Prove Fundamental Theorem Of Calculus

Proving the Fundamental Theorem of Calculus: A Comprehensive Guide

The Fundamental Theorem of Calculus is a cornerstone of integral calculus, elegantly bridging the seemingly disparate concepts of differentiation and integration. It essentially states that differentiation and integration are inverse operations. This powerful theorem simplifies the process of evaluating definite integrals and reveals a deep connection between the area under a curve and the function itself. This article provides a detailed exploration of the theorem's two parts and their rigorous proofs, aiming to provide a comprehensive understanding for students and enthusiasts alike.

Understanding the Two Parts of the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is actually comprised of two distinct but interconnected parts:

Part 1 (The First Fundamental Theorem of Calculus): This part connects differentiation and integration. It states that if F(x) is an antiderivative of f(x) (meaning F'(x) = f(x)), then the definite integral of f(x) from a to x is given by F(x) - F(a).

Part 2 (The Second Fundamental Theorem of Calculus): This part provides a method for evaluating definite integrals. It states that if f(x) is continuous on the interval [a, b], then the definite integral of f(x) from a to b is equal to F(b) - F(a), where F(x) is any antiderivative of f(x).

Let's delve into the proofs of each part, explaining the underlying concepts and mathematical reasoning.

Proof of Part 1: The First Fundamental Theorem of Calculus

To prove the first part, we start with the definition of the definite integral as a limit of Riemann sums:

∫_a^x f(t) dt = lim_n→∞ Σ_i=1ⁿ f(t_i*) Δt

where:

• a and x are the limits of integration.
• f(t) is the integrand (the function being integrated).
• t_i** is a sample point in the i-th subinterval.
• Δt = (x - a)/n is the width of each subinterval.

Now, let's consider the function G(x) defined as:

G(x) = ∫_a^x f(t) dt

We want to find the derivative of G(x). Using the limit definition of the derivative:

G'(x) = lim_h→0 [G(x + h) - G(x)] / h

Substituting the definition of G(x):

G'(x) = lim_h→0 [∫_a^x+h f(t) dt - ∫_a^x f(t) dt] / h

Using properties of definite integrals, we can simplify this to:

G'(x) = lim_h→0 [∫_x^x+h f(t) dt] / h

Now, we can use the Mean Value Theorem for Integrals, which states that there exists a value c in the interval [x, x + h] such that:

∫_x^x+h f(t) dt = f(c) * h

Substituting this back into the expression for G'(x):

G'(x) = lim_h→0 [f(c) * h] / h = lim_h→0 f(c)

Since c is between x and x + h, as h approaches 0, c approaches x. Assuming f(x) is continuous, we have:

G'(x) = f(x)

This proves that the derivative of the definite integral of f(x) from a to x is f(x). If F(x) is an antiderivative of f(x), then F'(x) = f(x). Since both G(x) and F(x) have the same derivative, they differ by a constant:

G(x) = F(x) + C

To find the constant C, we evaluate G(a):

G(a) = ∫_a^a f(t) dt = 0

Therefore, 0 = F(a) + C, implying C = -F(a). Thus:

G(x) = F(x) - F(a)

This completes the proof of the First Fundamental Theorem of Calculus.

Proof of Part 2: The Second Fundamental Theorem of Calculus

The second part builds upon the first. We assume f(x) is continuous on [a, b], and F(x) is an antiderivative of f(x). Let's define a function G(x) as before:

G(x) = ∫_a^x f(t) dt

From Part 1, we know that G'(x) = f(x). Since F(x) is also an antiderivative of f(x), we have:

G'(x) = F'(x)

This means that G(x) and F(x) differ by a constant:

G(x) = F(x) + C

To find C, we evaluate G(a):

G(a) = ∫_a^a f(t) dt = 0

Therefore, 0 = F(a) + C, and C = -F(a). So:

G(x) = F(x) - F(a)

Now, we can evaluate the definite integral of f(x) from a to b:

∫_a^b f(t) dt = G(b) = F(b) - F(a)

This completes the proof of the Second Fundamental Theorem of Calculus. It provides a direct and efficient method for evaluating definite integrals by finding an antiderivative and evaluating it at the limits of integration.

Illustrative Examples

Let's illustrate the theorems with examples.

Example 1 (Part 1):

Let f(x) = 2x. An antiderivative is F(x) = x². Then, according to Part 1:

d/dx [∫₀^x 2t dt] = 2x

This shows that the derivative of the integral is indeed the original function.

Example 2 (Part 2):

Evaluate ∫₁³ x² dx.

An antiderivative of x² is F(x) = (1/3)x³. Therefore, by Part 2:

∫₁³ x² dx = F(3) - F(1) = (1/3)(3)³ - (1/3)(1)³ = 9 - (1/3) = 26/3

Addressing Common Questions (FAQ)

• What if f(x) is not continuous? The Fundamental Theorem of Calculus, as stated, requires f(x) to be continuous on the interval [a, b]. If f(x) has discontinuities, the theorem may not hold, or the integral may need to be treated as an improper integral.

• How do I find an antiderivative? Finding antiderivatives can be challenging, and there's no single, universally applicable method. Techniques include using basic integration rules (power rule, trigonometric integrals, exponential integrals, etc.), substitution, integration by parts, and partial fraction decomposition. Some integrals may not have closed-form solutions and require numerical methods.

• What is the significance of the Fundamental Theorem of Calculus? The theorem is fundamental because it establishes a deep connection between differentiation and integration, two seemingly different operations. It simplifies the computation of definite integrals, which have broad applications in various fields such as physics, engineering, economics, and probability.

• Can the Fundamental Theorem be applied to multivariable calculus? Yes, extensions of the Fundamental Theorem of Calculus exist in multivariable calculus, such as the Fundamental Theorem for Line Integrals and Stokes' Theorem. These theorems generalize the core concepts to higher dimensions.

Conclusion

The Fundamental Theorem of Calculus is a truly remarkable result in mathematics. Its two parts elegantly link differentiation and integration, providing a powerful tool for evaluating definite integrals and deepening our understanding of the relationship between a function and its area under the curve. The proofs, while requiring some familiarity with limit definitions and the Mean Value Theorem, highlight the logical elegance and foundational importance of this theorem within the broader landscape of calculus. Understanding this theorem is crucial for anyone seeking a comprehensive grasp of calculus and its applications in diverse scientific and engineering disciplines. Mastering the proof not only enhances your understanding of the theorem itself, but also strengthens your overall problem-solving skills within the realm of mathematical analysis.