Fundamental theorem of calculus - Formal Theory and Applications

The Fundamental Theorem of Calculus: Formal Statements and Proofs Introduction The Fundamental Theorem of Calculus (FTC) is one of the most important results in mathematics. It establishes the deep connection between two seemingly different concepts: differentiation (finding rates of change) and integration (finding areas under curves). Rather than being two separate operations, the theorem shows that integration and differentiation are essentially inverse processes of each other. This is powerful because it gives us a practical way to compute definite integrals using antiderivatives—something we'll explore thoroughly. The Two Parts of the Fundamental Theorem The Fundamental Theorem of Calculus has two main parts. Let's state them clearly. Part 1: Integration Gives an Antiderivative Formal Statement: Let $f$ be a continuous function on $[a,b]$. Define $$A(x) = \inta^x f(t)\,dt$$ for $x \in [a,b]$. Then $A$ is differentiable on $(a,b)$, and $$A'(x) = f(x)$$ In words: if you take a function $f$ and integrate it from a fixed point $a$ to a variable point $x$, the derivative of that accumulated area with respect to $x$ gives you back the original function $f(x)$. Why this matters: This tells us that integration and differentiation are inverse operations. Taking the integral of $f$ produces a function whose derivative is $f$. Part 2: Evaluating Definite Integrals Using Antiderivatives Formal Statement: Let $f$ be a continuous function on $[a,b]$, and let $F$ be any antiderivative of $f$ (meaning $F'(x) = f(x)$). Then $$\inta^b f(x)\,dx = F(b) - F(a)$$ In words: to evaluate a definite integral, find an antiderivative $F$, then subtract its value at the left endpoint from its value at the right endpoint. Why this matters: This is the practical tool you use every time you evaluate a definite integral. Instead of computing the limit of Riemann sums (which is tedious), you just find an antiderivative and evaluate it at two points. Key observation: Part 2 explains why we can evaluate $\inta^b f(x)\,dx$ without directly constructing Riemann sums. This is the computational power of the FTC. Proof of Part 1: Why the Integral's Derivative is the Original Function Let's understand why $\frac{d}{dx}\left[\inta^x f(t)\,dt\right] = f(x)$. The key insight is to use the Mean Value Theorem for integrals. Here's the argument: Consider two points $x$ and $x + \Delta x$ in $[a,b]$. We want to find how $A(x)$ changes: $$A(x+\Delta x) - A(x) = \inta^{x+\Delta x} f(t)\,dt - \inta^x f(t)\,dt = \intx^{x+\Delta x} f(t)\,dt$$ Now apply the Mean Value Theorem for integrals: this integral equals the value of the function at some point $c$ (between $x$ and $x+\Delta x$) times the width: $$\intx^{x+\Delta x} f(t)\,dt = f(c) \cdot \Delta x$$ where $c$ is some point between $x$ and $x+\Delta x$. Therefore: $$\frac{A(x+\Delta x) - A(x)}{\Delta x} = f(c)$$ As $\Delta x \to 0$, the point $c$ must approach $x$ (since $c$ is squeezed between $x$ and $x+\Delta x$). By continuity of $f$, we have $f(c) \to f(x)$. Thus: $$A'(x) = \lim{\Delta x \to 0} \frac{A(x+\Delta x) - A(x)}{\Delta x} = f(x)$$ The image above visualizes this: the "excess" area (shown in red) between $x$ and $x+h$ is approximately $f(x) \cdot h$ for small $h$, which is exactly the derivative condition. Proof of Part 2: Evaluating with Antiderivatives (Riemann Sum Approach) This is the most intuitive proof. The idea is to use the Mean Value Theorem to connect Riemann sums to the change $F(b) - F(a)$. Step 1: Partition the interval Divide $[a,b]$ into $n$ subintervals: $$a = x0 < x1 < x2 < \cdots < xn = b$$ with widths $\Delta xi = xi - x{i-1}$. Step 2: Apply the Mean Value Theorem to $F$ For each subinterval $[x{i-1}, xi]$, the Mean Value Theorem guarantees a point $ci$ such that: $$F(xi) - F(x{i-1}) = F'(ci) \cdot \Delta xi = f(ci) \cdot \Delta xi$$ (Remember: $F' = f$ by definition of antiderivative.) Step 3: Sum over all subintervals Add up all the changes: $$F(b) - F(a) = \sum{i=1}^n [F(xi) - F(x{i-1})] = \sum{i=1}^n f(ci) \cdot \Delta xi$$ This is a Riemann sum for $f$ over the partition! Step 4: Take the limit As we refine the partition (making $\Delta xi$ smaller), the Riemann sum converges to the integral: $$\inta^b f(x)\,dx = \lim{\text{mesh} \to 0} \sum{i=1}^n f(ci) \cdot \Delta xi = F(b) - F(a)$$ Why this works: The left side doesn't depend on which partition we choose—it's always $F(b) - F(a)$. The right side is a Riemann sum that approaches the integral. Therefore they must be equal! A Useful Corollary: All Antiderivatives Differ by a Constant Statement: If $F$ and $G$ are both antiderivatives of the same function $f$ on $[a,b]$, then $F(x) - G(x) = C$ for some constant $C$. Proof: By assumption, $F'(x) = f(x) = G'(x)$. This means: $$(F - G)'(x) = F'(x) - G'(x) = 0$$ A function with zero derivative everywhere must be constant. Therefore $F(x) - G(x) = C$ for all $x$. To find $C$, evaluate at $x = a$: $C = F(a) - G(a)$. Why this matters: This tells us that when we write the indefinite integral as $\int f(x)\,dx = F(x) + C$, the "$+ C$" accounts for all possible antiderivatives. For definite integrals, the constant cancels when we compute $F(b) - F(a)$, so we don't need to worry about which antiderivative we use. How to Use the Fundamental Theorem in Practice The FTC gives us a systematic approach to evaluate definite integrals: Find an antiderivative $F$ such that $F'(x) = f(x)$ Evaluate at the endpoints: compute $F(b) - F(a)$ That's your answer: $\inta^b f(x)\,dx = F(b) - F(a)$ Example: To compute $\int0^3 2x\,dx$: An antiderivative of $2x$ is $F(x) = x^2$ (since $\frac{d}{dx}[x^2] = 2x$) Evaluate: $F(3) - F(0) = 9 - 0 = 9$ Therefore $\int0^3 2x\,dx = 9$ This is vastly simpler than computing the limit of Riemann sums! <extrainfo> Beyond the Basics: Connections to Other Areas The Fundamental Theorem is foundational for several advanced techniques: Differentiation under the integral sign: This technique allows you to differentiate an integral with respect to a parameter by moving the derivative inside. Line integrals and conservative vector fields: In multivariable calculus, the FTC generalizes to say that line integrals of gradient fields (derivatives in multiple dimensions) can be evaluated using potential functions. These applications extend the core idea: antiderivatives help us compute integrals. </extrainfo>