Riemann Integral - Fundamental Theorem of Calculus

Introduction to Riemann Integration

The Riemann integral is a formal definition of the definite integral, which represents the area under a curve. It was developed by Bernhard Riemann. The core idea is to approximate the area by dividing the region into a series of thin rectangles and summing their areas. As the width of these rectangles approaches zero, the sum of their areas converges to the exact area under the curve.

Definition of Riemann Integral

Let f be a bounded real-valued function defined on a closed interval [a, b]. A partition P of [a, b] is a finite sequence of points x₀, x₁, ..., x_n such that a = x₀ < x₁ < ... < x_n = b.

For each subinterval [x_i-1, x_i], let Δx_i = x_i - x_i-1 be its length. Let M_i = sup {f(x) : x ∈ [x_i-1, x_i]} and m_i = inf {f(x) : x ∈ [x_i-1, x_i]}.

The upper sum (or Riemann-Darboux upper sum) for the partition P is defined as: U(P, f) = Σ_i=1ⁿ M_i Δx_i

The lower sum (or Riemann-Darboux lower sum) for the partition P is defined as: L(P, f) = Σ_i=1ⁿ m_i Δx_i

The upper integral of f is defined as: ∫̄_a^b f(x) dx = inf {U(P, f) : P is a partition of [a, b]}

The lower integral of f is defined as: ∫_a^b f(x) dx = sup {L(P, f) : P is a partition of [a, b]}

A function f is said to be Riemann integrable on [a, b] if its lower integral equals its upper integral. In this case, the common value is called the Riemann integral (or definite integral) of f on [a, b], denoted by ∫_a^b f(x) dx.

In simpler terms, if we can make the difference between the sum of areas of rectangles from above and the sum of areas of rectangles from below arbitrarily small, then the function is Riemann integrable.

Conditions for Riemann Integrability

A bounded function f on [a, b] is Riemann integrable if and only if the set of its discontinuities has measure zero.

For example, a continuous function on [a, b] is always Riemann integrable. A function with a finite number of discontinuities on [a, b] is also Riemann integrable.

Properties of Riemann Integral

Let f and g be Riemann integrable functions on [a, b], and let c be a real number.

• Linearity: ∫_a^b (cf)(x) dx = c ∫_a^b f(x) dx and ∫_a^b (f + g)(x) dx = ∫_a^b f(x) dx + ∫_a^b g(x) dx.
• Additivity over interval: If c ∈ (a, b), then ∫_a^b f(x) dx = ∫_a^c f(x) dx + ∫_c^b f(x) dx.
• Monotonicity: If f(x) ≤ g(x) for all x ∈ [a, b], then ∫_a^b f(x) dx ≤ ∫_a^b g(x) dx.
• Boundedness: If m ≤ f(x) ≤ M for all x ∈ [a, b], then m(b - a) ≤ ∫_a^b f(x) dx ≤ M(b - a).
• Integral of zero function: ∫_a^b 0 dx = 0.
• Integral of constant function: ∫_a^b k dx = k(b - a).

The Fundamental Theorem of Calculus (FTC)

The Fundamental Theorem of Calculus is a cornerstone of calculus, establishing a profound link between differentiation and integration. It essentially states that differentiation and integration are inverse operations. There are two parts to the theorem.

Part 1: The First Fundamental Theorem of Calculus

This part states that if a function is continuous on a closed interval and we define a new function as the integral of the first function from a fixed point to a variable point, then the derivative of this new function is the original function itself.

Statement: Let f be a continuous real-valued function on a closed interval [a, b]. Define a function F on [a, b] by F(x) = ∫_a^x f(t) dt Then, F is continuous on [a, b] and differentiable on (a, b), and its derivative is F'(x) = f(x) for all x ∈ (a, b).

Explanation: Imagine f(t) represents the rate at which water is flowing into a tank at time t. F(x) represents the total amount of water in the tank at time x, assuming the tank was empty at time a. The first fundamental theorem says that the rate at which the amount of water is changing at time x (which is F'(x)) is exactly the rate at which water is flowing in at that moment (which is f(x)).

Proof Sketch: Consider F(x+h) - F(x) for a small h > 0. F(x+h) - F(x) = ∫_a^x+h f(t) dt - ∫_a^x f(t) dt Using the additivity property of integrals, this becomes: F(x+h) - F(x) = ∫_x^x+h f(t) dt Since f is continuous on [x, x+h], by the Mean Value Theorem for Integrals, there exists a c ∈ [x, x+h] such that: ∫_x^x+h f(t) dt = f(c) * ( (x+h) - x ) = f(c) * h So, (F(x+h) - F(x)) / h = f(c). As h approaches 0, c also approaches x (because c is between x and x+h). Since f is continuous, lim_h→0 f(c) = f(x). Therefore, F'(x) = lim_h→0 (F(x+h) - F(x)) / h = f(x). A similar argument holds for h < 0.

Part 2: The Second Fundamental Theorem of Calculus

This part provides a method for calculating definite integrals by finding an antiderivative of the integrand. It states that if F is any antiderivative of a continuous function f on [a, b], then the definite integral of f from a to b is simply the difference in the values of F at the endpoints.

Statement: Let f be a continuous real-valued function on a closed interval [a, b]. If F is any antiderivative of f on [a, b] (i.e., F'(x) = f(x) for all x ∈ (a, b)), then ∫_a^b f(x) dx = F(b) - F(a)

Explanation: This theorem is incredibly practical. Instead of using the cumbersome Riemann sum definition to find the area under a curve, we can find a function whose derivative is the function we want to integrate (an antiderivative) and then just evaluate this antiderivative at the limits of integration.

Proof: Let G(x) = ∫_a^x f(t) dt. By the First Fundamental Theorem of Calculus, G'(x) = f(x) for all x ∈ (a, b). We are given that F is also an antiderivative of f, so F'(x) = f(x). This means G'(x) = F'(x) for all x ∈ (a, b). Therefore, G(x) and F(x) must differ by a constant, say C. G(x) = F(x) + C Now, let's evaluate G(a): G(a) = ∫_a^a f(t) dt = 0. Substituting this into the equation: 0 = F(a) + C So, C = -F(a). Thus, G(x) = F(x) - F(a). Now, consider the definite integral ∫_a^b f(x) dx. By definition of G(x), this is G(b). G(b) = F(b) - F(a). Therefore, ∫_a^b f(x) dx = F(b) - F(a).

Notation: The difference F(b) - F(a) is often written as [F(x)]_a^b or F(x) |_a^b.

Example Applications of FTC

Example 1: Using FTC Part 1

Let f(x) = x². Find the derivative of the function G(x) = ∫₀^x t² dt.

Solution: Since f(t) = t² is continuous everywhere, we can apply the First Fundamental Theorem of Calculus. G'(x) = f(x) = x². So, the derivative of G(x) is simply x².

Example 2: Using FTC Part 2

Evaluate the definite integral ∫₁³ x² dx.

Solution: The integrand is f(x) = x², which is continuous on [1, 3]. First, find an antiderivative of f(x) = x². Using the power rule for integration (∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C), an antiderivative is F(x) = x³/3. Now, apply the Second Fundamental Theorem of Calculus: ∫₁³ x² dx = F(3) - F(1) = (3³/3) - (1³/3) = (27/3) - (1/3) = 9 - 1/3 = 27/3 - 1/3 = 26/3.

Example 3: Using FTC Part 2 with Trigonometric Functions

Evaluate the definite integral ∫₀^π/2 cos(x) dx.

Solution: The integrand is f(x) = cos(x), which is continuous on [0, π/2]. An antiderivative of cos(x) is F(x) = sin(x). Using the Second Fundamental Theorem of Calculus: ∫₀^π/2 cos(x) dx = F(π/2) - F(0) = sin(π/2) - sin(0) = 1 - 0 = 1.

Example 4: More Complex Antiderivative

Evaluate the definite integral ∫₀¹ (3x² + 2x + 1) dx.

Solution: The integrand is f(x) = 3x² + 2x + 1, which is continuous on [0, 1]. Find an antiderivative: F(x) = ∫(3x² + 2x + 1) dx = 3 * (x³/3) + 2 * (x²/2) + 1 * x = x³ + x² + x. (We can ignore the constant of integration C when evaluating definite integrals because it cancels out: (F(b)+C) - (F(a)+C) = F(b) - F(a)). Now, apply FTC Part 2: ∫₀¹ (3x² + 2x + 1) dx = F(1) - F(0) = (1³ + 1² + 1) - (0³ + 0² + 0) = (1 + 1 + 1) - (0) = 3.

The Net Change Theorem

The Net Change Theorem is a direct consequence of the Second Fundamental Theorem of Calculus. It states that the net change in a quantity is the integral of its rate of change.

Statement: If F'(x) is continuous on [a, b], then the net change in F(x) from a to b is given by: F(b) - F(a) = ∫_a^b F'(x) dx

Explanation: Suppose v(t) is the velocity of an object at time t. Then the position function s(t) has s'(t) = v(t). The net change in position (displacement) from time t=a to t=b is s(b) - s(a). According to the Net Change Theorem, this displacement is equal to the integral of the velocity function over that time interval: ∫_a^b v(t) dt.

Example: Net Change Theorem

A particle moves along a line such that its velocity at time t is given by v(t) = 3t² - 6t m/s. Find the net displacement of the particle from t=1 second to t=4 seconds.

Solution: The velocity v(t) is the derivative of the position function s(t). So, s'(t) = v(t) = 3t² - 6t. The net displacement is s(4) - s(1). Using the Net Change Theorem (which is FTC Part 2): Net Displacement = ∫₁⁴ v(t) dt = ∫₁⁴ (3t² - 6t) dt First, find an antiderivative of v(t): s(t) = ∫(3t² - 6t) dt = 3(t³/3) - 6(t²/2) = t³ - 3t². Now, evaluate: s(4) - s(1) = [t³ - 3t²]₁⁴ = (4³ - 3 * 4²) - (1³ - 3 * 1²) = (64 - 3 * 16) - (1 - 3) = (64 - 48) - (-2) = 16 + 2 = 18 meters. The net displacement is 18 meters.

Connection to Area Calculation

The Fundamental Theorem of Calculus is precisely what allows us to calculate the area under a curve y = f(x) from x=a to x=b. If f(x) ≥ 0, this area is given by the definite integral ∫_a^b f(x) dx. By FTC Part 2, this integral is computed as F(b) - F(a), where F is any antiderivative of f.

For instance, when we calculated ∫₁³ x² dx = 26/3, we found the exact area under the parabola y = x² between x=1 and x=3.

Important Considerations

The continuity of the integrand f is crucial for both parts of the Fundamental Theorem of Calculus. If f has discontinuities, the theorem might not apply directly, and one might need to use more advanced integration techniques or consider Lebesgue integration.

FTC Part 1 shows that every continuous function has an antiderivative (the integral function itself).

FTC Part 2 provides the practical tool for evaluating definite integrals, transforming a geometric problem (area) into an algebraic one (finding and evaluating an antiderivative).