Lecture: Fundamental Theorem of Calculus (Part 1)

Key Concepts

• Fundamental Theorem of Calculus (Part 1):
  • If ( g(x) ) is the integral ( \int_{a}^{x} f(t) , dt ), then ( g'(x) ) equals ( f(x) ).
  • The derivative of the integral from ( a ) to ( x ) of ( f(t) , dt ) equals ( f(x) ).
  • Essentially, this states that the derivative of the antiderivative returns the original function._

Example Problem 1

• Problem: Find the derivative of ( \int_{0}^{x} \sqrt{t^2 + 4} , dt ).
• Solution Approach:
  • Recognize that ( f(t) = \sqrt{t^2 + 4} ) and thus ( f(x) = \sqrt{x^2 + 4} ).
  • Directly apply the theorem: ( g'(x) = f(x) )._

Example Problem 2

• Problem: Evaluate ( \int_{x}^{4} \sqrt{t^3 + 5} , dt ).
• Solution Approach:
  • Define ( f(t) = \sqrt{t^3 + 5} ).
  • ( g(x) = F(4) - F(x) ).
  • Derivative: ( -f(x) ) since ( x ) is in the lower limit, resulting in ( -\sqrt{x^3 + 5} )._

Example Problem 3

• Problem: ( \int_{5}^{x^2} \sqrt{t^3 - 4} , dt ).
• Solution Approach:
  • Replace ( t ) with ( x^2 ) and multiply by the derivative of ( x^2 ) (chain rule).
  • Final result: ( 2x \sqrt{x^6 - 4} )._

Example Problem 4

• Problem: ( \int_{x^2}^{x^3} \sqrt{t^4 - 2} , dt ).
• Solution Approach:
  • Upper limit: ( 3x^2 \sqrt{x^{12} - 2} ) and multiply by the derivative of ( x^3 ).
  • Lower limit: ( -2x \sqrt{x^8 - 2} ) and multiply by the derivative of ( x^2 ).
  • Combine results: ( 3x^2 \sqrt{x^{12} - 2} - 2x \sqrt{x^8 - 2} )._

Summary

• The Fundamental Theorem of Calculus (Part 1) relates differentiation and integration, providing a straightforward method to compute derivatives of integrals.
• Practice with examples and utilizing the chain rule is essential for mastering this concept.
• Remember that the position of ( x ) (upper vs. lower limit) affects the sign of the result.