Lecture Notes: Fundamental Theorem of Calculus - Part 1

Key Concepts

• Fundamental Theorem of Calculus (Part 1): If ( g(x) ) is the definite integral of ( f(t) ) from ( a ) to ( x ), then ( g'(x) = f(x) ).
  • ( g ) is the antiderivative of ( f ), thus the derivative of ( g ) returns the original function ( f ).
  • Expressed as: ( \frac{d}{dx} \left( \int_{a}^{x} f(t) , dt \right) = f(x) )._

Problem Solving

• Differentiating an Integral:

  • To find ( \frac{d}{dx} \left( \int_{0}^{x} \sqrt{t^2 + 4} , dt \right) ):
    • The solution is ( \sqrt{x^2 + 4} ), using the process of replacing ( t ) with ( x ).
  • Key Insight: Integration of ( f(t) ) from ( a ) to ( x ) leads to antiderivative ( F(t) ) evaluated; its derivative provides ( f(x) ).

• Example with Different Limits:

  • For ( \int_{x}^{4} \sqrt{t^3 + 5} , dt ):
    • Define ( f(t) = \sqrt{t^3 + 5} ).
    • The result is (-f(x) = -\sqrt{x^3 + 5} ) due to limits order (upper ( \to ) lower).

Chain Rule Application

• Example with variable upper limit:
  • ( \int_{5}^{x^2} \sqrt{t^3 - 4} , dt ):
    • Replace ( t ) with ( x^2 ): ( \sqrt{(x^2)^3 - 4} ).
    • Multiply by derivative of ( x^2 ): ( 2x ).
    • Result: ( 2x \sqrt{x^6 - 4} )._

Combining Limits with Chain Rule

• Complex Example:
  • ( \int_{x^2}^{x^3} \sqrt{t^4 - 2} , dt ):
    • Top limit: ( \sqrt{(x^3)^4 - 2} \times 3x^2 ).
    • Bottom limit: ( \sqrt{(x^2)^4 - 2} \times 2x ).
    • Result: ( 3x^2 \sqrt{x^{12} - 2} - 2x \sqrt{x^8 - 2} )._

Conclusion

• The fundamental theorem connects differentiation and integration, allowing for evaluation of derivatives of integrals directly in terms of the integrand.