The Timpson Lyceum · Real Analysis (2020)

Lecture 21 (Real Analysis 2020)

Overview

This lecture proves the (First) Fundamental Theorem of Calculus: integrating a derivative over $[a,b]$ recovers the net change of the original function, $\int_a^b F'(x)\,dx = F(b) - F(a)$. Informally, differentiation and integration undo each other — an integral of a function can be expressed in terms of an antiderivative of that function. The proof rests on two pieces from prior work: a short lemma characterizing the value of an integral by a "sandwiching" condition, and the Mean Value Theorem applied to each subinterval of a partition. Together they show the antiderivative's net change is trapped between every lower and upper Darboux sum of the derivative, which forces it to equal the integral.

Key Concepts

• Antiderivative. A function $F$ is an antiderivative of $f$ if $F'(x) = f(x)$. Because the derivative of a constant is $0$, antiderivatives are unique only up to an additive constant: if $F$ is one antiderivative, so is $F + C$ for any constant $C$. The capital letter $F$ conventionally denotes the antiderivative of the lowercase $f$.

• Reverse power rule. Differentiation sends $x^n \mapsto n x^{n-1}$ (multiply by the exponent, lower it by one). Antidifferentiation undoes this — raise the exponent by one and divide: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1).$

• Indefinite vs. definite integral. An indefinite integral $\int f(x)\,dx$ has no bounds and returns a family of antiderivatives $F(x) + C$. Attaching bounds $a$ and $b$ pins down the constant: in $F(b)-F(a)$ the $+C$ cancels, so any single antiderivative works.

• Lemma (integral-value sandwich). Suppose $f:[a,b]\to\mathbb{R}$ is integrable and a number $A$ satisfies, for every partition $P$, $L(f,P) \le A \le U(f,P).$ Then $\int_a^b f = A$. Why: the lower integral (supremum of lower sums = least upper bound) is $\le A$, and the upper integral (infimum of upper sums = greatest lower bound) is $\ge A$. Since $f$ is integrable, lower and upper integrals coincide, so the common value is squeezed to equal $A$.

• Fundamental Theorem of Calculus (Theorem 6.22). Let $F:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$, with $F'$ bounded and continuous. Then $\int_a^b F'(x)\,dx = F(b) - F(a).$ $F'$ is integrable by the prior result (a continuous, bounded function on a closed bounded interval is integrable, and the integral does not depend on the endpoints).

• Proof strategy. By the lemma, it suffices to show $F(b)-F(a)$ lies between $L(F',P)$ and $U(F',P)$ for every partition. Fix a partition $a = x_0 < x_1 < \cdots < x_n = b$. On each subinterval $[x_{i-1}, x_i]$ apply the Mean Value Theorem to $F$: there exists $c_i \in (x_{i-1}, x_i)$ with $F'(c_i) = \frac{F(x_i) - F(x_{i-1})}{x_i - x_{i-1}}, \quad\text{i.e.}\quad F(x_i)-F(x_{i-1}) = F'(c_i)\,(x_i - x_{i-1}).$

• Bounding each piece. Let $m_i = \inf F'$ and $M_i = \sup F'$ on $[x_{i-1}, x_i]$. Since $c_i$ lies in the subinterval, $m_i \le F'(c_i) \le M_i$, so $m_i\,(x_i - x_{i-1}) \le F(x_i)-F(x_{i-1}) \le M_i\,(x_i - x_{i-1}).$ Summing over $i$, the middle column telescopes: $\sum_i \big(F(x_i)-F(x_{i-1})\big) = F(b)-F(a)$, while the outer columns are exactly $L(F',P)$ and $U(F',P)$. Hence $L(F',P) \le F(b)-F(a) \le U(F',P)$ for every $P$, and the lemma finishes the proof.

• Intuition (position and velocity). If $F(t)$ is position, then $F'(t)$ is velocity. The change in position from $t=a$ to $t=b$ equals $F(b)-F(a)$, and it also equals the area under the velocity curve, $\int_a^b F'(t)\,dt$. The two descriptions agree — that agreement is the theorem.

Worked Example

Compute the displacement of an object moving at constant velocity $2$ from $t = 1$ to $t = 3$, where position is $F(t) = 2t$ and velocity is $F'(t) = 2$. Apply the FTC, $\int_a^b F'(t)\,dt = F(b)-F(a)$:

The antiderivative of $2$ is $2t$ (its derivative is $2$); evaluating it at the bounds and subtracting gives the net change directly. Either reading — net change of position $F(3)-F(1)$, or area under the velocity graph — yields the same displacement of $4$.

Summary

• The First Fundamental Theorem of Calculus states $\int_a^b F'(x)\,dx = F(b) - F(a)$: integrating a derivative recovers the original function's net change.
• An antiderivative $F$ satisfies $F' = f$ and is unique only up to a constant $C$; bounds on a definite integral cancel that constant, so any antiderivative suffices.
• A supporting lemma says that if a number is sandwiched between every lower and upper Darboux sum of an integrable function, it must equal the integral.
• The proof partitions $[a,b]$, uses the Mean Value Theorem on each subinterval to write $F(x_i)-F(x_{i-1}) = F'(c_i)(x_i - x_{i-1})$, and bounds this between $m_i$ and $M_i$ times the width.
• Summing telescopes the middle to $F(b)-F(a)$ while the bounds become $L(F',P)$ and $U(F',P)$, trapping $F(b)-F(a)$ between them for every partition.
• With derivative and integral established as inverse operations, you can now evaluate an integral by finding an antiderivative of the integrand.