dx
• Integration involving inverse trigonometric functions
  • Example: Evaluating the integral of dx / √(9 - x^2)

Applications of Definite Integrals (cont.)

• Solving differential equations
  • Example: Solving the differential equation dy/dx = x^2 + 1
• Finding the arc length of a curve
  • Example: Finding the arc length of the curve y = x^2/2 between x = 0 and x = 2
• Calculating the surface area of revolution
  • Example: Finding the surface area of the solid generated by revolving y = x^2 about the x-axis between x = 1 and x = 3
• Evaluating improper integrals
  • Example: Evaluating the improper integral of 1 / (x^2 + 1) dx from 0 to ∞

Definite Integrals of Continuous Functions (cont.)

• The mean value theorem for integrals
  • Mean value theorem states that if f is continuous on [a, b], then there exists a number c in (a, b) such that the definite integral of f from a to b is equal to f(c) times (b - a)
• Computing the average value of a function
  • Example: Finding the average value of f(x) = x^3 on the interval [-1, 2]
• Finding the area between two curves
  • Example: Finding the area between the curves y = x^2 and y = 2x - 1 on the interval [0, 2]

Definite Integrals of Discontinuous Functions (cont.)

• Properties of integrable singularities
  • Example: Evaluating the integral of 1 / x dx from 0 to 1

Definite Integrals - Advanced Techniques

• Integration involving logarithmic functions
  • Example: Evaluating the integral of ln(x) dx using integration by parts
• Integration involving trigonometric functions
  • Example: Evaluating the integral of sin^2(x) cos^2(x) dx using trigonometric identities
• Integration involving hyperbolic functions
  • Example: Evaluating the integral of sinh(x) cosh(x) dx

Definite Integrals - Properties (cont.)

• Using the change of variables
  • Example: Evaluating the integral of x^2 sin(x^3) dx using the substitution u = x^3
• Evaluating definite integrals with infinite limits
  • Example: Evaluating the integral of 1 / (x^2 + 1) dx from -∞ to ∞
• Evaluating definite integrals with variable limits
  • Example: Evaluating the integral of (3x + 1) dx from x = 1 to x = 3

Definite Integrals - Applications (cont.)

• Calculating volumes of solids of revolution
  • Example: Finding the volume of the solid generated by revolving the region between y = x^2 and y = 3x around the y-axis
• Evaluating definite integrals as a limit of Riemann sums
  • Example: Evaluating the integral of x^3 dx from 0 to 1 using Riemann sums with n subintervals
• Solving motion problems with definite integrals
  • Example: Solving a velocity problem to find displacement using integrals

Summary

• Various techniques for evaluating definite integrals
• Applications of definite integrals in various fields
• Properties of definite integrals including linearity and change of variables
• Advanced techniques for evaluating definite integrals
• Solving problems involving discontinuous functions and integrable singularities

Practice Problems

• Evaluate the following definite integrals:
  • ∫(2x + 3) dx from 1 to 4
  • ∫sin(2x) dx from 0 to π/2
  • ∫(x^2 + 2x - 1) dx from -2 to 2
• Find the area between the curves y = x^2 and y = 3 - x^2 on the interval [-1, 1]
• Calculate the volume of the solid generated by revolving y = e^x between x = 0 and x = 1 around the x-axis

References

• Stewart, J. (2015). Calculus: Early Transcendentals. Cengage Learning.

Definite Integrals - Techniques for Evaluating (cont.)

• Integration by trigonometric substitution
  • Example: Evaluating the integral of √(9 - x^2) dx using trigonometric substitution
    • Solution: Let x = 3sinθ, then dx = 3cosθdθ
    • The integral becomes ∫√(9 - 9sin^2θ) * 3cosθdθ
    • Simplifying further, we get ∫3cos^2θdθ
    • Using the trigonometric identity cos^2θ = (1 + cos2θ)/2, the integral becomes ∫(3/2)(1 + cos2θ)dθ
    • Finally, integrating with respect to θ, we get (3/2)(θ + (1/2)sin2θ) + C
• Integration involving rational functions
  • Example: Evaluating the integral of (3x^2 + 2x - 1) / (x^3 + x^2) dx using partial fractions
    • Solution: First, perform the partial fraction decomposition, which gives us (3x^2 + 2x - 1) / (x^3 + x^2) = A/x + B/(x + 1) + C/(x^2)
    • The integral becomes ∫A/x dx + ∫B/(x + 1) dx + ∫C/(x^2) dx
    • We can then solve for the constants A, B, and C using algebraic methods
    • After finding the values of A, B, and C, we can integrate each term separately to find the final result
• Integration involving exponential and logarithmic functions
  • Example: Evaluating the integral of e^x / (e^x + 1) dx
    • Solution: Let u = e^x + 1, then du = e^x dx
    • The integral becomes ∫(1/u)du = ln|u| + C = ln|e^x + 1| + C

Definite Integrals - Techniques for Evaluating (cont.)

• Integration involving inverse trigonometric functions
  • Example: Evaluating the integral of dx / √(9 - x^2)
    • Solution: Let x = 3sinθ, then dx = 3cosθdθ
    • The integral becomes ∫(3cosθ) / (3cosθ) dθ = θ + C
    • Since x = 3sinθ, we can use arcsin(x/3) as the substitution for θ
    • Therefore, the final result is arcsin(x/3) + C
• Solving differential equations
  • Example: Solving the differential equation dy/dx = x^2 + 1
    • Solution: We can solve this differential equation by integrating both sides with respect to x
    • ∫dy = ∫(x^2 + 1) dx
    • y = (1/3)x^3 + x + C, where C is the constant of integration
• Finding the arc length of a curve
  • Example: Finding the arc length of the curve y = x^2/2 between x = 0 and x = 2
    • Solution: We can use the formula for arc length: L = ∫√(1 + (dy/dx)^2) dx
    • Plugging in the given function, we get L = ∫√(1 + (x^2)^2) dx = ∫√(1 + x^4) dx
    • Evaluating the integral between x = 0 and x = 2 gives us the arc length of the curve

Definite Integrals - Techniques for Evaluating (cont.)

• Calculating the surface area of revolution
  • Example: Finding the surface area of the solid generated by revolving y = x^2 about the x-axis between x = 1 and x = 3
    • Solution: We can use the formula for surface area of revolution: S = 2π∫y√(1 + (dy/dx)^2) dx
    • Plugging in the given function, we get S = 2π∫x^2√(1 + (2x)^2) dx
    • Evaluating the integral between x = 1 and x = 3 gives us the surface area of the solid
• Evaluating improper integrals
  • Example: Evaluating the improper integral of 1 / (x^2 + 1) dx from 0 to ∞
    • Solution: We can evaluate this improper integral by taking the limit as the upper limit of integration approaches infinity
    • ∫(1 / (x^2 + 1)) dx = lim[x->∞] ∫(1 / (x^2 + 1)) dx = lim[x->∞] arctan(x) = π/2
• The mean value theorem for integrals
  • Mean value theorem states that if f is continuous on [a, b], then there exists a number c in (a, b) such that the definite integral of f from a to b is equal to f(c) times (b - a)

Definite Integrals of Continuous Functions (cont.)

• Computing the average value of a function
  • Example: Finding the average value of f(x) = x^3 on the interval [-1, 2]
    • Solution: We can calculate the average value of a function by taking the definite integral of the function over the interval and dividing it by the length of the interval
    • Average value of f(x) = (1 / (2 - (-1))) * ∫(x^3) dx = (1/3) * (1/4)x^4 | from -1 to 2 = (1/3) * (2^4 - (-1)^4) = 9/3 = 3
• Finding the area between two curves
  • Example: Finding the area between the curves y = x^2 and y = 2x - 1 on the interval [0, 2]
    • Solution: To find the area between two curves, we need to calculate the definite integral of the difference between the curves over the given interval
    • Area = ∫[(2x - 1) - x^2] dx = ∫(2x - x^2 - 1) dx = x^2 - (1/3)x^3 - x | from 0 to 2 = 2^2 - (1/3)(2^3) - 2 - (0^2 - (1/3)(0^3) - 0) = 10/3
• Solving motion problems with definite integrals
  • Example: Solving a velocity problem to find displacement using integrals
    • Solution: Given the velocity function v(t), we can find the displacement by evaluating the definite integral of the velocity function over the given time interval
    • Displacement = ∫v(t) dt

Definite Integrals of Discontinuous Functions (cont.)

• Types of discontinuities
  • Removable discontinuity: A point where the function has a hole but can be filled in to make the function continuous
  • Jump discontinuity: A point where the function has a sudden jump in value
  • Infinite discontinuity: A point where the function approaches infinity or negative infinity
• Jump discontinuities
  • Example: Evaluating the integral of f(x) = {x-1, x < 1; x^2 + 1, x ≥ 1} from 0 to 2
    • Solution: We need to split the integral into two parts at x = 1, where the function has a jump discontinuity
    • ∫f(x) dx = ∫[x-1, 0 to 1] + ∫[x^2 + 1, 1 to 2] = [(1/2)x^2 - x | from 0 to 1] + [(1/3)x^3 + x | from 1 to 2]
• Infinite discontinuities
  • Example: Evaluating the integral of f(x) = 1 / x from 0 to 1
    • Solution: The function f(x) has an infinite discontinuity at x = 0
    • We can evaluate the integral by splitting it into two parts as ∫(1 / x) dx = ∫[1 / x, 0 to 1] + ∫[1 / x, 1 to 1+ε] + ∫[1 / x, 1+ε to 1] for a small positive ε
• Discontinuous functions with integrable singularities
  • Example: Evaluating the integral of f(x) = 1 / x from 0 to 1
    • Solution: The function f(x) has an integrable singularity at x = 0
    • We can evaluate the integral as a limit as x approaches 0 as ∫(1 / x) dx = lim[x->0] ∫(1 / x) dx = lim[x->0] ln|x| = -∞

Definite Integrals - Advanced Techniques

• Integration involving logarithmic functions
  • Example: Evaluating the integral of ln(x) dx using integration by parts
    • Solution: We can use integration by parts to evaluate this integral
    • ∫ln(x) dx = xln(x) - ∫(1/x) x dx = xln(x) - ∫dx = xln(x) - x + C
• Integration involving trigonometric functions
  • Example: Evaluating the integral of sin^2(x) cos^2(x) dx using trigonometric identities
    • Solution: We can use the double angle identity for cosine to simplify this integral
    • sin^2(x) cos^2(x) = (1 - cos^2(x)) cos^2(x) = cos^2(x) - cos^4(x)
    • Integrating term by term, we get ∫(cos^2(x) - cos^4(x)) dx = (1/2)x + (1/3)sin(2x) + C
• Integration involving hyperbolic functions
  • Example: Evaluating the integral of sinh(x) cosh(x) dx
    • Solution: We can use the identity sinh(x) cosh(x) = (1/2)sinh(2x)
    • The integral becomes ∫(1/2)sinh(2x) dx = (1/4) cosh(2x) + C

Definite Integrals - Properties (cont.)

• Using the change of variables
  • Example: Evaluating the integral of x^2 sin(x^3) dx using the substitution u = x^3
    • Solution: Let u = x^3, then du = 3x^2 dx
    • The integral becomes ∫(1/3)sin(u)