Fundamentals of Calculus

Overview

Definition: Limits help to evaluate a function at a point where it may be undefined or indeterminate.
Example: Evaluating ( f(2) ) for ( f(x) = \frac{x^2 - 4}{x - 2} ).
- Direct substitution yields ( \frac{0}{0} ) (indeterminate).
- As ( x ) approaches 2, ( f(x) ) approaches 4.
- Limit Expression: ( \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4 )
- Factoring gives: ( \lim_{x \to 2} (x + 2) = 4 )

Definition: Derivatives represent the slope of the original function at a specific value.
Notation: If ( f(x) ), the derivative is ( f'(x) ).
Power Rule: ( \frac{d}{dx}(x^n) = nx^{n-1} )
- Example:
  - Derivative of ( x^2 ) is ( 2x )
  - Derivative of ( x^3 ) is ( 3x^2 )
Tangent vs. Secant Lines:
- Tangent: touches curve at one point (slope = derivative).
- Secant: touches curve at two points (average slope).
Example Calculation:
- For ( f(x) = x^3 ), ( f'(x) = 3x^2 ).
- Slope at ( x = 2 ): ( f'(2) = 12 )

Definition: Integration is the process of finding the area under the curve (anti-derivative).
Relationship to Derivatives:
- If ( f'(x) = g(x) ), then ( \int g(x) ,dx = f(x) + C )
Anti-Derivative Example:
- Derivative of ( x^4 ) is ( 4x^3 ), so integral of ( 4x^3 ) is ( x^4 + C )
Key Concepts:
- Definite Integrals yield a number (area under the curve) while Indefinite Integrals yield a function.

Function: ( A(t) = 0.01t^2 + 0.5t + 100 )
Evaluate at specific times to find water volume.
Rate of change at ( t = 10 ): find derivative ( A'(t) )
- ( A'(t) = 0.02t + 0.5 )
- At ( t = 10 ): 0.7 gallons/minute.

Rate function: ( R(t) = 0.5t + 20 )
Accumulation from ( t = 20 ) to ( t = 100 ): Use definite integral.
- Calculate area under curve using geometry (rectangle + triangle).
- Result: 4000 gallons accumulated.