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Lecture 4.9.2: Antiderivatives

Nov 1, 2024

Antiderivatives and Indefinite Integrals

Definition of Antiderivative

  • Antiderivative: A function ( F ) is an antiderivative of ( f ) if the derivative of ( F ) is ( f ).
  • General form: ( F(x) + C ), where ( C ) is an arbitrary constant.
  • Without additional information, a function has infinite possible antiderivatives.

Indefinite Integral

  • Notation: Integral sign represents finding the antiderivative.
  • Indefinite Integral Interpretation: Considered as the antiderivative of the expression.

Examples

  • Example 1: ( \int \frac{3}{\sqrt{1-x^2}} + \frac{2 \sqrt{1-x^2}}{\sqrt{1-x^2}} , dx )

    • Simplify and find antiderivative: ( 3 \sin^{-1}(x) + 2x + C )
    • Alternative expression: (-3 \cos^{-1}(x) + 2x + C )
  • Example 2: ( \int (t^2 + 1)(2t - 5) , dt )

    • Expand and find antiderivative: ( \frac{1}{2}t^4 - \frac{5}{3}t^3 + t^2 - 5t + C )
  • Example 3: ( \int \frac{\cos x - 1}{\sin^2 x} , dx )

    • Rewrite and find antiderivative: (- \csc x + \cot x + C )

Differential Equations

Definition

  • Differential Equation: An equation involving an unknown function and its derivatives.
  • Example: ( 4x^2 y'' + 12xy' + 3y = 0 )
  • Solution: Function that satisfies equation, yields result upon substitution.

Solving Example

  • Equation: ( 2y' - 3 = \cos t )
    • Rearrange & solve: ( y' = \frac{1}{2} \cos t + \frac{3}{2} )
    • Antiderivative: ( y = \frac{1}{2} \sin t + \frac{3}{2} t + C )

Initial Value Problem

  • If y(0) = 3, find particular solution.
  • Example: ( e^x y' - 1 = e^{2x}, y(0) = 3 )
    • Solve for function ( y ) and constant.
    • General solution: ( y = e^x + x + C )
    • Particular solution with initial condition: ( C = 2 )

Motion Problems

Concepts

  • Position Function: Describes location.
  • Velocity: First derivative of position.
  • Acceleration: Second derivative of position.

Process

  • Reverse from acceleration to velocity to position using antiderivatives.

Example

  • Given: ( a(t) = -32 ) ft/s², ( v(0) = 2 ) ft/s, ( s(0) = 10 ) ft.
    • Find velocity: ( v(t) = -32t + 2 )
    • Find position: ( s(t) = -16t^2 + 2t + 10 )

Summary

  • Antiderivatives are foundational in calculus for solving differential equations and understanding motion.
  • Antiderivatives offer multiple representations, often involving inverse trigonometric functions.
  • Differential equations require finding a function, not just a number, that balances the equation.