2.1 Riemann Sums and the Fundamental Theorem of Calculus

2.1 Riemann Sums and the Fundamental Theorem of Calculus

Riemann integration forms the theoretical bedrock of accumulation and area in single-variable calculus. On the GRE Mathematics Subject Test, questions in this domain evaluate both theoretical real analysis principles (Darboux sums, partition refinements, and Lebesgue's integrability theorem) and advanced computational techniques (Leibniz differentiation of parameter-dependent integrals and summation limit conversions). Navigating this section successfully requires understanding how continuity, integrability, and differentiability interlock.

Partitions, Darboux Sums, and Integrability

Let $[a, b] \subset \mathbb{R}$ be a closed, bounded interval. A partition $P$ of $[a, b]$ is a finite ordered set of points: $P = \{a = x_0 < x_1 < x_2 < \dots < x_{n-1} < x_n = b\}$ The width of the $i$-th subinterval $[x_{i-1}, x_i]$ is denoted $\Delta x_i = x_i - x_{i-1}$, and the mesh (or norm) of the partition is $|P| = \max_{1 \le i \le n} \Delta x_i$.

Let $f: [a, b] \to \mathbb{R}$ be a bounded function. On each subinterval $[x_{i-1}, x_i]$, define the local infimum and supremum: $m_i = \inf_{x \in [x_{i-1}, x_i]} f(x), \quad M_i = \sup_{x \in [x_{i-1}, x_i]} f(x)$ The Lower Darboux Sum $L(f, P)$ and Upper Darboux Sum $U(f, P)$ are defined by: $L(f, P) = \sum_{i=1}^n m_i \Delta x_i, \quad U(f, P) = \sum_{i=1}^n M_i \Delta x_i$ For any partition $P$, $m(b - a) \le L(f, P) \le U(f, P) \le M(b - a)$, where $m = \inf_{[a, b]} f$ and $M = \sup_{[a, b]} f$. A partition $P'$ is a refinement of $P$ if $P \subseteq P'$. Adding refinement points can only increase lower sums and decrease upper sums: $L(f, P) \le L(f, P') \le U(f, P') \le U(f, P)$ Consequently, for any two arbitrary partitions $P_1$ and $P_2$, $L(f, P_1) \le U(f, P_2)$.

The lower Darboux integral and upper Darboux integral are defined respectively by: $\underline{\int_a^b} f(x)\,dx = \sup_P L(f, P), \quad \overline{\int_a^b} f(x)\,dx = \inf_P U(f, P)$ A bounded function $f$ is defined to be Riemann integrable on $[a, b]$ (denoted $f \in \mathcal{R}[a, b]$) if and only if: $\underline{\int_a^b} f(x)\,dx = \overline{\int_a^b} f(x)\,dx$ When this equality holds, the common value is the definite Riemann integral, denoted $\int_a^b f(x),dx$.

The Darboux Integrability Criterion

A bounded function $f$ is Riemann integrable on $[a, b]$ if and only if for every $\varepsilon > 0$, there exists a partition $P$ of $[a, b]$ such that: $U(f, P) - L(f, P) = \sum_{i=1}^n (M_i - m_i) \Delta x_i < \varepsilon$

Lebesgue's Criterion for Riemann Integrability

A bounded real-valued function $f$ on $[a, b]$ is Riemann integrable if and only if its set of discontinuities $D_f = {x \in [a, b] : f \text{ is discontinuous at } x}$ has Lebesgue measure zero.

• Every continuous function on $[a, b]$ is Riemann integrable.
• Every monotonic function on $[a, b]$ has at most countably many jump discontinuities (which has measure zero) and is therefore Riemann integrable.
• Functions with countably many discontinuities (such as Thomae's popcorn function) are Riemann integrable.
• The Dirichlet indicator function $\mathbf{1}_{\mathbb{Q}}$ is discontinuous at every point in $[a, b]$. Because $\mathbb{R} \setminus \mathbb{Q}$ and $\mathbb{Q}$ are both dense, $U(f, P) = b - a$ and $L(f, P) = 0$ for every partition $P$, making it not Riemann integrable.

Fundamental Properties of the Definite Integral

For any functions $f, g \in \mathcal{R}[a, b]$ and constants $\alpha, \beta \in \mathbb{R}$:

1. Linearity: $\int_a^b (\alpha f(x) + \beta g(x)),dx = \alpha \int_a^b f(x),dx + \beta \int_a^b g(x),dx$.
2. Interval Additivity: For any $c \in [a, b]$, $\int_a^b f(x),dx = \int_a^c f(x),dx + \int_c^b f(x),dx$.
3. Monotonicity: If $f(x) \le g(x)$ for all $x \in [a, b]$, then $\int_a^b f(x),dx \le \int_a^b g(x),dx$.
4. Triangle Inequality: $\left| \int_a^b f(x),dx \right| \le \int_a^b |f(x)|,dx$.

The Fundamental Theorem of Calculus

FTC Part 1: Differentiation of Accumulation Functions

Let $f \in \mathcal{R}[a, b]$ and define the accumulation function $F: [a, b] \to \mathbb{R}$ by: $F(x) = \int_a^x f(t)\,dt$

1. $F$ is uniformly continuous on $[a, b]$. If $|f(t)| \le M$, then $|F(x) - F(y)| \le M|x - y|$, proving $F$ is Lipschitz continuous.
2. If $f$ is continuous at a point $x_0 \in (a, b)$, then $F$ is differentiable at $x_0$, and $F'(x_0) = f(x_0)$.

Leibniz Integral Rule (Variable Limits and Parameters)

When the integration limits are differentiable functions $u(x)$ and $v(x)$, applying the chain rule to $F(x) = \int_{u(x)}^{v(x)} f(t),dt$ yields: $\frac{d}{dx} \int_{u(x)}^{v(x)} f(t)\,dt = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x)$ If the integrand also depends explicitly on $x$, Leibniz's general formula incorporates differentiation under the integral sign: $\frac{d}{dx} \int_{u(x)}^{v(x)} f(x, t)\,dt = f(x, v(x)) v'(x) - f(x, u(x)) u'(x) + \int_{u(x)}^{v(x)} \frac{\partial f}{\partial x}(x, t)\,dt$

FTC Part 2: The Evaluation Theorem

If $f: [a, b] \to \mathbb{R}$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ on $[a, b]$ (meaning $F'(x) = f(x)$ for all $x \in (a, b)$), then: $\int_a^b f(x)\,dx = F(b) - F(a) = [F(x)]_a^b$

The Mean Value Theorem for Definite Integrals

If $f: [a, b] \to \mathbb{R}$ is continuous, there exists at least one point $c \in (a, b)$ such that: $f(c) = \frac{1}{b - a} \int_a^b f(x)\,dx$ The value $f_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x),dx$ defines the average value of $f$ on $[a, b]$.

GRE Exam Pattern: Limits of Riemann Sums

A classic GRE Subject Test question presents a discrete limit of sums and requires converting it into an equivalent Riemann integral over $[0, 1]$: $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n g\left(\frac{k}{n}\right) = \int_0^1 g(x)\,dx$ To recognize this structure:

1. Factor out $\frac{1}{n}$ from the general term to represent the differential $dx$.
2. Group the remaining expression in terms of $\frac{k}{n}$, which corresponds to the continuous variable $x$.
3. Set the limits of integration to $0$ and $1$ (for $k = 1, \dots, n$) and evaluate the resulting definite integral.

Integrability Classes Comparison

Worked Problem: Leibniz Rule with Variable Limits

Problem: Find $G'(1)$ where $G(x) = \int_{x}^{x^2} \sqrt{1 + t^3},dt$. Solution: Here $f(t) = \sqrt{1 + t^3}$, lower limit $u(x) = x$, and upper limit $v(x) = x^2$. By Leibniz's rule: $G'(x) = f(v(x)) v'(x) - f(u(x)) u'(x) = \sqrt{1 + (x^2)^3} \cdot (2x) - \sqrt{1 + x^3} \cdot (1)$ $G'(x) = 2x \sqrt{1 + x^6} - \sqrt{1 + x^3}$ Evaluating at $x = 1$: $G'(1) = 2(1) \sqrt{1 + 1^6} - \sqrt{1 + 1^3} = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$

Common GRE Traps & Pitfalls

• Omission of the inner derivative: When applying Leibniz's rule to $\int_0^{x^2} e^{-t^2},dt$, writing $e^{-x^4}$ instead of $2x e^{-x^4}$ overlooks the chain rule factor $v'(x)$.
• Blindly applying FTC Part 2 across singularities: Evaluating $\int_{-1}^1 \frac{1}{x^2},dx = \left[-\frac{1}{x}\right]_{-1}^1 = -1 - (1) = -2$ produces a negative value for a strictly positive function because the integrand has an infinite discontinuity at $x = 0$, violating the continuity hypothesis of FTC.
• Overlooking integrand variable dependence: When differentiating $\int_0^x x \cos(t),dt$, one must either pull $x$ outside the integral first or apply the partial derivative term in Leibniz's general rule.