📈 The Fundamental Theorem of Calculus

• Core overview of the theorem and its dual nature
• Intuitive understanding through geometric and physical interpretations
• Formal statements of the First and Second parts, including the Corollary
• Rigorous mathematical proofs for each part
• Relationship and distinctions between the two parts
• Practical and theoretical examples demonstrating application

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💡 Core Concepts and Overview

The fundamental theorem of calculus is a cornerstone mathematical theorem that links two seemingly distinct operations:

• Differentiating a function: Calculating its slopes or rate of change at every point on its domain.
• Integrating a function: Calculating the area under its graph or the cumulative effect of small contributions.

Roughly speaking, the two operations can be thought of as inverses of each other.

The theorem is presented in two primary parts:

1. First Part: States that for a continuous function $f$, an antiderivative or indefinite integral $F$ can be obtained as the integral of $f$ over an interval with a variable upper bound.
2. Second Part: States that the integral of a function $f$ over a fixed interval equals the net change of any antiderivative $F$ between the endpoints of the interval. This vastly simplifies definite integral calculations by avoiding numerical integration, provided an antiderivative can be found via symbolic integration.

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🔍 Intuitive Understanding

Geometric Interpretation (The First Part)

• Given a continuous function $y = f(x)$ plotted as a curve, we define an area function $x \mapsto A(x)$ representing the area beneath the curve between $0$ and $x$.
• The area of a small "strip" between $x$ and $x + h$ can be estimated in two ways:
  • Subtracting areas: $A(x + h) - A(x)$
  • Multiplying width by height (rectangle approximation): $f(x) \cdot h$
• Setting these approximations equal yields: $A(x + h) - A(x) \approx f(x) \cdot h$
• Dividing by $h$ and taking the limit as $h \to 0$ gives: $f(x) = \lim_{h \to 0} \frac{A(x + h) - A(x)}{h} \mathrel{\stackrel{\text{def}}{=}} A'(x)$
• Thus, the derivative of the area function equals the original function, confirming that differentiation and integration are inverse operations.

Physical Interpretation (The Second Part)

• Imagine traveling in a car where you can observe your velocity on the speedometer but cannot look outside to track your absolute position.
• Each time interval $\Delta t$, the distance traveled is approximately $\text{velocity} \times \text{time interval} = v_t \times \Delta t$.
• Summing these small steps approximates the total distance traveled: $\text{distance traveled} = \sum v_t \times \Delta t$
• As $\Delta t$ becomes infinitesimally small, this sum becomes an integral. Therefore, the integral of the velocity function (the derivative of position) computes the net change in position.

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📜 Formal Statements

First Part (First Fundamental Theorem)

Let $f$ be a continuous real-valued function defined on a closed interval $[a, b]$. Let $F$ be the function defined for all $x$ in $[a, b]$ by: $F(x) = \int_{a}^{x} f(t) \, dt$ Then:

• $F$ is uniformly continuous on $[a, b]$.
• $F$ is differentiable on the open interval $(a, b)$.
• $F'(x) = f(x)$ for all $x$ in $(a, b)$, making $F$ an antiderivative of $f$.

Corollary (Computation of Definite Integrals)

If $f$ is a real-valued continuous function on $[a, b]$ and $F$ is an antiderivative of $f$ on $[a, b]$, then: $\int_{a}^{b} f(t) \, dt = F(b) - F(a)$

Second Part (Second Fundamental Theorem / Newton–Leibniz Theorem)

Let $f$ be a real-valued function on a closed interval $[a, b]$ and $F$ a continuous function on $[a, b]$ which is an antiderivative of $f$ in $(a, b)$ such that: $F'(x) = f(x)$ If $f$ is Riemann integrable on $[a, b]$, then: $\int_{a}^{b} f(x) \, dx = F(b) - F(a)$

Key Distinction: The second part is stronger than the corollary because it does not strictly require $f$ to be continuous; it only requires $f$ to be Riemann integrable and possess an antiderivative.

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📐 Mathematical Proofs

Proof of the First Part

1. Define $F(x) = \int_{a}^{x} f(t) \, dt$.
2. For two numbers $x_1$ and $x_1 + \Delta x$ in $[a, b]$: $F(x_1 + \Delta x) - F(x_1) = \int_{x_1}^{x_1 + \Delta x} f(t) \, dt$
3. By the mean value theorem for integration, there exists a real number $c \in [x_1, x_1 + \Delta x]$ such that: $\int_{x_1}^{x_1 + \Delta x} f(t) \, dt = f(c) \cdot \Delta x$
4. Dividing by $\Delta x$: $\frac{F(x_1 + \Delta x) - F(x_1)}{\Delta x} = f(c)$
5. Taking the limit as $\Delta x \to 0$ (noting that $c \to x_1$ and applying the squeeze theorem and continuity of $f$): $F'(x_1) = \lim_{\Delta x \to 0} \frac{F(x_1 + \Delta x) - F(x_1)}{\Delta x} = \lim_{\Delta x \to 0} f(c) = f(x_1)$

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Proof of the Corollary

1. Let $F$ be an antiderivative of a continuous function $f$ on $[a, b]$, and define: $G(x) = \int_{a}^{x} f(t) \, dt$
2. By the first part, $G$ is also an antiderivative of $f$.
3. Since $F' - G' = 0$, the mean value theorem implies $F - G$ is a constant function ($G(x) = F(x) + c$).
4. Evaluating at $x = a$: $F(a) + c = G(a) = \int_{a}^{a} f(t) \, dt = 0 \implies c = -F(a)$
5. Therefore, $G(x) = F(x) - F(a)$, which yields: $\int_{a}^{b} f(x) \, dx = G(b) = F(b) - F(a)$

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Proof of the Second Part (Riemann Sum Limit Proof)

1. Recall the mean value theorem: If $F$ is continuous on $[a, b]$ and differentiable on $(a, b)$, there exists $c \in (a, b)$ such that $F'(c)(b - a) = F(b) - F(a)$.
2. Partition the interval $[a, b]$ such that $a = x_0 < x_1 < \cdots < x_n = b$.
3. Express $F(b) - F(a)$ using telescoping sums: $F(b) - F(a) = \sum_{i=1}^{n} [F(x_i) - F(x_{i-1})]$
4. Apply the mean value theorem to each subinterval $[x_{i-1}, x_i]$ using some $c_i \in (x_{i-1}, x_i)$: $F(x_i) - F(x_{i-1}) = F'(c_i)(x_i - x_{i-1}) = f(c_i) \Delta x_i$
5. Summing these terms yields: $F(b) - F(a) = \sum_{i=1}^{n} f(c_i) \Delta x_i$
6. Taking the limit as the norm of the partitions approaches zero ($\|\Delta x_i\| \to 0$) produces the Riemann integral: $F(b) - F(a) = \lim_{\|\Delta x_i\| \to 0} \sum_{i=1}^{n} f(c_i) \Delta x_i = \int_{a}^{b} f(x) \, dx$

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🔄 Relationship Between the Parts

Theorem Part	Primary Function	Key Requirement
First Part	Proves existence of antiderivatives and links differentiation to integration	$f$ must be continuous
Second Part	Evaluates definite integrals via antiderivatives	$f$ must be Riemann integrable and possess an antiderivative

• Dependency Note: While a weaker version of the second part follows from the first, the reverse path requires knowing that continuous functions always have antiderivatives—a fact established precisely by the first part.
• Elementary Antiderivatives: Not all integrable functions have elementary antiderivatives (e.g., $f(x) = e^{-x^2}$), and not all functions with antiderivatives are Riemann integrable. Therefore, the second part should not be viewed merely as the definition of the integral.

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📝 Practical and Theoretical Examples

1. Computing a Particular Definite Integral

Calculate $\int_{2}^{5} x^2 \, dx$:

• Let $f(x) = x^2$. An antiderivative is $F(x) = \frac{1}{3}x^3$.
• Applying the corollary: $\int_{2}^{5} x^2 \, dx = F(5) - F(2) = \frac{5^3}{3} - \frac{2^3}{3} = \frac{125}{3} - \frac{8}{3} = \frac{117}{3} = 39$

2. Using the First Part

Calculate $\frac{d}{dx} \int_{0}^{x} t^3 \, dt$:

• Using the first part directly with $f(t) = t^3$: $\frac{d}{dx} \int_{0}^{x} t^3 \, dt = f(x) = x^3$

3. An Integral Where the Simple Corollary Fails

Consider the discontinuous function: $f(x) = \begin{cases} \sin\left(\frac{1}{x}\right) - \frac{1}{x}\cos\left(\frac{1}{x}\right) & x \neq 0 \\ 0 & x = 0 \end{cases}$

• Because $\lim_{x \to 0} f(x)$ does not exist, the standard corollary cannot be used directly.
• However, the function $F(x) = \begin{cases} x\sin\left(\frac{1}{x}\right) & x \neq 0 \\ 0 & x = 0 \end{cases}$ is continuous on $[0,1]$ and satisfies $F'(x) = f(x)$ on $(0,1)$.
• Applying the second part: $\int_{0}^{1} f(x) \, dx = F(1) - F(0) = \sin(1) - 0 = \sin(1)$