The Fundamental Theorem of Calculus: Fresh Take

  • Understand the Mean Value Theorem for Integrals and both components of the Fundamental Theorem of Calculus
  • Use the Fundamental Theorem of Calculus to find derivatives of integral functions and calculate definite integrals
  • Describe how differentiation and integration are interconnected

The Mean Value Theorem for Integrals

The Main Idea 

  • Theorem Statement:
    • If [latex]f(x)[/latex] is continuous on [latex][a,b][/latex], then there exists at least one point [latex]c \in [a,b][/latex] such that: [latex]f(c) = \frac{1}{b-a} \int_a^b f(x) dx[/latex]
  • Geometric Interpretation:
    • [latex]f(c)[/latex] equals the average value of the function over [latex][a,b][/latex]
    • The area of the rectangle with base [latex]b-a[/latex] and height [latex]f(c)[/latex] equals the area under the curve
  • Alternative Form:
    • [latex]\int_a^b f(x) dx = f(c)(b-a)[/latex]
  • Significance:
    • Guarantees the existence of a point where the function takes on its average value
    • Bridges discrete and continuous concepts of average
  • The Mean Value Theorem for Integrals provides the theoretical basis for the average value of a function

Find the average value of the function [latex]f(x)=\dfrac{x}{2}[/latex] over the interval [latex]\left[0,6\right][/latex] and find [latex]c[/latex] such that [latex]f(c)[/latex] equals the average value of the function over [latex]\left[0,6\right].[/latex]

Given [latex]{\displaystyle\int }_{0}^{3}(2{x}^{2}-1)dx=15,[/latex] find [latex]c[/latex] such that [latex]f(c)[/latex] equals the average value of [latex]f(x)=2{x}^{2}-1[/latex] over [latex]\left[0,3\right].[/latex]

Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives

The Main Idea 

  • Theorem Statement:
    • If [latex]f(x)[/latex] is continuous on [latex][a,b][/latex], and [latex]F(x) = \int_a^x f(t) dt[/latex], then [latex]F'(x) = f(x)[/latex] for all [latex]x \in [a,b][/latex]
  • Key Implications:
    • Integration and differentiation are inverse processes
    • Every continuous function has an antiderivative
    • Provides a way to evaluate definite integrals without using Riemann sums
  • Interpretation of [latex]F(x)[/latex]:
    • [latex]F(x)[/latex] is a function that gives the area under the curve of [latex]f(t)[/latex] from [latex]a[/latex] to [latex]x[/latex]
    • The rate of change of this area function is the original function [latex]f(x)[/latex]
  • Connection to Anti-differentiation:
    • [latex]F(x)[/latex] is an antiderivative of [latex]f(x)[/latex]
    • All antiderivatives of [latex]f(x)[/latex] differ by a constant
  • Applications:
    • Simplifies the process of finding antiderivatives
    • Allows for the evaluation of complex definite integrals
    • Forms the basis for many advanced calculus techniques

Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of [latex]g(r)={\displaystyle\int }_{0}^{r}\sqrt{{x}^{2}+4}dx.[/latex]

Let [latex]F(x)={\displaystyle\int }_{1}^{{x}^{3}} \cos tdt.[/latex] Find [latex]{F}^{\prime }(x).[/latex]

Let [latex]F(x)={\displaystyle\int }_{x}^{{x}^{2}} \cos tdt.[/latex] Find [latex]{F}^{\prime }(x).[/latex]

Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem

The Main Idea 

  • Theorem Statement:
    • If [latex]f[/latex] is continuous on [latex][a,b][/latex] and [latex]F(x)[/latex] is any antiderivative of [latex]f(x)[/latex], then: [latex]\int_a^b f(x)dx = F(b) - F(a)[/latex]
  • Key Implications:
    • Provides a simple method to evaluate definite integrals
    • Connects the concept of area under a curve to the difference of antiderivative values
    • Eliminates the need for limits of Riemann sums in many cases
  • Notation:
    • [latex]F(b) - F(a)[/latex] is often written as [latex]F(x)|_a^b[/latex]
  • Important Points:
    • Any antiderivative can be used (constant of integration cancels out)
    • The theorem applies to continuous functions on a closed interval
    • The result can be positive, negative, or zero, depending on the function and interval
  • Applications:
    • Simplifies calculations in physics, engineering, and economics
    • Allows for quick evaluation of accumulated change over an interval

Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:

[latex]{\displaystyle\int }_{1}^{9}\dfrac{x-1}{\sqrt{x}}dx.[/latex]

Suppose James and Kathy have a rematch, but this time the official stops the contest after only [latex]3[/latex] sec. Does this change the outcome?