The Riemann Integral & FTC

A partition $P$ of $[a, b]$ is a finite ordered set $P = \{x_0, x_1, \ldots, x_n\}$ with $a = x_0 < x_1 < \cdots < x_n = b$. The mesh (or norm) of $P$ is $\|P\| = \max_{1 \le i \le n} \Delta x_i$ where $\Delta x_i = x_i - x_{i-1}$. A partition $Q$ is a refinement of $P$ if $P \subseteq Q$ (every point of $P$ is also a point of $Q$).

Given a bounded function $f: [a, b] \to \mathbb{R}$, a partition $P = \{x_0, \ldots, x_n\}$, and sample points $t_i \in [x_{i-1}, x_i]$ for each $i$, the Riemann sum is

$S(f, P, \{t_i\}) = \sum_{i=1}^{n} f(t_i) \, \Delta x_i.$

Specific choices of sample points yield named rules:

• Left rule: $t_i = x_{i-1}$ (left endpoint of each subinterval)
• Right rule: $t_i = x_i$ (right endpoint)
• Midpoint rule: $t_i = \frac{x_{i-1} + x_i}{2}$ (midpoint)

Take the uniform partition with $n = 4$ subintervals, so $\Delta x = 1/4$ and the partition points are $\{0, 1/4, 1/2, 3/4, 1\}$.

Left sum: $S_L = \frac{1}{4}\bigl(f(0) + f(1/4) + f(1/2) + f(3/4)\bigr) = \frac{1}{4}\bigl(0 + \tfrac{1}{16} + \tfrac{4}{16} + \tfrac{9}{16}\bigr) = \frac{14}{64} \approx 0.219.$

Right sum: $S_R = \frac{1}{4}\bigl(f(1/4) + f(1/2) + f(3/4) + f(1)\bigr) = \frac{1}{4}\bigl(\tfrac{1}{16} + \tfrac{4}{16} + \tfrac{9}{16} + 1\bigr) = \frac{30}{64} \approx 0.469.$

The exact integral is $\int_0^1 x^2 \, dx = 1/3 \approx 0.333$, which is trapped between the left and right sums (as expected for this increasing function). As $n$ grows, both sums converge to $1/3$.

Average the left and right Riemann sums to get the trapezoidal rule:

$T(f, P) = \frac{1}{2}[S_L(f, P) + S_R(f, P)] = \sum_{i=1}^n \frac{f(x_{i-1}) + f(x_i)}{2} \, \Delta x_i.$

Geometrically, each rectangle is replaced by a trapezoid whose top edge connects $(x_{i-1}, f(x_{i-1}))$ to $(x_i, f(x_i))$. This typically gives a better approximation than either endpoint rule alone, because the trapezoid’s area is the average of the over-estimate and under-estimate.

For a bounded function $f$ on $[a, b]$ and a partition $P = \{x_0, \ldots, x_n\}$, define

$M_i = \sup_{x \in [x_{i-1}, x_i]} f(x) \quad \text{and} \quad m_i = \inf_{x \in [x_{i-1}, x_i]} f(x).$

The upper Darboux sum is $U(f, P) = \sum_{i=1}^n M_i \, \Delta x_i$ and the lower Darboux sum is $L(f, P) = \sum_{i=1}^n m_i \, \Delta x_i$.

For any choice of sample points, $L(f, P) \le S(f, P, \{t_i\}) \le U(f, P)$.

If $Q$ is a refinement of $P$, then

$L(f, P) \le L(f, Q) \le U(f, Q) \le U(f, P).$

Refining a partition increases lower sums and decreases upper sums — the bounds get tighter.

It suffices to show the result when $Q$ is obtained by adding a single point $x^* \in (x_{k-1}, x_k)$ to $P$, since the general case follows by induction on the number of added points.

On the subinterval $[x_{k-1}, x_k]$, the upper sum contributes $M_k \cdot \Delta x_k$ where $M_k = \sup_{[x_{k-1}, x_k]} f$. Splitting at $x^*$ gives two subintervals $[x_{k-1}, x^*]$ and $[x^*, x_k]$ with suprema

$M_k' = \sup_{[x_{k-1}, x^*]} f \le M_k \quad \text{and} \quad M_k'' = \sup_{[x^*, x_k]} f \le M_k.$

So the new contribution is $M_k'(x^* - x_{k-1}) + M_k''(x_k - x^*) \le M_k(x^* - x_{k-1}) + M_k(x_k - x^*) = M_k \cdot \Delta x_k.$

All other subintervals are unchanged, so $U(f, Q) \le U(f, P)$. The argument for $L(f, P) \le L(f, Q)$ is analogous (infima on smaller subintervals are $\ge$ the infimum on the larger subinterval).

For any partitions $P$ and $Q$ (not necessarily related by refinement), $L(f, P) \le U(f, Q)$.

The common refinement $P \cup Q$ refines both $P$ and $Q$. By Proposition 1:

$L(f, P) \le L(f, P \cup Q) \le U(f, P \cup Q) \le U(f, Q).$

The middle inequality $L(f, P \cup Q) \le U(f, P \cup Q)$ holds because $m_i \le M_i$ on every subinterval.

A bounded function $f: [a, b] \to \mathbb{R}$ is Riemann integrable if the upper and lower integrals agree:

$\underline{\int_a^b} f = \sup_P L(f, P) = \inf_P U(f, P) = \overline{\int_a^b} f.$

The common value is written $\int_a^b f(x) \, dx$ and called the Riemann integral of $f$ on $[a, b]$.

$f$ is Riemann integrable if and only if for every $\varepsilon > 0$, there exists a partition $P$ such that $U(f, P) - L(f, P) < \varepsilon$. This is the criterion we’ll use in all integrability proofs: show that the gap between upper and lower sums can be made as small as we like.

If $f: [a, b] \to \mathbb{R}$ is continuous, then $f$ is Riemann integrable.

Since $[a, b]$ is compact (Heine-Borel, Completeness & Compactness), $f$ is uniformly continuous on $[a, b]$ (Heine-Cantor theorem, also from Topic 3).

Given $\varepsilon > 0$, choose $\delta > 0$ such that

$|x - y| < \delta \implies |f(x) - f(y)| < \frac{\varepsilon}{b - a}.$

Take any partition $P$ with mesh $\|P\| < \delta$. On each subinterval $[x_{i-1}, x_i]$, the length is $\Delta x_i < \delta$, so for any $x, y \in [x_{i-1}, x_i]$ we have $|f(x) - f(y)| < \varepsilon/(b - a)$. In particular:

$M_i - m_i = \sup_{[x_{i-1}, x_i]} f - \inf_{[x_{i-1}, x_i]} f < \frac{\varepsilon}{b - a}.$

Therefore:

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

Every step is explicit. The chain compactness $\to$ uniform continuity $\to$ integrability is fully traced.

If $f: [a, b] \to \mathbb{R}$ is monotone (either increasing or decreasing throughout), then $f$ is Riemann integrable.

Assume $f$ is increasing (the decreasing case is analogous). For any partition $P$, the supremum on $[x_{i-1}, x_i]$ is $M_i = f(x_i)$ and the infimum is $m_i = f(x_{i-1})$ — since $f$ is increasing.

With the uniform partition $\Delta x_i = (b - a)/n$:

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

The sum telescopes: $\sum_{i=1}^n [f(x_i) - f(x_{i-1})] = f(b) - f(a)$.

So $U - L = \frac{(b - a)(f(b) - f(a))}{n}$. Given $\varepsilon > 0$, choose $n > \frac{(b - a)(f(b) - f(a))}{\varepsilon}$.

Define $\mathbf{1}_{\mathbb{Q}}(x) = 1$ if $x \in \mathbb{Q}$ and $\mathbf{1}_{\mathbb{Q}}(x) = 0$ if $x \notin \mathbb{Q}$. On any subinterval $[x_{i-1}, x_i]$:

• $M_i = 1$ because the rationals are dense — every subinterval contains rational numbers.
• $m_i = 0$ because the irrationals are dense — every subinterval contains irrational numbers.

So $U(f, P) = \sum M_i \, \Delta x_i = b - a$ and $L(f, P) = \sum m_i \, \Delta x_i = 0$ for every partition $P$. The upper and lower integrals disagree: $\overline{\int} = b - a \neq 0 = \underline{\int}$, so $f$ is not Riemann integrable.

This example matters for ML: the Dirichlet function is a legitimate measurable function, and the Lebesgue integral can handle it ($\int \mathbf{1}_{\mathbb{Q}} = 0$). The failure of Riemann integration for such functions is precisely why measure-theoretic probability (→ formalML: Measure-Theoretic Probability) requires the more powerful Lebesgue integral.

If $f$ is bounded on $[a, b]$ and has only finitely many points of discontinuity, then $f$ is still Riemann integrable. The idea: cover the discontinuities with subintervals of arbitrarily small total length (contributing at most $\varepsilon$ to $U - L$), and handle the continuous portions with Theorem 1. Discontinuities on a set of “measure zero” don’t spoil integrability — a fact that foreshadows the Lebesgue criterion.

If $f$ and $g$ are Riemann integrable on $[a, b]$ and $c \in \mathbb{R}$:

(a) Linearity:

$\int_a^b [f(x) + g(x)] \, dx = \int_a^b f(x) \, dx + \int_a^b g(x) \, dx \quad \text{and} \quad \int_a^b c \, f(x) \, dx = c \int_a^b f(x) \, dx.$

(b) Monotonicity: If $f(x) \le g(x)$ for all $x \in [a, b]$, then $\int_a^b f \le \int_a^b g$.

(c) Additivity over intervals: For any $c \in (a, b)$:

$\int_a^b f = \int_a^c f + \int_c^b f.$

(d) Triangle inequality:

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

(a) Linearity (additivity). For any partition $P$, the infimum of a sum satisfies $\inf(f + g) \ge \inf f + \inf g$ on each subinterval. Therefore $L(f + g, P) \ge L(f, P) + L(g, P)$. Similarly, $U(f + g, P) \le U(f, P) + U(g, P)$. Taking the supremum of lower sums and infimum of upper sums gives

$\int_a^b f + \int_a^b g \le \underline{\int_a^b} (f + g) \le \overline{\int_a^b} (f + g) \le \int_a^b f + \int_a^b g.$

So all four quantities are equal. The scalar multiplication $\int cf = c \int f$ follows by similar reasoning (with separate cases for $c \ge 0$ and $c < 0$).

(b) Monotonicity. If $f(x) \le g(x)$ on $[a, b]$, then $L(f, P) \le L(g, P)$ for every partition $P$ (the infimum of $f$ on each subinterval is $\le$ the infimum of $g$). Taking $\sup_P$ gives $\int f \le \int g$.

(c) Additivity. Use a common refinement that includes the point $c$. On such a refinement, the Riemann sums over $[a, b]$ split exactly into sums over $[a, c]$ and $[c, b]$.

(d) Triangle inequality. Since $-|f(x)| \le f(x) \le |f(x)|$ for all $x$, apply monotonicity: $-\int |f| \le \int f \le \int |f|$, which gives $|\int f| \le \int |f|$.

We compute the integral directly as a limit of Riemann sums — the “hard” way — to see the definition in action.

Using the right-endpoint sum with the uniform partition ($\Delta x = 1/n$, sample points $t_i = i/n$):

$S_n = \sum_{i=1}^n \left(\frac{i}{n}\right)^2 \cdot \frac{1}{n} = \frac{1}{n^3} \sum_{i=1}^n i^2 = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2}.$

As $n \to \infty$:

$\lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{6n^2} = \frac{2}{6} = \frac{1}{3}.$

So $\int_0^1 x^2 \, dx = 1/3$. This is the hard way — the FTC (coming in §7) will make it a one-line computation.

Right-endpoint uniform partition: $S_n = \frac{1}{n} \sum_{i=1}^n e^{i/n} = \frac{1}{n} \cdot e^{1/n} \cdot \frac{e^{n \cdot (1/n)} - 1}{e^{1/n} - 1} = \frac{e^{1/n}(e - 1)}{n(e^{1/n} - 1)}$ (geometric series).

As $n \to \infty$, using $e^{1/n} - 1 \approx 1/n$: $S_n \to e - 1$. This matches $[e^x]_0^1 = e - 1$.

A function $F: [a, b] \to \mathbb{R}$ is an antiderivative of $f$ on $(a, b)$ if $F'(x) = f(x)$ for all $x \in (a, b)$.

If $F$ is an antiderivative of $f$, then so is $F + C$ for any constant $C$ — and these are the only antiderivatives. This follows from Corollary 1 of the MVT (Mean Value Theorem & Taylor Expansion): if $(F - G)' = 0$ on $(a, b)$, then $F - G$ is constant.

If $f$ is continuous on $[a, b]$, then the function $F: [a, b] \to \mathbb{R}$ defined by

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

is continuous on $[a, b]$, differentiable on $(a, b)$, and $F'(x) = f(x)$ for all $x \in (a, b)$.

Fix $x \in (a, b)$. For $h \neq 0$ with $x + h \in [a, b]$, the additivity of the integral gives:

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

Since $f$ is continuous at $x$: given $\varepsilon > 0$, choose $\delta > 0$ such that $|t - x| < \delta \implies |f(t) - f(x)| < \varepsilon$. For $|h| < \delta$, every $t$ between $x$ and $x + h$ satisfies $|t - x| \le |h| < \delta$, so $|f(t) - f(x)| < \varepsilon$. Therefore:

$\left|\frac{F(x+h) - F(x)}{h} - f(x)\right| = \left|\frac{1}{h} \int_x^{x+h} [f(t) - f(x)] \, dt\right| \le \frac{1}{|h|} \int_x^{x+h} |f(t) - f(x)| \, dt < \frac{1}{|h|} \cdot \varepsilon \cdot |h| = \varepsilon.$

The integral bound uses $|f(t) - f(x)| < \varepsilon$ for all $t$ in the interval of length $|h|$, and the triangle inequality for integrals (Theorem 3(d)).

This shows $\lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = f(x)$, i.e., $F'(x) = f(x)$.

$\frac{d}{dx} \int_0^x \sin(t^2) \, dt = \sin(x^2).$

No antiderivative of $\sin(t^2)$ exists in elementary functions (the integral defines the Fresnel function $S(x)$). But the FTC Part 1 tells us the derivative of the area function anyway. This illustrates the power of FTC1: we can differentiate integrals even when we cannot compute them in closed form.

$\frac{d}{dx} \int_0^{x^2} e^{-t^2} \, dt = e^{-x^4} \cdot 2x.$

Here the upper limit is $g(x) = x^2$ rather than $x$ itself. By the chain rule (The Derivative & Chain Rule, Theorem 6): if $G(u) = \int_0^u e^{-t^2} \, dt$, then $G'(u) = e^{-u^2}$ by FTC1, and $\frac{d}{dx} G(x^2) = G'(x^2) \cdot 2x = e^{-x^4} \cdot 2x$.

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ on $[a, b]$ (i.e., $F'(x) = f(x)$ for all $x \in (a, b)$), then

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

Let $P = \{x_0, x_1, \ldots, x_n\}$ be any partition of $[a, b]$. Write $F(b) - F(a)$ as a telescoping sum:

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

By the Mean Value Theorem (Mean Value Theorem & Taylor Expansion, Theorem 2), applied to $F$ on each subinterval $[x_{i-1}, x_i]$: there exists $c_i \in (x_{i-1}, x_i)$ such that

$F(x_i) - F(x_{i-1}) = F'(c_i)(x_i - x_{i-1}) = f(c_i) \, \Delta x_i.$

So the telescoping sum becomes a Riemann sum:

$F(b) - F(a) = \sum_{i=1}^n f(c_i) \, \Delta x_i.$

Since $f$ is continuous (hence Riemann integrable by Theorem 1), as $\|P\| \to 0$ every Riemann sum converges to $\int_a^b f$. But the left side $F(b) - F(a)$ is independent of the partition. Therefore:

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

$\int_0^1 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3} - 0 = \frac{1}{3}.$

Compare with Example 3: the same answer, in one line instead of a page of algebra. The FTC transforms a limit evaluation into an antiderivative evaluation.

$\int_0^\pi \sin(x) \, dx = [-\cos(x)]_0^\pi = -\cos(\pi) + \cos(0) = 1 + 1 = 2.$

The area under one arch of the sine curve is exactly 2.

Part 1 says: start with $f$ continuous $\to$ construct $F(x) = \int_a^x f$ $\to$ get $F' = f$. Part 2 says: start with any antiderivative $F$ of $f$ $\to$ $\int_a^b f = F(b) - F(a)$. They are complementary, not equivalent. Part 1 is an existence result (continuous functions have antiderivatives); Part 2 is a computation result (integrals can be evaluated via antiderivatives).

If $g: [a, b] \to \mathbb{R}$ is continuously differentiable and $f$ is continuous on the range of $g$, then

$\int_a^b f(g(x)) \, g'(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, du.$

Let $F$ be an antiderivative of $f$ (exists by FTC Part 1 since $f$ is continuous). Then by the chain rule (The Derivative & Chain Rule, Theorem 6):

$(F \circ g)'(x) = F'(g(x)) \cdot g'(x) = f(g(x)) \cdot g'(x).$

So $F \circ g$ is an antiderivative of $f(g(x)) \cdot g'(x)$. By FTC Part 2:

$\int_a^b f(g(x)) \, g'(x) \, dx = F(g(b)) - F(g(a)) = \int_{g(a)}^{g(b)} f(u) \, du.$

Let $u = x^2$, so $du = 2x \, dx$. The bounds transform: $u(0) = 0$, $u(1) = 1$. Therefore:

$\int_0^1 2x \, e^{x^2} \, dx = \int_0^1 e^u \, du = [e^u]_0^1 = e - 1.$

If $u$ and $v$ are continuously differentiable on $[a, b]$, then

$\int_a^b u(x) \, v'(x) \, dx = [u(x) \, v(x)]_a^b - \int_a^b u'(x) \, v(x) \, dx.$

By the product rule (The Derivative & Chain Rule, Theorem 4): $(uv)' = u'v + uv'$. Integrate both sides from $a$ to $b$ using FTC Part 2:

$u(b) \, v(b) - u(a) \, v(a) = \int_a^b u'(x) \, v(x) \, dx + \int_a^b u(x) \, v'(x) \, dx.$

Rearrange to get the integration by parts formula.

Let $u = x$ and $dv = e^x \, dx$, so $du = dx$ and $v = e^x$. Then:

$\int_0^1 x \, e^x \, dx = [x \, e^x]_0^1 - \int_0^1 e^x \, dx = (1 \cdot e - 0) - [e^x]_0^1 = e - (e - 1) = 1.$

Substitution is the chain rule $(f \circ g)' = (f' \circ g) \cdot g'$ read from right to left. Integration by parts is the product rule $(uv)' = u'v + uv'$ rearranged. Every differentiation rule has an integration counterpart — they are the same theorems, applied in opposite directions.

If $f$ is $(n+1)$-times continuously differentiable on an interval containing $a$ and $x$, then

$f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^k + R_n(x)$

where the remainder is given by

$R_n(x) = \frac{1}{n!} \int_a^x (x - t)^n \, f^{(n+1)}(t) \, dt.$

By induction on $n$.

Base case ($n = 0$): We need $f(x) = f(a) + \int_a^x f'(t) \, dt$. This is exactly FTC Part 2 applied to $F = f$: $\int_a^x f'(t) \, dt = f(x) - f(a)$.

Inductive step: Assume the result holds for $n - 1$:

$R_{n-1}(x) = \frac{1}{(n-1)!} \int_a^x (x - t)^{n-1} f^{(n)}(t) \, dt.$

Apply integration by parts with

$u = f^{(n)}(t), \quad dv = \frac{(x - t)^{n-1}}{(n-1)!} \, dt, \quad \text{so} \quad v = -\frac{(x - t)^n}{n!}.$

Then:

$R_{n-1}(x) = \left[-\frac{(x - t)^n}{n!} f^{(n)}(t)\right]_a^x + \frac{1}{n!} \int_a^x (x - t)^n f^{(n+1)}(t) \, dt.$

Evaluating the boundary term: at $t = x$, $(x - t)^n = 0$; at $t = a$, the term is $-(-1) \cdot \frac{(x - a)^n}{n!} f^{(n)}(a) = \frac{f^{(n)}(a)}{n!}(x - a)^n$. So:

$R_{n-1}(x) = \frac{f^{(n)}(a)}{n!}(x - a)^n + R_n(x).$

Absorbing $\frac{f^{(n)}(a)}{n!}(x - a)^n$ into the Taylor polynomial of degree $n$ gives the result.

The Lagrange form $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1}$ (from Mean Value Theorem & Taylor Expansion, Theorem 4) follows from the integral form by the Mean Value Theorem for Integrals. The integral form is stronger: it gives the exact remainder as an integral, not just a bound involving an unknown intermediate point $c$.

For $f$ with sufficient smoothness on $[a, b]$ with $n$ subintervals of width $h = (b - a)/n$:

(a) Left/Right endpoint rule: $|E_n| \le \frac{(b - a)^2}{2n} \max_{[a,b]} |f'|$ — error $O(h)$.

(b) Trapezoidal rule: $|E_n| \le \frac{(b - a)^3}{12n^2} \max_{[a,b]} |f''|$ — error $O(h^2)$.

(c) Simpson’s rule: $|E_n| \le \frac{(b - a)^5}{180n^4} \max_{[a,b]} |f^{(4)}|$ — error $O(h^4)$.

Approximate $\int_0^1 e^x \, dx = e - 1 \approx 1.71828$ with $n = 10$ subintervals:

| Rule | Approximation | $|E_{10}|$ | Bound | |------|---------------|-----------|-------| | Left | $\approx 1.5819$ | $\approx 0.136$ | $\le 0.136$ | | Trapezoidal | $\approx 1.7197$ | $\approx 0.00136$ | $\le 0.00227$ | | Simpson’s | $\approx 1.718282$ | $\approx 1.5 \times 10^{-6}$ | $\le 1.5 \times 10^{-6}$ |

Simpson’s rule achieves 6 digits of accuracy with just 10 subintervals. The convergence rates — $O(1/n)$, $O(1/n^2)$, $O(1/n^4)$ — explain why higher-order methods dominate in practice.

Simpson’s rule fits a parabola through every three consecutive points $(x_{i-1}, f(x_{i-1}))$, $(x_i, f(x_i))$, $(x_{i+1}, f(x_{i+1}))$, then integrates the parabola exactly. It is exact for polynomials up to degree 3 — not just degree 2 — because of a symmetry cancellation. This is why the error involves $f^{(4)}$, not $f^{(3)}$.