Fundamentals of Calculus

Key Areas of Calculus

1. Limits

   • Evaluate a function as x approaches a specific value.
   • Useful when the function is undefined at a specific point.

2. Derivatives

   • Provide the slope of the original function at some value.
   • The derivative function of f(x) is denoted as f'(x).
   • Useful for calculating rates of change.

3. Integration

   • Opposite of derivatives, finds the area under the curve.
   • Useful for calculating accumulation over time.
   • Integral of f'(x) returns the original function f(x).
   • Associated with anti-differentiation.

Limits

• Example Problem: Evaluating Limits
  • Function: ( \frac{x^2 - 4}{x - 2} )
  • Direct substitution at x = 2 yields ( \frac{0}{0} ) (indeterminate).
  • Use limit to find: ( \lim_{{x \to 2}} \frac{x^2 - 4}{x - 2} )
  • Factor form: ( \frac{(x + 2)(x - 2)}{x - 2} )
  • Simplifies to: ( \lim_{{x \to 2}} (x + 2) = 4 )
  • Conclusion: As x gets closer to 2, f(x) approaches 4.

Derivatives

• Power Rule: ( \frac{d}{dx} [x^n] = nx^{n-1} )

  • Example: ( \frac{d}{dx} [x^2] = 2x )

• Tangent Line

  • Tangent line touches the curve at one point.
  • Slope of tangent line equals the derivative at a specific value.
  • Example: For ( f(x) = x^3 ), ( f'(x) = 3x^2 ).
  • Slope at x = 2: ( f'(2) = 12 ).

• Secant Line

  • Touches curve at two points.
  • Slope using formula: ( \frac{y_2 - y_1}{x_2 - x_1} ).

• Limit Process for Derivatives

  • Example function: ( f(x) = x^3 )
  • Using limits to find the slope: ( \lim_{{x \to 2}} \frac{x^3 - 8}{x - 2} )
  • Factoring: ( x^3 - 8 = (x - 2)(x^2 + 2x + 4) )
  • Result: 12 (consistent with derivative calculation).

Integration

• Anti-Differentiation

  • Opposite process of differentiation.
  • Example: Integral of ( 4x^3 ) yields ( x^4 + C ).

• Comparison with Derivatives

  • Derivatives: Calculate slope (rate of change).
  • Integration: Calculate area under the curve (accumulation).
  • Derivatives involve division (rise over run), integration involves multiplication.

• Example Problem: Water in a tank

  • Function: ( A(t) = 0.01t^2 + 0.5t + 100 )
  • Amount of water at specific times by substitution.
  • Rate of change at t = 10 using derivative: ( A'(t) = 0.02t + 0.5 ).
  • Plug in t = 10: ( A'(10) = 0.7 ) (gallons per minute).

• Accumulation of Water

  • Rate function: ( R(t) = 0.5t + 20 ).
  • Calculate total water from t = 20 to t = 100 using definite integral.
  • Integral calculation yields 4000 gallons.
  • Graphical interpretation: Area under the curve.
  • Area of rectangle and triangle to verify result.

Summary

• Limits: Evaluate function as x approaches a value.
• Derivatives: Calculate instantaneous rate of change (slope).
• Integration: Calculate total accumulation (area under the curve).

Additional Resources

• Links to more problems and video playlist for practice.