Mathematics / Analysis Integral Calculus & Series 100% Free Open Access
Chapter 2 โ€ข Theory & Derivations

Riemann Sums, Definite Integrals & Fundamental Theorems

ยง2.1 Partitions, Darboux Sums & The Rigorous Riemann Integral

1. Partitions and Mesh Size

Let $[a, b] \subset \mathbb{R}$ be a compact interval. A partition $\mathcal{P}$ of $[a, b]$ is a finite ordered sequence of points:

$$\mathcal{P} = \{ a = x_0 < x_1 < x_2 < \dots < x_{n-1} < x_n = b \}$$

The $k$-th subinterval is $\Delta x_k = x_k - x_{k-1}$. The mesh (or norm) of $\mathcal{P}$ is the maximal subinterval width:

$$\|\mathcal{P}\| = \max_{1 \le k \le n} \Delta x_k$$

2. Darboux Upper and Lower Sums

For a bounded function $f: [a, b] \to \mathbb{R}$, define the infimum and supremum on each subinterval $[x_{k-1}, x_k]$:

$$m_k = \inf_{x \in [x_{k-1}, x_k]} f(x), \qquad M_k = \sup_{x \in [x_{k-1}, x_k]} f(x)$$

The Lower Darboux Sum $L(f, \mathcal{P})$ and Upper Darboux Sum $U(f, \mathcal{P})$ are:

$$L(f, \mathcal{P}) = \sum_{k=1}^n m_k \Delta x_k, \qquad U(f, \mathcal{P}) = \sum_{k=1}^n M_k \Delta x_k$$

For any partition $\mathcal{P}$, $L(f, \mathcal{P}) \le U(f, \mathcal{P})$. Furthermore, if $\mathcal{P}^*$ is a refinement of $\mathcal{P}$ ($\mathcal{P} \subseteq \mathcal{P}^*$), adding partition points only increases lower sums and decreases upper sums:

$$L(f, \mathcal{P}) \le L(f, \mathcal{P}^*) \le U(f, \mathcal{P}^*) \le U(f, \mathcal{P})$$

3. The Riemann Integrability Criterion

Define the Lower and Upper Darboux Integrals over all possible partitions $\mathscr{P}$ of $[a, b]$:

$$\underline{\int_a^b} f(x) \, dx = \sup_{\mathcal{P} \in \mathscr{P}} L(f, \mathcal{P}), \qquad \overline{\int_a^b} f(x) \, dx = \inf_{\mathcal{P} \in \mathscr{P}} U(f, \mathcal{P})$$
$$\mathbf{\text{Riemann Integrability: } f \text{ is Riemann integrable on } [a, b] \iff \underline{\int_a^b} f(x) \, dx = \overline{\int_a^b} f(x) \, dx = \int_a^b f(x) \, dx}$$

Cauchy-Riemann $\varepsilon$-Criterion: A bounded function $f$ is Riemann integrable on $[a, b]$ if and only if for every $\varepsilon > 0$, there exists a partition $\mathcal{P}_\varepsilon$ such that:

$$U(f, \mathcal{P}_\varepsilon) - L(f, \mathcal{P}_\varepsilon) = \sum_{k=1}^n (M_k - m_k) \Delta x_k < \varepsilon$$

Theorem: Every continuous function $f \in C([a, b])$ is Riemann integrable (by Heine-Cantor Uniform Continuity on compact intervals).

ยง2.2 Riemann Sum Approximations: Left, Right, Midpoint & Trapezoidal Rules

1. Tagged Partitions and General Riemann Sums

Choosing an arbitrary evaluation tag $c_k \in [x_{k-1}, x_k]$ inside each subinterval yields the general Riemann sum:

$$S(f, \mathcal{P}, \{c_k\}) = \sum_{k=1}^n f(c_k) \Delta x_k$$

The definite integral is the strict analytical limit as mesh size vanishes:

$$\int_a^b f(x) \, dx = \lim_{\|\mathcal{P}\| \to 0} \sum_{k=1}^n f(c_k) \Delta x_k$$

2. Standard Uniform Partitions ($\Delta x = \frac{b - a}{n}$)

Dividing $[a, b]$ into $n$ equal subintervals with $x_k = a + k \Delta x$ generates classical numerical rules:

  • Left Riemann Sum: $c_k = x_{k-1} \implies L_n = \sum_{k=1}^n f(x_{k-1}) \Delta x$
  • Right Riemann Sum: $c_k = x_k \implies R_n = \sum_{k=1}^n f(x_k) \Delta x$
  • Midpoint Rule: $c_k = \frac{x_{k-1} + x_k}{2} \implies M_n = \sum_{k=1}^n f\left(x_{k-1/2}\right) \Delta x$
  • Trapezoidal Rule: The average of Left and Right sums: $$T_n = \frac{L_n + R_n}{2} = \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + \dots + 2f(x_{n-1}) + f(x_n) \right]$$

3. Asymptotic Error Order

Using Taylor expansions, the truncation error $E(f) = \int_a^b f(x)dx - \text{Approximation}$ scales as:

$$|E_{\text{Left}}| \le \frac{(b-a)^2}{2n} M_1, \qquad |E_{\text{Midpoint}}| \le \frac{(b-a)^3}{24n^2} M_2, \qquad |E_{\text{Trap}}| \le \frac{(b-a)^3}{12n^2} M_2$$

where $M_1 = \max |f'(x)|$ and $M_2 = \max |f''(x)|$. The Midpoint and Trapezoidal rules achieve second-order convergence $\mathcal{O}(1/n^2)$.

ยง2.3 Fundamental Properties of Definite Integrals & The Integral Mean Value Theorem

1. Algebraic and Order Properties

For Riemann integrable functions $f, g$ and scalars $\alpha, \beta \in \mathbb{R}$:

  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. Subinterval 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. Integral Triangle Inequality: $$\left| \int_a^b f(x) \, dx \right| \le \int_a^b |f(x)| \, dx$$

2. The Cauchy-Schwarz Inequality for Integrals

For any two real square-integrable functions $f, g \in L^2([a, b])$:

$$\left( \int_a^b f(x) g(x) \, dx \right)^2 \le \left( \int_a^b f(x)^2 \, dx \right) \left( \int_a^b g(x)^2 \, dx \right)$$

Proof: Consider the quadratic polynomial in $\lambda \in \mathbb{R}$: $P(\lambda) = \int_a^b (\lambda f(x) + g(x))^2 \, dx \ge 0$. Expanding $P(\lambda) = A \lambda^2 + 2B \lambda + C \ge 0$, the discriminant $\Delta = 4(B^2 - AC) \le 0 \implies B^2 \le AC$.

3. The Mean Value Theorem for Definite Integrals

$$\mathbf{\text{Theorem: If } f: [a, b] \to \mathbb{R} \text{ is continuous, there exists at least one } c \in (a, b) \text{ such that:}}$$
$$f(c) = \frac{1}{b - a} \int_a^b f(x) \, dx$$

Proof: By the Extreme Value Theorem, $f$ attains minimum $m$ and maximum $M$ on $[a, b]$. Integrating $m \le f(x) \le M$ gives $m(b - a) \le \int_a^b f(x) \, dx \le M(b - a) \implies m \le \frac{1}{b-a}\int_a^b f(x)\,dx \le M$. By Bolzano's Intermediate Value Theorem, $f$ must attain this average value at some point $c \in (a, b)$.

ยง2.4 The Fundamental Theorems of Calculus (FTC 1 & 2): Line-by-Line Proofs

1. The First Fundamental Theorem of Calculus (FTC-1: Differentiation of Accumulation)

$$\mathbf{\text{Theorem (FTC-1): Let } f: [a, b] \to \mathbb{R} \text{ be continuous. Define the accumulation function } F(x) = \int_a^x f(t) \, dt. \text{ Then } F \text{ is differentiable on } (a, b) \text{ and } F'(x) = f(x).}$$

Rigorous Proof: Form the Newton difference quotient for $h \ne 0$:

$$\frac{F(x + h) - F(x)}{h} = \frac{1}{h} \left( \int_a^{x+h} f(t) \, dt - \int_a^x f(t) \, dt \right) = \frac{1}{h} \int_x^{x+h} f(t) \, dt$$

Subtract $f(x) = \frac{1}{h} \int_x^{x+h} f(x) \, dt$ from both sides:

$$\left| \frac{F(x + h) - F(x)}{h} - f(x) \right| = \left| \frac{1}{h} \int_x^{x+h} \Big( f(t) - f(x) \Big) dt \right| \le \frac{1}{|h|} \left| \int_x^{x+h} |f(t) - f(x)| dt \right|$$

Since $f$ is continuous at $x$, for any $\varepsilon > 0$ there exists $\delta > 0$ such that $|t - x| < \delta \implies |f(t) - f(x)| < \varepsilon$. When $0 < |h| < \delta$, every $t$ in the integration interval satisfies $|t - x| \le |h| < \delta$. Therefore:

$$\left| \frac{F(x + h) - F(x)}{h} - f(x) \right| < \frac{1}{|h|} \int_{\min(x, x+h)}^{\max(x, x+h)} \varepsilon \, dt = \frac{1}{|h|} \cdot \varepsilon |h| = \varepsilon$$

Taking the limit as $h \to 0$ proves that $F'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = f(x)$. $\blacksquare$

2. The Second Fundamental Theorem of Calculus (FTC-2: Evaluation Theorem)

$$\mathbf{\text{Theorem (FTC-2): If } f \in C([a, b]) \text{ and } G \text{ is ANY antiderivative of } f \text{ (i.e. } G'(x) = f(x)\text{), then:}}$$
$$\int_a^b f(x) \, dx = G(b) - G(a)$$

Proof: From FTC-1, $F(x) = \int_a^x f(t)\,dt$ is an antiderivative of $f$. Since any two antiderivatives differ by a constant on a connected interval, $G(x) = F(x) + C$ for some $C \in \mathbb{R}$.

Evaluating at $x = a$: $G(a) = F(a) + C = \int_a^a f(t)\,dt + C = 0 + C = C$.

Evaluating at $x = b$: $G(b) = F(b) + C = \int_a^b f(t)\,dt + G(a)$.

Subtracting $G(a)$ from both sides establishes: $\int_a^b f(t)\,dt = G(b) - G(a)$. $\blacksquare$

ยง2.5 The Leibniz Integral Rule: Differentiation Under the Integral Sign

1. Variable Limits and Parameter-Dependent Integrals

Let $f(x, t)$ and its partial derivative $\frac{\partial f}{\partial x}$ be continuous in both variables, and let $u(x), v(x)$ be continuously differentiable functions. Consider the accumulation integral:

$$I(x) = \int_{u(x)}^{v(x)} f(x, t) \, dt$$

2. The Full Leibniz Integral Formula

$$\mathbf{\frac{d}{dx} \left[ \int_{u(x)}^{v(x)} f(x, t) \, dt \right] = f(x, v(x)) \cdot v'(x) - f(x, u(x)) \cdot u'(x) + \int_{u(x)}^{v(x)} \frac{\partial f}{\partial x}(x, t) \, dt}$$

3. Analytical Derivation via Multi-Variable Chain Rule

Define $H(x, u, v) = \int_u^v f(x, t) \, dt$. The total derivative of $I(x) = H(x, u(x), v(x))$ with respect to $x$ is:

$$\frac{dI}{dx} = \frac{\partial H}{\partial x} + \frac{\partial H}{\partial u} \frac{du}{dx} + \frac{\partial H}{\partial v} \frac{dv}{dx}$$

By FTC-1:

$$\frac{\partial H}{\partial v} = \frac{\partial}{\partial v}\int_u^v f(x, t) dt = f(x, v)$$
$$\frac{\partial H}{\partial u} = \frac{\partial}{\partial u}\left( -\int_v^u f(x, t) dt \right) = -f(x, u)$$

Differentiating the integral with respect to parameter $x$ inside the constant limits $[u, v]$ allows passing the derivative under the integral sign: $\frac{\partial H}{\partial x} = \int_u^v \frac{\partial f}{\partial x}(x, t) dt$. Summing these three terms recovers the Leibniz rule.

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Example 2.1: Limit of a Riemann Sum via Definite Integration

Evaluate the limit of the sequence by expressing it as a definite Riemann integral:

$$L = \lim_{n \to \infty} \sum_{k=1}^n \frac{k^3}{n^4 + k^4}$$

Step 1: Normalize into a Standard Riemann Sum Form

To convert $\lim_{n \to \infty} \sum_{k=1}^n f(k/n) \frac{1}{n}$ into $\int_0^1 f(x) \, dx$, factor $n^4$ from the denominator:

$$\frac{k^3}{n^4 + k^4} = \frac{k^3}{n^4 \left( 1 + \frac{k^4}{n^4} \right)} = \frac{1}{n} \cdot \frac{(k/n)^3}{1 + (k/n)^4}$$

Step 2: Recognize the Riemann Sum Components

With $\Delta x = \frac{1}{n}$ and evaluation points $x_k = \frac{k}{n} \in [0, 1]$:

$$L = \lim_{n \to \infty} \sum_{k=1}^n \frac{(x_k)^3}{1 + (x_k)^4} \Delta x = \int_0^1 \frac{x^3}{1 + x^4} \, dx$$

Step 3: Evaluate the Definite Integral

Substitute $u = 1 + x^4 \implies du = 4x^3 \, dx \implies x^3 \, dx = \frac{1}{4} du$.

Limits: when $x = 0$, $u = 1$; when $x = 1$, $u = 2$.

$$L = \int_1^2 \frac{1/4}{u} \, du = \frac{1}{4} \Big[ \ln|u| \Big]_1^2 = \frac{1}{4} \Big( \ln(2) - \ln(1) \Big) = \frac{\ln(2)}{4}$$
Final Answer & Physical Insight

$L = \frac{\ln(2)}{4}$

Tier 2: Intermediate Exam Example 2.2: Leibniz Integral Rule with Variable Limits

Compute the derivative $F'(x)$ for $x > 0$ where:

$$F(x) = \int_{x}^{x^2} \frac{\cos(xt)}{t} \, dt$$

Step 1: Identify Leibniz Rule Components

For $F(x) = \int_{u(x)}^{v(x)} f(x, t) \, dt$ with $u(x) = x$, $v(x) = x^2$, and $f(x, t) = \frac{\cos(xt)}{t}$:

$$u'(x) = 1, \qquad v'(x) = 2x$$

The partial derivative with respect to parameter $x$ is:

$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( \frac{\cos(xt)}{t} \right) = \frac{-t \sin(xt)}{t} = -\sin(xt)$$

Step 2: Apply the Leibniz Formula

$$F'(x) = f(x, v(x)) \cdot v'(x) - f(x, u(x)) \cdot u'(x) + \int_{u(x)}^{v(x)} \frac{\partial f}{\partial x} \, dt$$
$$F'(x) = \left( \frac{\cos(x \cdot x^2)}{x^2} \right) (2x) - \left( \frac{\cos(x \cdot x)}{x} \right) (1) + \int_x^{x^2} (-\sin(xt)) \, dt$$

Step 3: Evaluate the Boundary Terms and the Residual Integral

Boundary terms:

$$\frac{2x \cos(x^3)}{x^2} - \frac{\cos(x^2)}{x} = \frac{2\cos(x^3) - \cos(x^2)}{x}$$

Residual integral with respect to $t$:

$$\int_x^{x^2} (-\sin(xt)) \, dt = \left[ \frac{\cos(xt)}{x} \right]_x^{x^2} = \frac{\cos(x^3) - \cos(x^2)}{x}$$

Step 4: Combine All Terms

$$F'(x) = \frac{2\cos(x^3) - \cos(x^2)}{x} + \frac{\cos(x^3) - \cos(x^2)}{x} = \frac{3\cos(x^3) - 2\cos(x^2)}{x}$$
Final Answer & Physical Insight

$F'(x) = \frac{3\cos(x^3) - 2\cos(x^2)}{x}$

Tier 3: Honors / Proof Challenge Example 2.3: Strict Integrability of Composition & Triangle Inequality

Let $f: [a, b] \to \mathbb{R}$ be bounded and Riemann integrable. (a) Prove that the absolute value function $|f|: [a, b] \to \mathbb{R}$ is also Riemann integrable. (b) Prove the integral triangle inequality:

$$\left| \int_a^b f(x) \, dx \right| \le \int_a^b |f(x)| \, dx$$

(c) Provide a counterexample showing that the converse of (a) is generally false.

Part (a): Proof of Integrability of $|f|$

Let $\mathcal{P} = \{x_0, x_1, \dots, x_n\}$ be any partition of $[a, b]$. For any subinterval $[x_{k-1}, x_k]$, define:

$$M_k = \sup_{x \in I_k} f(x), \quad m_k = \inf_{x \in I_k} f(x), \quad M_k^* = \sup_{x \in I_k} |f(x)|, \quad m_k^* = \inf_{x \in I_k} |f(x)|$$

For any two points $x, y \in I_k$, the reverse triangle inequality states:

$$||f(x)| - |f(y)|| \le |f(x) - f(y)| \le M_k - m_k$$

Taking the supremum over all $x, y \in I_k$ yields the key oscillation inequality:

$$M_k^* - m_k^* \le M_k - m_k$$

Multiplying by $\Delta x_k$ and summing over $k = 1, \dots, n$:

$$U(|f|, \mathcal{P}) - L(|f|, \mathcal{P}) = \sum_{k=1}^n (M_k^* - m_k^*) \Delta x_k \le \sum_{k=1}^n (M_k - m_k) \Delta x_k = U(f, \mathcal{P}) - L(f, \mathcal{P})$$

Since $f$ is Riemann integrable, for any $\varepsilon > 0$ there exists a partition $\mathcal{P}_\varepsilon$ such that $U(f, \mathcal{P}_\varepsilon) - L(f, \mathcal{P}_\varepsilon) < \varepsilon$. Thus $U(|f|, \mathcal{P}_\varepsilon) - L(|f|, \mathcal{P}_\varepsilon) < \varepsilon$, establishing that $|f|$ is Riemann integrable by the Cauchy criterion. $\blacksquare$

Part (b): Proof of the Integral Triangle Inequality

For all $x \in [a, b]$, the definitions of absolute value imply:

$$-|f(x)| \le f(x) \le |f(x)|$$

By the monotonicity property of the Riemann integral:

$$-\int_a^b |f(x)| \, dx \le \int_a^b f(x) \, dx \le \int_a^b |f(x)| \, dx$$

This is equivalent to the statement:

$$\mathbf{\left| \int_a^b f(x) \, dx \right| \le \int_a^b |f(x)| \, dx} \quad \blacksquare$$

Part (c): Counterexample to the Converse

Consider Dirichlet's Modified Function on $[0, 1]$:

$$f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \cap [0, 1] \\ -1 & \text{if } x \in (\mathbb{R} \setminus \mathbb{Q}) \cap [0, 1] \end{cases}$$

Then $|f(x)| = 1$ for all $x \in [0, 1]$, which is a constant function and trivially Riemann integrable with $\int_0^1 |f(x)|dx = 1$.

However, for any partition $\mathcal{P}$ of $[0, 1]$, every subinterval contains both rationals and irrationals, so $M_k = 1$ and $m_k = -1$. Hence $U(f, \mathcal{P}) = 1$ and $L(f, \mathcal{P}) = -1$. Since $U \ne L$, $f(x)$ is NOT Riemann integrable. This proves the converse does not hold.

Final Answer & Physical Insight

$|f|$ is Riemann integrable and satisfies $|\int f| \le \int |f|$; converse is refuted by the modified Dirichlet function.