Riemann Integration | Mathematics

6.1 Definition

Let $f : [a,b] \to \mathbb{R}$ be bounded. A partition of $[a,b]$ is a finite set $P = \{x_0, x_1, \ldots, x_n\}$ with $a = x_0 \lt x_1 \lt \cdots \lt x_n = b$ .

The upper sum and lower sum of $f$ with respect to $P$ are:

$U(f, P) = \sum_{i=1}^{n} M_i \Delta x_i, \quad L(f, P) = \sum_{i=1}^{n} m_i \Delta x_i$

Where $M_i = \sup\{f(x) : x \in [x_{i-1}, x_i]\}$ , $m_i = \inf\{f(x) : x \in [x_{i-1}, x_i]\}$ And $\Delta x_i = x_i - x_{i-1}$ .

The mesh of $P$ is $\|P\| = \max_{1 \leq i \leq n} \Delta x_i$ .

Definition. $f$ is Riemann integrable on $[a,b]$ if the upper and lower integrals are equal:

$\overline{\int_a^b} f(x)\, dx = \underline{\int_a^b} f(x)\, dx$

Where $\overline{\int_a^b} f = \inf\{U(f,P) : P \mathrm{\ is\ a\ partition}\}$ and $\underline{\int_a^b} f = \sup\{L(f,P) : P \mathrm{\ is\ a\ partition}\}$ .

The common value is denoted $\int_a^b f(x)\, dx$ .

6.2 Integrability Criteria

Theorem 6.1 (Riemann Integrability Criterion). A bounded function $f : [a,b] \to \mathbb{R}$ is Riemann integrable if and only if for every $\varepsilon > 0$ There exists a partition $P$ such that

$U(f,P) - L(f,P) \lt \varepsilon$

Theorem 6.2. Every continuous function on $[a,b]$ is Riemann integrable.

Proof. Let $f$ be continuous on $[a,b]$ . By the Heine-Cantor theorem, $f$ is uniformly continuous. Given $\varepsilon > 0$ Choose $\delta > 0$ such that $|x - y| \lt \delta$ implies $|f(x) - f(y)| \lt \varepsilon/(b-a)$ .

Let $P$ be any partition with $\|P\| \lt \delta$ . On each subinterval $[x_{i-1}, x_i]$ By the Extreme Value Theorem, $f$ attains its maximum $M_i$ and minimum $m_i$ . By uniform continuity: $M_i - m_i \lt \varepsilon/(b-a)$ . Therefore:

$U(f,P) - L(f,P) = \sum_{i=1}^{n}(M_i - m_i)\Delta x_i \lt \frac{\varepsilon}{b-a} \sum_{i=1}^{n} \Delta x_i = \varepsilon$

By the Riemann integrability criterion, $f$ is integrable. $\blacksquare$

Theorem 6.3. Every monotone function on $[a,b]$ is Riemann integrable.

Proof. Assume $f$ is increasing (the decreasing case is analogous). Given $\varepsilon > 0$ Let $P_n$ be the uniform partition with $n$ subintervals of length $(b-a)/n$ . On $[x_{i-1}, x_i]$ : $M_i = f(x_i)$ and $m_i = f(x_{i-1})$ . Then:

$U(f, P_n) - L(f, P_n) = \sum_{i=1}^{n} [f(x_i) - f(x_{i-1})] \cdot \frac{b-a}{n} = [f(b) - f(a)] \cdot \frac{b-a}{n}$

Choose $n$ large enough that $[f(b) - f(a)](b-a)/n \lt \varepsilon$ . $\blacksquare$

Theorem 6.4. A bounded function with finitely many discontinuities on $[a,b]$ is Riemann integrable.

Proof (sketch). Let $f$ have discontinuities at $d_1, \ldots, d_m \in [a,b]$ . Given $\varepsilon > 0$ Enclose each $d_j$ in a small interval $I_j$ of total length $\varepsilon/(2M)$ Where $M = \sup_{[a,b]} |f|$ . On the remaining set (a finite union of closed intervals), $f$ is continuous, Hence uniformly continuous. Choose a partition fine enough that the oscillation of $f$ on each Subinterval outside the $I_j$ is less than $\varepsilon/(2(b-a))$ . Then:

$U(f, P) - L(f, P) \leq \frac{\varepsilon}{2(b-a)} \cdot (b - a) + 2M \cdot \frac{\varepsilon}{2M} = \varepsilon$

$\blacksquare$

Proposition 6.4a. The set of Riemann integrable functions on $[a,b]$ forms a vector space, and If $f$ and $g$ are integrable, then so are $|f|$ , $f^2$ And $\max(f, g)$ .

Theorem 6.4b (Lebesgue”s Criterion for Riemann Integrability). A bounded function $f : [a,b] \to \mathbb{R}$ Is Riemann integrable if and only if the set of its discontinuities has (Lebesgue) measure zero.

Remark. A set has measure zero if it can be covered by countably many intervals of arbitrarily Small total length. In particular, every countable set has measure zero. This means:

Every continuous function is integrable (empty set of discontinuities).
Every function with countably many discontinuities is integrable (Theorem 6.4 is a special case).
The Dirichlet function $f(x) = 1$ for $x \in \mathbb{Q}$ and $f(x) = 0$ for $x \notin \mathbb{Q}$ is discontinuous everywhere (set of discontinuities = $[a,b]$ Measure $> 0$ ), hence not integrable.
Thomae’s function $f(x) = 1/q$ if $x = p/q$ in lowest terms, and $f(x) = 0$ if $x$ is irrational, is continuous at every irrational and discontinuous at every rational. Since $\mathbb{Q}$ is countable (measure zero), Thomae’s function is Riemann integrable, with $\int_0^1 f = 0$ .

6.3 Properties of the Integral

Theorem 6.5 (Linearity). If $f$ and $g$ are integrable on $[a,b]$ and $\alpha, \beta \in \mathbb{R}$ :

$\int_a^b (\alpha f + \beta g) = \alpha \int_a^b f + \beta \int_a^b g$

Theorem 6.6 (Monotonicity). If $f(x) \leq g(x)$ for all $x \in [a,b]$ Then $\int_a^b f \leq \int_a^b g$ .

Theorem 6.7 (Triangle Inequality). $\left|\int_a^b f\right| \leq \int_a^b |f|$ .

6.4 The Fundamental Theorem of Calculus

Theorem 6.8 (FTC Part 1). If $f$ is continuous on $[a,b]$ Then the function

$F(x) = \int_a^x f(t)\, dt$

Is differentiable on $(a,b)$ and $F'(x) = f(x)$ .

Proof. Let $h > 0$ (the case $h \lt 0$ is similar). By the Mean Value Theorem for Integrals (which follows from the EVT), there exists $\xi \in [x, x+h]$ such that

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

As $h \to 0^+$ We have $\xi \to x^+$ (since $\xi \in [x, x+h]$ ). By continuity of $f$ $f(\xi) \to f(x)$ . Hence $F'_+(x) = f(x)$ . A similar argument gives $F'_-(x) = f(x)$ . $\blacksquare$

Theorem 6.9 (FTC Part 2). If $F$ is differentiable on $[a,b]$ with $F' = f$ (and $f$ is integrable), Then

$\int_a^b f(x)\, dx = F(b) - F(a)$

Proof. Let $P = \{x_0, \ldots, x_n\}$ be any partition of $[a,b]$ . By the Mean Value Theorem, For each $i$ there exists $\xi_i \in [x_{i-1}, x_i]$ with $F(x_i) - F(x_{i-1}) = f(\xi_i)\Delta x_i$ . Summing:

$F(b) - F(a) = \sum_{i=1}^{n} [F(x_i) - F(x_{i-1})] = \sum_{i=1}^{n} f(\xi_i) \Delta x_i$

The right-hand side is a Riemann sum for $\int_a^b f$ . As $\|P\| \to 0$ This converges to the Integral. Hence $F(b) - F(a) = \int_a^b f(x)\, dx$ . $\blacksquare$

6.5 Worked Examples

Problem. Compute $\int_0^1 x^2\, dx$ from the definition.

Solution. Let $P_n = \{0, 1/n, 2/n, \ldots, 1\}$ . On $[x_{i-1}, x_i] = [(i-1)/n, i/n]$ , $f(x) = x^2$ Has $M_i = (i/n)^2$ and $m_i = ((i-1)/n)^2$ .

$U(f, P_n) = \sum_{i=1}^{n} \frac{i^2}{n^2} \cdot \frac{1}{n} = \frac{1}{n^3} \sum_{i=1}^{n} i^2 = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}$

As $n \to \infty$ : $\lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \frac{2}{6} = \frac{1}{3}$ .

Similarly, $L(f, P_n) \to 1/3$ . So $\int_0^1 x^2\, dx = 1/3$ . $\blacksquare$

Worked Example: Compute $\int_0^1 \sqrt{x}\, dx$ from the definition

Solution. Let $P_n = \{0, 1/n, 2/n, \ldots, 1\}$ . On $[(i-1)/n, i/n]$ , $f(x) = \sqrt{x}$ has $M_i = \sqrt{i/n}$ and $m_i = \sqrt{(i-1)/n}$ .

$U(f, P_n) = \sum_{i=1}^{n} \sqrt{\frac{i}{n}} \cdot \frac{1}{n} = \frac{1}{n^{3/2}} \sum_{i=1}^{n} \sqrt{i}$

Using $\sum_{i=1}^{n} \sqrt{i} = \frac{2}{3} n^{3/2} + O(n^{1/2})$ (obtained from comparing with $\int_0^n \sqrt{x}\, dx$ ):

$\lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} \frac{1}{n^{3/2}} \cdot \frac{2}{3}n^{3/2} = \frac{2}{3}$

Similarly $L(f, P_n) \to 2/3$ Confirming $\int_0^1 \sqrt{x}\, dx = 2/3$ . $\blacksquare$

6.6 Improper Integrals

Definition. An improper integral is a Riemann integral where either the interval of integration Is unbounded or the integrand is unbounded.

Type I (Infinite Intervals). If $f$ is Riemann integrable on $[a, b]$ for every $b > a$ Define:

$\int_a^{\infty} f(x)\, dx = \lim_{b \to \infty} \int_a^b f(x)\, dx$

The integral converges if this limit exists as a finite number; otherwise it diverges.

Type II (Unbounded Integrands). If $f$ is unbounded near $a$ but integrable on $[c, b]$ for every $c \in (a, b]$ :

$\int_a^b f(x)\, dx = \lim_{c \to a^+} \int_c^b f(x)\, dx$

Theorem 6.10 (Comparison Test for Improper Integrals). If $0 \leq f(x) \leq g(x)$ for $x \geq a$ :

If $\int_a^{\infty} g$ converges, then $\int_a^{\infty} f$ converges.
If $\int_a^{\infty} f$ diverges, then $\int_a^{\infty} g$ diverges.

Theorem 6.11 (Absolute Convergence). If $\int_a^{\infty} |f(x)|\, dx$ converges, then $\int_a^{\infty} f(x)\, dx$ converges.

Theorem 6.12 ( $p$ -Test for Improper Integrals).

Type I: $\int_1^{\infty} \frac{1}{x^p}\, dx$ converges if and only if $p > 1$ .
Type II: $\int_0^1 \frac{1}{x^p}\, dx$ converges if and only if $p < 1$ .

Proof. For Type I with $p \neq 1$ :

$\int_1^{\infty} x^{-p}\, dx = \lim_{b \to \infty} \left[\frac{x^{1-p}}{1-p}\right]_1^b = \lim_{b \to \infty} \frac{b^{1-p} - 1}{1-p}$

This converges when $1 - p < 0$ I.e., $p > 1$ . For $p = 1$ : $\int_1^{\infty} 1/x\, dx = \lim_{b \to \infty} \ln b = \infty$ .

For Type II: $\int_0^1 x^{-p}\, dx = \lim_{c \to 0^+} \frac{1 - c^{1-p}}{1-p}$ . This converges when $1 - p > 0$ I.e., $p < 1$ . $\blacksquare$

Remark. The $p$ -test for Type I integrals mirrors the $p$ -series test: $\sum 1/n^p$ converges Iff $p > 1$ . This is not a coincidence --- the integral test establishes the connection.

Worked Example: Evaluate $\int_0^{\infty} e^{-x}\, dx$

Solution. This is a Type I improper integral:

$\int_0^{\infty} e^{-x}\, dx = \lim_{b \to \infty} \int_0^b e^{-x}\, dx = \lim_{b \to \infty} \left[-e^{-x}\right]_0^b = \lim_{b \to \infty} (-e^{-b} + 1) = 1$

So the integral converges to $1$ . $\blacksquare$

Worked Example: Does $\int_1^{\infty} \frac{\sin x}{x}\, dx$ converge?

Solution. The integral $\int_1^{\infty} \left|\frac{\sin x}{x}\right|\, dx$ diverges (compare with $\int_1^{\infty} \frac{|\sin x|}{x}\, dx \geq \sum_{k=1}^{\infty} \int_{k\pi}^{(k+1)\pi} \frac{|\sin x|}{x}\, dx \geq \sum_{k=1}^{\infty} \frac{2}{(k+1)\pi}$ , which diverges by comparison with the harmonic series).

However, $\int_1^{\infty} \frac{\sin x}{x}\, dx$ converges by Dirichlet’s test for integrals. Let $F(b) = \int_1^b \sin x\, dx = \cos 1 - \cos b$ Which is bounded by $|\cos 1 - \cos b| \leq 2$ . Since $1/x$ decreases to $0$ By integration by parts:

$\int_1^b \frac{\sin x}{x}\, dx = \frac{-\cos x}{x}\bigg|_1^b - \int_1^b \frac{\cos x}{x^2}\, dx$

As $b \to \infty$ The boundary term $\cos b / b \to 0$ and $\int_1^{\infty} \frac{|\cos x|}{x^2}\, dx \leq \int_1^{\infty} \frac{1}{x^2}\, dx = 1$ , so the improper integral converges (conditionally). $\blacksquare$

Worked Example: Evaluate $\int_0^1 \frac{1}{\sqrt{x}}\, dx$ (Type II improper integral)

Solution. The integrand $f(x) = 1/\sqrt{x}$ is unbounded as $x \to 0^+$ . Compute:

$\int_0^1 \frac{1}{\sqrt{x}}\, dx = \lim_{c \to 0^+} \int_c^1 x^{-1/2}\, dx = \lim_{c \to 0^+} \left[2\sqrt{x}\right]_c^1 = \lim_{c \to 0^+} (2 - 2\sqrt{c}) = 2$

The improper integral converges to $2$ . Note that $\int_0^1 x^{-p}\, dx$ converges for $p \lt 1$ and Diverges for $p \geq 1$ . $\blacksquare$

Worked Example: Determine convergence of $\int_1^{\infty} \frac{1}{x^p}\, dx$ for various $p$

Solution. By the $p$ -test (Theorem 6.12): $\int_1^{\infty} x^{-p}\, dx$ converges iff $p > 1$ .

Specifically:

$p = 2$ : $\int_1^{\infty} 1/x^2\, dx = \lim_{b \to \infty} [-1/x]_1^b = 0 - (-1) = 1$ . Converges.
$p = 1$ : $\int_1^{\infty} 1/x\, dx = \lim_{b \to \infty} \ln b = \infty$ . Diverges.
$p = 1/2$ : $\int_1^{\infty} 1/\sqrt{x}\, dx = \lim_{b \to \infty} [2\sqrt{x}]_1^b = \infty$ . Diverges.

This mirrors the $p$ -series test: $\sum 1/n^p$ converges iff $p > 1$ . $\blacksquare$

Worked Example: Evaluate $\int_0^{\infty} x e^{-x}\, dx$

Solution. This integral requires both a Type I and Type II limit:

$\int_0^{\infty} x e^{-x}\, dx = \lim_{a \to 0^+} \lim_{b \to \infty} \int_a^b x e^{-x}\, dx$

Integrate by parts with $u = x$ , $dv = e^{-x}\, dx$ So $du = dx$ , $v = -e^{-x}$ :

$\int x e^{-x}\, dx = -xe^{-x} + \int e^{-x}\, dx = -xe^{-x} - e^{-x} = -(x+1)e^{-x}$

Evaluating: $\lim_{b \to \infty} [-(b+1)e^{-b}] - \lim_{a \to 0^+} [-(a+1)e^{-a}] = 0 - (-1) = 1$ .

So $\int_0^{\infty} x e^{-x}\, dx = 1$ . This equals $\Gamma(2) = 1! = 1$ . $\blacksquare$

6.7 Common Mistakes

Mistake 1: Confusing Riemann integrability with Lebesgue integrability Every Riemann integrable function is Lebesgue integrable, but not vice versa. The Dirichlet function is not Riemann integrable but is Lebesgue integrable (with integral zero). The Riemann integral requires the set of discontinuities to have measure zero, while the Lebesgue integral handles a much broader class of functions.

Mistake 2: Interchanging limits and integrals without justification Swapping $\lim$ and $\int$ requires a convergence theorem such as the dominated convergence theorem or the monotone convergence theorem. Pointwise convergence alone is insufficient: the sequence $f_n = n\chi_{(0, 1/n)}$ converges pointwise to $0$ on $[0, 1]$ , yet $\int_0^1 f_n = 1$ for all $n$ . Always verify the hypotheses of a convergence theorem before interchanging limits and integrals.

Mistake 3: Applying the Fundamental Theorem of Calculus to discontinuous integrands The FTC Part 1 requires $f$ to be continuous. If $f$ has a jump discontinuity, the integral function $F(x) = \int_a^x f(t)\,dt$ is continuous but not differentiable at the discontinuity. The FTC Part 2 requires $F' = f$ everywhere on $[a, b]$ , which fails if $f$ is discontinuous. Check continuity before applying the FTC.

Intuition

The Riemann integral answers the question: how do we add up infinitely many infinitely thin slices? Consider the area under a curve f(x) from a to b. The Riemann sum approximates this by dividing the interval into subintervals and summing f(x) times the width of each subinterval — a collection of rectangles whose total area approaches the true area as the partition becomes finer. This is the same logic as computing distance by summing velocity over time, or charge by summing current over time: the integral accumulates a quantity that varies continuously.

The Fundamental Theorem of Calculus reveals the deep connection between accumulation (integration) and rate of change (differentiation). If F(x) represents the accumulated area from a to x, then F’(x) = f(x): the rate at which area accumulates is exactly the height of the curve. Conversely, integrating a rate function recovers the total change. This duality is why the FTC is the most important theorem in calculus: it converts the hard problem of computing areas into the easy problem of finding antiderivatives. The improper integral extends this to infinite intervals or unbounded functions by taking limits, but the core intuition — accumulation of slices — remains the same.

intervals. For unbounded Functions or infinite intervals, one must use the improper Riemann integral. A common error is Applying the FTC directly to improper integrals without taking the limit. Also, conditional Convergence of improper integrals behaves differently from absolute convergence: rearranging the “terms” (subintervals) of a conditionally convergent improper integral can change its value.

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    A[6_Riemann Integration] --> B[Key Concepts]
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Cross-References

Continuity: The Heine-Cantor theorem guarantees that continuous functions on closed bounded intervals are uniformly continuous, which is used to prove Riemann integrability.
Differentiability: The fundamental theorem of calculus connects the Riemann integral with antiderivatives and differentiation.
Lebesgue Integration: Lebesgue integration generalises Riemann integration and handles functions with more complex discontinuity patterns.

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