5.3

5.3 Fundamental Theorem of Calculus

Overview

The Fundamental Theorem of Calculus (FTC) connects differentiation and integration, showing the relationship between the antiderivative and the definite integral.

Example 1 (Page 368)

  • Graph Information: The graph of function f is provided with areas of various regions marked.
  • Area Functions Defined:
    • Let A(x) = \int_{-1}^{x} f(t) \, dt
    • Let B(x) = \int_{3}^{x} f(t) \, dt
  • Evaluation Requests:
    • (a) Evaluate A(3) and B(3)
    • (b) Evaluate A(5) and B(5)
    • (c) Evaluate A(9) and B(9)

Integration

Fundamental Theorem of Calculus Part 1
  • Statement: If f is continuous on the interval [a, b], the area function defined as: A(x) = \int_{a}^{x} f(t) \, dt for a \le x \le b, is:
    • Continuous: A(x) is continuous on [a, b].
    • Differentiable: A(x) is differentiable on (a, b).
    • The derivative of the area function satisfies:
      A'(x) = f(x)
  • Equivalence:
    \frac{d}{dx} A(x) = \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x)
  • This implies that the area function of f serves as an antiderivative of f across [a, b].
Example Explanation: The Chain Rule Case
  • Specific Scenario:
    • If g(x) is a differentiable function, the area function can be expressed as:
      A(x) = \int_{a}^{g(x)} f(t) \, dt
    • The derivative becomes:
      A'(x) = f(g(x)) \cdot g'(x)
  • Example: If A(x) = \int_{0}^{2x} t \, dt, then:
    • Result: A'(x) = (2x) \cdot 2 = 4x
Fundamental Theorem of Calculus Part 2
  • Statement: If f is continuous on [a, b] and F is an antiderivative of f on [a, b], then:
    \int_{a}^{b} f(x) \, dx = F(b) - F(a)
  • Components:
    • Limits of integration: a and b define the interval.
    • Antiderivative: F is evaluated at the boundaries a and b.
    • The expression captures the net area under the curve between points a and b.

Example 3 (Page 372)

  • Objective: Evaluate definite integrals using the Fundamental Theorem of Calculus, Part 2.
  • Geometric Interpretation: Each result should be interpreted in a geometric context (e.g., area under the curve).

Example 4: Net Areas and Definite Integrals

  • Function: Given f(x) = 6x(x + 1)(x - 2)
  • Bounded Regions Defined:
    • Region R_1 is on [-1, 0]
    • Region R_2 is on [0, 2]
  • Tasks:
    • (a) Find the net area of the region between the curve and the x-axis on the interval [-1, 2].
    • (b) Find the total area of the region between the curve and the x-axis on the interval [-1, 2].

Example 5: Derivatives of Integrals

  • Using Part 1: Simplify the following expressions:
    • \frac{d}{dx} \left[ \int_{a}^{x} f(t) \, dt \right] = f(x)
    • Evaluate for cases where limits are functions of x, e.g., using \sin(x) or \cos(x) as limits.

Example 6: Working with Area Functions

  • Function Visualization:
    • The graph of a function f is provided, with its area function A(x) = \int_{0}^{x} f(t) \, dt.
  • Properties of Regions: Assume regions R1, R2, R3, and R4 have equal areas.
  • Tasks:
    • (a) Identify zeros of A on the interval [0, 17].
    • (b) Determine points at which A attains local maxima or minima.
    • (c) Sketch the function A for the interval from 0 to 17.

Example 7: The Sine Integral Function

  • Function Definition:
    Si(x) = \int_{0}^{x} \frac{\sin(t)}{t} \, dt
  • Note: The integrand behaves like 1 at t=0.

Section 5.3 Exercises (Pages 377-381)

  • Focus on the application of FTC Part 1 (differentiation of integrals) and Part 2 (evaluation of definite integrals).