---
title: Properties of the Integral
module: The Riemann Integral
moduleNumber: 6
lessonNumber: 3
order: 603
summary: >
  The integral is a linear, order-preserving, additive operator on the
  integrable functions. It splits across subintervals, respects inequalities,
  bounds the size of a function by the integral of its absolute value, and
  preserves products. The mean value theorem for integrals identifies the
  integral with an attained average height on a fixed rectangle.
topics: [The Riemann Integral]
draft: false
sources:
  - book: Lebl
    ref: "§5.2 Properties of the integral (additivity, linearity, monotonicity)"
  - book: Shkoller
    ref: "§1.8 The Riemann integral is linear; §1.9 Further properties of the Riemann integral"
---

The integral behaves like a sum: it adds across pieces of the
domain, scales and adds across functions, and preserves order. Each property is
inherited from the corresponding fact about the Darboux sums that define it, and
together they turn $\int_a^b$ into a linear operator on the space
$\mathcal{R}[a,b]$ of [integrable functions](/real-analysis/riemann-integration/integrability-classes).

## Additivity over subintervals

Cutting the domain at an interior point splits the integral into two, and
subdividing at a point does not affect integrability.

> **Lemma (Additivity of Darboux integrals).** For $a < b < c$ and bounded
> $f : [a,c] \to \mathbb{R}$,
> $$
> \underline{\int_a^c} f = \underline{\int_a^b} f + \underline{\int_b^c} f,
> \qquad
> \overline{\int_a^c} f = \overline{\int_a^b} f + \overline{\int_b^c} f.
> $$

> **Proof.** A partition of $[a,c]$ that contains $b$ splits into partitions of
> $[a,b]$ and $[b,c]$, and its lower sum splits accordingly. Restricting the
> supremum to partitions containing $b$ loses nothing, since inserting $b$ is a
> refinement that only raises lower sums.[^lebl-52] The upper case is symmetric.

> **Proposition (Additivity).** For $a < b < c$, a function is integrable on
> $[a,c]$ if and only if it is integrable on $[a,b]$ and on $[b,c]$, and then
> $$
> \int_a^c f = \int_a^b f + \int_b^c f.
> $$

$$
% caption: Additivity: the area over $[a,c]$ is the area over $[a,b]$ plus the
% caption: area over $[b,c]$. Cutting at the interior point $b$ neither creates nor
% caption: destroys integrability.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% region under curve, split at b
\fill[acc!16] (0,0) -- (0,1.1) .. controls (1.2,2.0) and (1.6,1.9) .. (2.4,1.7) -- (2.4,0) -- cycle;
\fill[acc!8] (2.4,0) -- (2.4,1.7) .. controls (3.4,1.4) and (4.0,1.9) .. (4.8,2.2) -- (4.8,0) -- cycle;
\draw[acc, very thick] (0,1.1) .. controls (1.2,2.0) and (1.6,1.9) .. (2.4,1.7) .. controls (3.4,1.4) and (4.0,1.9) .. (4.8,2.2);
\draw[black] (2.4,0) -- (2.4,1.7);
\draw[->] (-0.15,0) -- (5.1,0);
\draw[->] (0,-0.15) -- (0,2.7);
\foreach \x/\lab in {0/a, 2.4/b, 4.8/c}
  \draw (\x,0.07) -- (\x,-0.07) node[below, font=\footnotesize] {$\lab$};
\node[acc, font=\footnotesize] at (1.2,0.6) {$\int_a^b f$};
\node[black, font=\footnotesize] at (3.6,0.6) {$\int_b^c f$};
\end{tikzpicture}
$$

> **Proof.** The equivalence runs both ways. Integrability on $[a,c]$ forces
> integrability on each piece, because the lower and upper integrals of the whole
> are additive and already equal, which pins the pieces' lower and upper integrals
> together; conversely, integrability on both pieces makes the whole integral
> additive.[^lebl-52]

Two conventions extend the formula to any ordering of the limits, so additivity
holds for arbitrary $\alpha, \beta, \gamma \in [a,b]$:[^lebl-52]
$$
\int_b^a f \coloneqq -\int_a^b f,
\qquad
\int_a^a f \coloneqq 0,
\qquad
\int_\alpha^\gamma f = \int_\alpha^\beta f + \int_\beta^\gamma f.
$$

> **Worked example (A piecewise integrand).** Additivity and linearity together
> evaluate piecewise integrands. For
> $$
> f(x) = \begin{cases} 2x & 0 \le x \le 1, \\ 3 - x & 1 < x \le 2, \end{cases}
> $$
> split at $x = 1$ and integrate each polynomial piece:
> $$
> \int_0^2 f = \int_0^1 2x\,\d x + \int_1^2 (3 - x)\,\d x
> = 1 + \big(3 - \tfrac32\big) = \tfrac52,
> $$
> using the
> [fundamental theorem](/real-analysis/riemann-integration/fundamental-theorem)
> on each continuous piece.

## Linearity

The integral is linear: it commutes with scaling and with addition. On a fixed
partition the Darboux sums already scale and split, and the operations are
preserved in the passage to $\sup$ and $\inf$.

> **Proposition (Linearity).** If $f, g \in \mathcal{R}[a,b]$ and
> $\alpha \in \mathbb{R}$, then $\alpha f$ and $f + g$ are integrable and
> $$
> \int_a^b \alpha f = \alpha \int_a^b f,
> \qquad
> \int_a^b (f + g) = \int_a^b f + \int_a^b g.
> $$

> **Proof.** For $\alpha \ge 0$, multiplication by $\alpha$ passes through both the
> infimum and the supremum defining $m_i$ and $M_i$, so
> $L(P, \alpha f) = \alpha L(P,f)$ and likewise for $U$; the case $\alpha < 0$
> swaps the roles of $\inf$ and $\sup$ and is handled by proving
> $\int (-f) = -\int f$ separately.[^lebl-52] For the sum, the tagged-sum form is
> immediate — $S(P, f+g) = S(P,f) + S(P,g)$ for every partition — and passing to
> the limit gives additivity.[^shk-lin] For Darboux sums directly, the
> inequalities $\overline{\int}(f+g) \le \overline{\int} f + \overline{\int} g$
> and $\underline{\int}(f+g) \ge \underline{\int} f + \underline{\int} g$ pinch to
> equality when $f$ and $g$ are integrable.

## Order and monotonicity

Larger integrands give larger integrals. On each subinterval $f \le g$ forces
$m_i(f) \le m_i(g)$, so the lower sums compare, and the ordering is preserved by
the supremum.

> **Proposition (Monotonicity).** If $f, g \in \mathcal{R}[a,b]$ and
> $f(x) \le g(x)$ for all $x$, then
> $$
> \int_a^b f \;\le\; \int_a^b g.
> $$

Two consequences are used constantly. Taking $g \equiv M$ recovers the crude bound
$\int_a^b f \le M(b-a)$; taking $f \equiv m$ gives the lower bound. And applied to
a nonnegative integrand, monotonicity says $\int_a^b f \ge 0$ whenever $f \ge 0$.

$$
% caption: Monotonicity of the integral. With $f \le g$ pointwise, the region
% caption: under $f$ sits inside the region under $g$, so $\int f \le \int g$; the
% caption: shaded band between the graphs is the nonnegative difference.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\definecolor{sec}{HTML}{7A6A9A}
\fill[acc!16] (0,0) -- (0,0.6) .. controls (1.4,1.2) and (3.0,0.7) .. (4.4,1.1) -- (4.4,0) -- cycle;
\fill[sec!16] (0,0.6) .. controls (1.4,1.2) and (3.0,0.7) .. (4.4,1.1)
  -- (4.4,2.0) .. controls (3.0,1.7) and (1.4,2.3) .. (0,1.5) -- cycle;
\draw[acc, very thick] (0,0.6) .. controls (1.4,1.2) and (3.0,0.7) .. (4.4,1.1);
\draw[sec, very thick] (0,1.5) .. controls (1.4,2.3) and (3.0,1.7) .. (4.4,2.0);
\draw[->] (-0.15,0) -- (4.9,0);
\draw[->] (0,-0.15) -- (0,2.7);
\node[sec, font=\footnotesize, anchor=west] at (4.45,2.0) {$g$};
\node[acc, font=\footnotesize, anchor=west] at (4.45,1.1) {$f$};
\node[sec, font=\footnotesize] at (2.1,1.5) {gap};
\foreach \x/\lab in {0/a, 4.4/b}
  \draw (\x,0.07) -- (\x,-0.07) node[below, font=\footnotesize] {$\lab$};
\end{tikzpicture}
$$

## The integral triangle inequality

The size of an integral is controlled by the integral of the size. First, taking
absolute values preserves integrability, because the oscillation of $|f|$ never
exceeds that of $f$.

> **Proposition (Absolute value).** If $f \in \mathcal{R}[a,b]$ then
> $|f| \in \mathcal{R}[a,b]$ and
> $$
> \left| \int_a^b f \right| \;\le\; \int_a^b |f|.
> $$

> **Proof.** The oscillation bound is pointwise: on each subinterval
> $\sup |f| - \inf |f| \le \sup f - \inf f$, so
> $U(P, |f|) - L(P, |f|) \le U(P,f) - L(P,f)$, and the Cauchy criterion transfers.
> The inequality itself follows from $-|f| \le f \le |f|$ and
> monotonicity.[^shk-abs]

Geometrically, $\int_a^b f$ is a **signed** area — lobes below the axis subtract —
while $\int_a^b |f|$ folds those lobes up and adds them, so cancellation can only
shrink the magnitude.

$$
% caption: Signed versus unsigned area. On the left $\int f$ counts the lower lobe
% caption: negatively, so magnitudes cancel; on the right $|f|$ reflects that lobe
% caption: above the axis and everything adds, giving the larger value.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\definecolor{neg}{HTML}{9C4A4A}
% ---- left: signed ----
\begin{scope}
  \draw[->] (-0.15,0) -- (4.4,0);
  \draw[->] (0,-1.2) -- (0,1.6);
  \fill[acc!16] (0,0) .. controls (0.6,1.3) and (1.4,1.3) .. (2.0,0) -- cycle;
  \fill[neg!18] (2.0,0) .. controls (2.6,-1.0) and (3.4,-1.0) .. (4.0,0) -- cycle;
  \draw[acc, very thick] (0,0) .. controls (0.6,1.3) and (1.4,1.3) .. (2.0,0)
      .. controls (2.6,-1.0) and (3.4,-1.0) .. (4.0,0);
  \node[acc, font=\footnotesize] at (1.0,0.55) {adds};
  \node[neg, font=\footnotesize] at (3.0,-0.5) {subtracts};
  \node[font=\footnotesize, anchor=north] at (2.0,-1.25) {$\int_a^b f$};
\end{scope}
% ---- right: unsigned ----
\begin{scope}[xshift=5.4cm]
  \draw[->] (-0.15,0) -- (4.4,0);
  \draw[->] (0,-1.2) -- (0,1.6);
  \fill[acc!16] (0,0) .. controls (0.6,1.3) and (1.4,1.3) .. (2.0,0) -- cycle;
  \fill[acc!16] (2.0,0) .. controls (2.6,1.0) and (3.4,1.0) .. (4.0,0) -- cycle;
  \draw[acc, very thick] (0,0) .. controls (0.6,1.3) and (1.4,1.3) .. (2.0,0)
      .. controls (2.6,1.0) and (3.4,1.0) .. (4.0,0);
  \node[acc, font=\footnotesize] at (1.0,0.55) {adds};
  \node[acc, font=\footnotesize] at (3.0,0.45) {adds};
  \node[font=\footnotesize, anchor=north] at (2.0,-1.25) {$\int_a^b |f|$};
\end{scope}
\end{tikzpicture}
$$

> **Worked example (Bounding an integral).** The triangle inequality and
> monotonicity bound an integral without evaluating it. Since $|\cos x| \le 1$,
> $$
> \left| \int_0^1 \frac{\cos x}{1 + x^2}\, \d x \right|
> \;\le\; \int_0^1 \frac{|\cos x|}{1 + x^2}\, \d x
> \;\le\; \int_0^1 \frac{1}{1 + x^2}\, \d x = \arctan 1 = \frac{\pi}{4}.
> $$
> The integral has magnitude at most $\tfrac{\pi}{4} \approx 0.785$, and is in fact
> positive because its integrand is.

## Products and the algebra of integrable functions

The integrable functions are closed under multiplication, not only addition, a
fact used whenever an integrand is a product.

> **Proposition (Products).** If $f, g \in \mathcal{R}[a,b]$, then $f^2$ and
> $fg$ are integrable.

> **Proof.** For $f^2$, the identity
> $f(x)^2 - f(y)^2 = (f(x) - f(y))(f(x) + f(y))$ with the bound $|f| \le A$ gives
> $U(P, f^2) - L(P, f^2) \le 2A\,\big(U(P,f) - L(P,f)\big)$, so the Cauchy
> criterion carries over. For the product, polarize:
> $fg = \tfrac14\big((f+g)^2 - (f-g)^2\big)$, and each square is integrable by the
> previous case.[^shk-prod]

Closure under products gives the integral form of the Cauchy–Schwarz inequality,
proved exactly as in a finite sum.

> **Proposition (Cauchy–Schwarz for integrals).** For $f, g \in \mathcal{R}[a,b]$,
> $$
> \left( \int_a^b f g \right)^2 \;\le\; \left( \int_a^b f^2 \right)\left( \int_a^b g^2 \right).
> $$

> **Proof.** The quadratic
> $t \mapsto \int_a^b (f + t g)^2 = t^2\int g^2 + 2t\int fg + \int f^2$ is
> nonnegative for all real $t$ (its integrand is a square), so its discriminant is
> nonpositive, giving the stated inequality.

## The mean value theorem for integrals

A continuous function attains, somewhere on the interval, exactly its average
value. Equivalently, the integral equals a rectangle whose height is a genuine
function value.

> **Theorem (Integral mean value theorem).** If $f : [a,b] \to \mathbb{R}$ is
> continuous, there exists $c \in [a,b]$ with
> $$
> \int_a^b f = f(c)\,(b-a),
> \qquad\text{equivalently}\qquad
> f(c) = \frac{1}{b-a}\int_a^b f.
> $$

> **Proof.** Let $m$ and $M$ be the min and max of $f$ on the compact interval.
> Monotonicity gives $m(b-a) \le \int_a^b f \le M(b-a)$, so the average
> $\tfrac{1}{b-a}\int_a^b f$ lies in $[m, M]$. The
> [intermediate value theorem](/real-analysis/continuity/evt-ivt) then supplies a
> $c$ where $f(c)$ equals that average.[^shk-imvt]

$$
% caption: The integral mean value theorem: the shaded area under $f$ equals the
% caption: area of a rectangle of the same width whose height $f(c)$ is an attained
% caption: value — the average height of $f$ over $[a,b]$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% rectangle of average height
\fill[acc!10] (0,0) rectangle (4.4,1.5);
\draw[black] (0,1.5) -- (4.4,1.5);
% curve crossing the average level twice
\draw[acc, very thick] (0,0.7) .. controls (1.0,2.2) and (1.8,2.1) .. (2.6,1.2)
    .. controls (3.3,0.6) and (3.9,0.9) .. (4.4,1.4);
\draw[->] (-0.15,0) -- (4.9,0);
\draw[->] (0,-0.15) -- (0,2.5);
\draw[black, dashed] (1.35,0) -- (1.35,1.5);
\draw (1.35,0.07) -- (1.35,-0.07) node[below, font=\footnotesize] {$c$};
\foreach \x/\lab in {0/a, 4.4/b}
  \draw (\x,0.07) -- (\x,-0.07) node[below, font=\footnotesize] {$\lab$};
\node[black, font=\footnotesize, anchor=east] at (0,1.5) {$f(c)$};
\end{tikzpicture}
$$

Two refinements matter. The first is a strict-positivity statement.

> **Proposition (Vanishing integral of a nonnegative continuous function).** If
> $f : [a,b] \to \mathbb{R}$ is continuous, $f \ge 0$, and $\int_a^b f = 0$, then
> $f \equiv 0$.

> **Proof.** If $f(x_0) > 0$ at some point, continuity gives an interval around
> $x_0$ of length $\ell$ on which $f \ge \tfrac{f(x_0)}{2}$; monotonicity then
> forces $\int_a^b f \ge \tfrac{f(x_0)}{2}\,\ell > 0$, a contradiction.

The second
refinement is a weighted mean value theorem: for continuous $f$ and nonnegative
integrable $w$, some $c \in [a,b]$ gives $\int_a^b f w = f(c)\int_a^b w$, since the
average of $f$ weighted by $w$ still lies between $\min f$ and $\max f$.

## Invariance under finite edits

Integration ignores what happens at finitely many points.

> **Proposition (Finite edits).** If $f \in \mathcal{R}[a,b]$ and $g$ agrees with
> $f$ except on a finite set, then $g \in \mathcal{R}[a,b]$ and
> $\int_a^b g = \int_a^b f$.

> **Proof.** Each altered point can be enclosed in a strip of arbitrarily small
> width whose contribution to any sum is bounded by (edit height) $\times$ (width)
> and hence vanishes.[^lebl-52]

This is why the value of an integrand at a jump (such as the $\tfrac12$ at the
discontinuity of a step function) never affects its integral.

| Property | Statement | Source of the proof |
| --- | --- | --- |
| Additivity | $\int_a^c f = \int_a^b f + \int_b^c f$ | partitions split at $b$ |
| Linearity | $\int(\alpha f + \beta g) = \alpha\int f + \beta\int g$ | sums scale and add |
| Monotonicity | $f \le g \Rightarrow \int f \le \int g$ | $m_i(f) \le m_i(g)$ |
| Triangle | $\lvert\int f\rvert \le \int \lvert f\rvert$ | $-\lvert f\rvert \le f \le \lvert f\rvert$ |
| Products | $fg \in \mathcal{R}[a,b]$ | polarization of squares |
| Mean value | $\int_a^b f = f(c)(b-a)$ | bounds plus IVT |

[^lebl-52]: **Lebl**, _Basic Analysis I_, §5.2 — additivity (Lemma 5.2.1, Prop. 5.2.2), linearity (Prop. 5.2.4), monotonicity (Prop. 5.2.6), and invariance under changing finitely many values (Prop. 5.2.10), together with the $\int_b^a = -\int_a^b$ conventions.
[^shk-lin]: **Shkoller**, _MAT125B Lecture Notes_, §1.8 — linearity of the Riemann integral (Thm. 1.22), reduced to the linearity $S_\delta(f + g) = S_\delta(f) + S_\delta(g)$ of Riemann sums.
[^shk-abs]: **Shkoller**, _MAT125B Lecture Notes_, §1.9 — integrability of $|f|$ and the bound $\lvert\int f\rvert \le \int|f|$ (Thm. 1.30), via $U_\delta(|f|) - L_\delta(|f|) \le U_\delta(f) - L_\delta(f)$.
[^shk-prod]: **Shkoller**, _MAT125B Lecture Notes_, §1.13 (Problems 1.2–1.3) — $f \in \mathcal{R}(a,b) \Rightarrow f^2 \in \mathcal{R}(a,b)$ and the polarization identity $(f-g)^2 - (f+g)^2 = -4fg$ giving $fg \in \mathcal{R}(a,b)$.
[^shk-imvt]: **Shkoller**, _MAT125B Lecture Notes_, §1.9 — the intermediate value (mean value) theorem for integrals (Thm. 1.28): a continuous $f$ attains its average value; cf. Lebl §5.2 Exercise 5.2.4.