CommeUnJeu · L1 PCSI
Integration on a segment
Chapter overview
This chapter establishes the integral \(\int_a^b f(t)\,dt\) of a function \(f\) on a segment \([a, b]\), valued in \(K = \mathbb{R}\) or \(\mathbb{C}\), and its main properties: linearity, positivity, the triangle inequality, the Chasles relation, the mean value, Riemann sums, the link between integral and primitive (fundamental theorem of calculus, integration by parts, change of variable), and finally the global Taylor formulas. The central object is a continuous function on a segment; the simplest case --- a step function, constant on each piece of a subdivision --- gives the integral its first, purely geometric meaning (« base \(\times\) height »), and the extension from step functions to continuous functions is admitted here. Building the integral fully is not our aim at this level; the point is to understand how it is approached and to use its properties fluently.
Convention on \(K\). Throughout the chapter, \(K\) denotes \(\mathbb{R}\) or \(\mathbb{C}\). Statements involving order (\(\le\), positivity, monotonicity) are real-valued and flagged explicitly.
Convention on \(K\). Throughout the chapter, \(K\) denotes \(\mathbb{R}\) or \(\mathbb{C}\). Statements involving order (\(\le\), positivity, monotonicity) are real-valued and flagged explicitly.
I
Uniform continuity
Why uniform continuity?
A continuous function \(f\) on an interval \(I\) satisfies: at each point \(y \in I\) and each tolerance \(\varepsilon > 0\), there is a radius \(\alpha_{y, \varepsilon} > 0\) such that \(|f(x) - f(y)| < \varepsilon\) as long as \(|x - y| < \alpha_{y, \varepsilon}\). The radius \(\alpha_{y, \varepsilon}\) depends on \(y\) --- it can shrink as \(y\) moves. Uniform continuity demands that one radius works for all \(y\) simultaneously. On a segment \([a, b]\), this stronger property holds for free (Heine's theorem), and this is precisely what powers the uniform approximation of a continuous function by step functions in the next section, Step and piecewise continuous functions.
Definition — Uniform continuity
Let \(f : I \to K\) be a function on an interval \(I\). We say that \(f\) is uniformly continuous on \(I\) if $$ \textcolor{colordef}{\forall \varepsilon > 0, \ \exists \alpha > 0, \ \forall x, y \in I, \quad |x - y| < \alpha \Rightarrow |f(x) - f(y)| < \varepsilon}. $$ The radius \(\alpha\) depends on \(\varepsilon\) but not on the points \(x, y\) --- the same \(\alpha\) works for every pair of points within distance \(\alpha\).
Figure --- uniform continuity
A schematic comparison: at a steep point, the required \(\alpha\) shrinks; uniform continuity requires one \(\alpha\) valid for every point of the interval. 
Example
The identity \(f(x) = x\) is uniformly continuous on \(\mathbb{R}\).
For every \(\varepsilon > 0\), take \(\alpha = \varepsilon\). Then \(|x - y| < \alpha \Rightarrow |f(x) - f(y)| = |x - y| < \alpha = \varepsilon\).
Example
The function \(f(x) = x^2\) is continuous on \(\mathbb{R}\) but not uniformly continuous on \(\mathbb{R}\).
For every \(\alpha > 0\), take \(x_n = n\) and \(y_n = n + \frac{\alpha}{2}\). Then \(|x_n - y_n| = \alpha/2 < \alpha\), but $$ |f(x_n) - f(y_n)| = \left| n^2 - \left(n + \frac{\alpha}{2}\right)^2 \right| = n \alpha + \frac{\alpha^2}{4} \xrightarrow[n \to +\infty]{} +\infty. $$ Hence no fixed \(\alpha\) can ensure \(|f(x) - f(y)| < 1\) for all \(x, y\) at distance \(< \alpha\). So \(f\) is not uniformly continuous on \(\mathbb{R}\).
Proposition — Lipschitz \(\Rightarrow\) uniformly continuous
If \(f : I \to K\) is \(K_0\)-Lipschitz on \(I\) for some \(K_0 > 0\) (i.e. \(|f(x) - f(y)| \le K_0 |x - y|\) for all \(x, y \in I\)), then \(f\) is uniformly continuous on \(I\).
Let \(\varepsilon > 0\). Set \(\alpha = \varepsilon / K_0\). Then for all \(x, y \in I\) with \(|x - y| < \alpha\): $$ |f(x) - f(y)| \le K_0 |x - y| < K_0 \alpha = \varepsilon. $$ Hence \(f\) is uniformly continuous on \(I\).
Theorem — Heine's theorem
Every function continuous on a segment \([a, b]\) is uniformly continuous on \([a, b]\): $$ \textcolor{colorprop}{f \in C([a, b], K) \Rightarrow f \text{ uniformly continuous on } [a, b]}. $$
Proof of Heine --- admitted
The proof is not required by the programme. It proceeds by contraposition + Bolzano-Weierstrass: if \(f\) is not uniformly continuous, build two sequences \((x_n), (y_n) \in [a, b]^{\mathbb{N}}\) with \(|x_n - y_n| \to 0\) but \(|f(x_n) - f(y_n)| \ge \varepsilon\), then extract a convergent subsequence with the same limit for both \(x_{\varphi(n)}\) and \(y_{\varphi(n)}\) to contradict continuity at the limit.
Method — Using Heine in practice
On a segment \([a, b]\), when given a continuous \(f\) and a tolerance \(\varepsilon > 0\): - Apply Heine to obtain a uniform radius \(\alpha > 0\) working at every point of \([a, b]\).
- This radius depends only on \(\varepsilon\) and on \(f\), not on the point: it can be used inside a subdivision or a partition argument without worrying about variation of \(\alpha\).
Skills to practice
- Establishing uniform continuity from the definition
- Disproving uniform continuity
- Using Heine's theorem in practice
II
Step and piecewise continuous functions
The building blocks
The construction of the integral is staged. Step functions --- piecewise constant functions on a subdivision --- have a geometric integral defined by « base \(\times\) height » summation. Piecewise continuous functions extend this to a richer class containing all continuous functions, and the integral will extend by uniform approximation. This section introduces both classes and their algebraic structure.
Convention. Unless otherwise stated, the construction on a segment \([a, b]\) assumes \(a < b\). The degenerate case \(a = b\) is handled later by the convention \(\int_a^a f = 0\) (Definition D3.4).
Convention. Unless otherwise stated, the construction on a segment \([a, b]\) assumes \(a < b\). The degenerate case \(a = b\) is handled later by the convention \(\int_a^a f = 0\) (Definition D3.4).
Definition — Subdivision of a segment
A subdivision of \([a, b]\) is a finite family \(\sigma = (x_0, x_1, \dots, x_p)\) of points of \([a, b]\) with \(a = x_0 < x_1 < \dots < x_p = b\). The real number $$ \textcolor{colordef}{\operatorname{pas}(\sigma) := \max_{0 \le k \le p-1} (x_{k+1} - x_k) > 0} $$ is called the mesh of \(\sigma\). Example
\(\sigma = (0, 1, 2, 3)\) is a subdivision of \([0, 3]\) with mesh \(\operatorname{pas}(\sigma) = 1\). The family \((0, 0.5, 1, 3)\) is another subdivision of \([0, 3]\), with mesh \(2\) (the largest gap is \(3 - 1 = 2\)). Definition — Step function
A function \(f : [a, b] \to K\) is a step function (« en escalier ») if there is a subdivision \(\sigma = (x_0, \dots, x_p)\) of \([a, b]\), called a subdivision adapted to \(f\), such that $$ \textcolor{colordef}{f \text{ is constant on each open subinterval } ]x_k, x_{k+1}[ \text{ for every } k \in \{0, \dots, p-1\}}. $$ The values of \(f\) at the subdivision points \(x_0, x_1, \dots, x_p\) are not constrained by this definition. Example
Any constant function \(f \equiv c\) on \([a, b]\) is a step function: take the trivial subdivision \((a, b)\) and \(f\) is constant on the single open subinterval \(\,]a, b[\). The floor function \(\lfloor \cdot \rfloor\) on \([0, 3]\) is a step function with adapted subdivision \((0, 1, 2, 3)\) and values \(0, 1, 2\) on the three open subintervals.
Figure --- step function
A step function on \([a, b]\) with subdivision \((x_0, x_1, x_2, x_3, x_4)\) and constant values \(y_0, y_1, y_2, y_3\) on the open subintervals. 
Proposition — Stability of step functions
Let \(f, g : [a, b] \to K\) be step functions and \(\lambda, \mu \in K\). The union of a subdivision adapted to \(f\) and a subdivision adapted to \(g\) is adapted to both. Moreover, \(|f|\), \(\operatorname{Re}(f)\), \(\operatorname{Im}(f)\), \(\lambda f + \mu g\), and \(f g\) are step functions on \([a, b]\). In particular, the set of step functions on \([a, b]\) valued in \(K\) is a \textcolor{colorprop}{sub-algebra of \(K^{[a, b]}\)}.
On the union of two adapted subdivisions, both \(f\) and \(g\) are constant on each open subinterval; hence any algebraic combination of constant values is constant on that subinterval. The same holds for \(|f|\), \(\operatorname{Re}\), \(\operatorname{Im}\) (applied to constant values). The sub-algebra claim follows.
Definition — Piecewise continuous function
A function \(f : [a, b] \to K\) is piecewise continuous (« continue par morceaux ») on \([a, b]\) if there is a subdivision \((x_0, \dots, x_p)\) of \([a, b]\), called adapted to \(f\), such that: - \(f\) is continuous on each open subinterval \(\,]x_k, x_{k+1}[\);
- \(f\) admits a finite continuous extension at \(x_k^+\) and at \(x_{k+1}^-\) on each open subinterval.
Figure --- piecewise continuous function
A piecewise continuous function with jump discontinuities at the subdivision points \(x_1, x_2\). 
Example
The floor function \(\lfloor \cdot \rfloor\) is piecewise continuous on \([0, 3]\) (it is in fact a step function, with subdivision \((0, 1, 2, 3)\)). Example
The function \(f(x) = 1/x\) on \(\,]0, 1]\) extended by any value at \(0\) is not piecewise continuous on \([0, 1]\): it is continuous on \(\,]0, 1]\) but does not admit a finite continuous extension at \(0^+\) (since \(1/x \to +\infty\)). Proposition — Piecewise continuous functions form a sub-algebra
The set \(\mathcal{CM}([a, b], K)\) is a sub-algebra of \(K^{[a, b]}\): \(|f|\), \(\operatorname{Re}(f)\), \(\operatorname{Im}(f)\), \(\lambda f + \mu g\), and \(f g\) are piecewise continuous on \([a, b]\) whenever \(f, g\) are. It contains all step functions on \([a, b]\) and all continuous functions on \([a, b]\).
On the union of two adapted subdivisions, the continuous extensions of \(f\) and \(g\) on each open subinterval are well-defined; any algebraic combination of these continuous extensions is itself continuously extendable. The step function case (everywhere constant on each open subinterval) trivially fits, and a globally continuous function uses the trivial subdivision \((a, b)\).
Proposition — Piecewise continuous \(\Rightarrow\) bounded
Every \(f \in \mathcal{CM}([a, b], K)\) is \textcolor{colorprop}{bounded} on \([a, b]\); in particular, its infinity norm \(\|f\|_\infty = \sup_{x \in [a, b]} |f(x)|\) is finite. Note that \(|f|\) need not attain its supremum (the supremum can be attained only as a one-sided limit at a subdivision point).
Let \((x_0, \dots, x_p)\) be an adapted subdivision. On each open subinterval \(\,]x_k, x_{k+1}[\), the continuous extension \(\tilde f_k\) of \(f\) is continuous on the closed subinterval \([x_k, x_{k+1}]\); by the extreme value theorem (from Limits and continuity), \(|\tilde f_k|\) is bounded by some \(M_k\). The maximum of \(M_0, \dots, M_{p-1}\) together with the values \(|f(x_0)|, \dots, |f(x_p)|\) bounds \(|f|\) on \([a, b]\).
Theorem — Uniform approximation by step functions
Every \(f \in \mathcal{CM}([a, b], K)\) is the \textcolor{colorprop}{uniform limit} of a sequence of step functions on \([a, b]\). In particular, for every \(\varepsilon > 0\), there exists a step function \(\varphi\) such that \(\|f - \varphi\|_\infty < \varepsilon\).
Two cases, by an adapted subdivision argument.
- Case 1 --- \(f\) continuous on \([a, b]\). For every \(n \in \mathbb{N}^*\), set \(x_{n, k} = a + k (b - a)/n\) for \(k \in \{0, \dots, n\}\), and let \(\varphi_n\) be the step function defined by $$ \varphi_n(x) = f(x_{n, k}) \text{ for } x \in [x_{n, k}, x_{n, k+1}[, \quad \varphi_n(b) = f(b). $$ By Heine's theorem (T1.1), \(f\) is uniformly continuous on \([a, b]\): for \(\varepsilon > 0\) given, there is \(\alpha > 0\) with \(|f(x) - f(y)| < \varepsilon\) whenever \(|x - y| < \alpha\). Pick \(N \in \mathbb{N}^*\) such that \((b - a)/N < \alpha\). For \(n \ge N\) and \(x \in [a, b[\), \(x\) lies in some \([x_{n, k}, x_{n, k+1}[\) with \(|x - x_{n, k}| < (b - a)/n \le (b - a)/N < \alpha\), hence \(|\varphi_n(x) - f(x)| = |f(x_{n, k}) - f(x)| < \varepsilon\). Thus \(\|\varphi_n - f\|_\infty \le \varepsilon\), so \(\varphi_n \to f\) uniformly.
- Case 2 --- \(f\) piecewise continuous on \([a, b]\). Let \((x_0, \dots, x_p)\) be an adapted subdivision. On each \([x_k, x_{k+1}]\), the continuous extension \(\tilde f_k\) of \(f\) on the open part is continuous on a closed segment, so by Case 1 there is a sequence \((\varphi_{k, n})_n\) of step functions on \([x_k, x_{k+1}]\) uniformly approximating \(\tilde f_k\). Patch them: define \(\varphi_n(x) = \varphi_{k, n}(x)\) for \(x \in \,]x_k, x_{k+1}[\) and \(\varphi_n(x_k) = f(x_k)\) at each subdivision point. Then \(\varphi_n\) is a step function on \([a, b]\), and \(\|\varphi_n - f\|_\infty = \max_k \|\varphi_{k, n} - \tilde f_k\|_{\infty, \,]x_k, x_{k+1}[} \to 0\).
Figure --- uniform approximation
A continuous \(f\) on \([a, b]\) and a step function \(\varphi_n\) taking the value \(f(x_{n, k})\) on each \([x_{n, k}, x_{n, k+1}[\) of a uniform partition: \(\|\varphi_n - f\|_\infty \to 0\) as \(n \to +\infty\) by Heine. 
Skills to practice
- Identifying step and piecewise continuous functions
- Constructing subdivisions adapted to a function
- Using the sub-algebra structure of piecewise continuous functions
- Approximating a continuous function by step functions
III
Integral of a piecewise continuous function
Strategy
We first define the integral \(\int_{[a, b]} f\) for a step function \(f\) as the geometric sum « base \(\times\) height » over its plateaus. This definition is independent of the adapted subdivision (refinement argument). For a general piecewise continuous \(f\), we approach \(f\) uniformly by step functions \(\varphi_n\) (Theorem T2.1) and define \(\int_{[a, b]} f\) as the limit of \(\int_{[a, b]} \varphi_n\). The bulk of the work is to verify that this limit exists and does not depend on the choice of \((\varphi_n)\). The standard properties (linearity, triangle inequality, Chasles) then follow by passing to the limit from the step-function case.
Definition — Integral of a step function
Let \(f : [a, b] \to K\) be a step function with adapted subdivision \((x_0, \dots, x_p)\), and let \(y_k \in K\) be the constant value of \(f\) on \(\,]x_k, x_{k+1}[\) for \(k \in \{0, \dots, p - 1\}\). The number $$ \textcolor{colordef}{\sum_{k=0}^{p - 1} y_k (x_{k+1} - x_k)} $$ depends only on \(f\) and not on the choice of the adapted subdivision. It is called the integral of \(f\) on \([a, b]\), and is denoted $$ \textcolor{colordef}{\int_{[a, b]} f \quad \text{or} \quad \int_{[a, b]} f(t)\,dt}. $$ The values of \(f\) at the subdivision points \(x_0, x_1, \dots, x_p\) do not appear in this sum, hence do not affect the integral. Example
Compute \(\int_{[0, 3]} f\) for the step function \(f\) defined by \(f = 2\) on \(\,]0, 1[\), \(f = -1\) on \(\,]1, 3[\), with arbitrary values at \(0, 1, 3\).
The adapted subdivision is \((0, 1, 3)\). By D3.1: $$ \int_{[0, 3]} f = 2 \cdot (1 - 0) + (-1) \cdot (3 - 1) = 2 - 2 = 0. $$ The values of \(f\) at \(0\), \(1\), \(3\) do not enter the computation.
Proposition — Linearity and triangle inequality for step functions
Let \(f, g : [a, b] \to K\) be step functions and \(\lambda, \mu \in K\). Then - (i) \textcolor{colorprop}{Linearity}: \(\int_{[a, b]} (\lambda f + \mu g) = \lambda \int_{[a, b]} f + \mu \int_{[a, b]} g\).
- (ii) \textcolor{colorprop}{Triangle inequality}: \(\left| \int_{[a, b]} f \right| \le \int_{[a, b]} |f| \le (b - a) \|f\|_\infty\).
On a common adapted subdivision (union of subdivisions adapted to \(f\) and to \(g\)), denote the constant values \(y_k\) for \(f\) and \(z_k\) for \(g\) on \(\,]x_k, x_{k+1}[\).
- (i) The constant value of \(\lambda f + \mu g\) on \(\,]x_k, x_{k+1}[\) is \(\lambda y_k + \mu z_k\), so $$ \int_{[a, b]} (\lambda f + \mu g) = \sum_k (\lambda y_k + \mu z_k)(x_{k+1} - x_k) = \lambda \sum_k y_k(x_{k+1} - x_k) + \mu \sum_k z_k(x_{k+1} - x_k). $$
- (ii) The triangle inequality on sums gives \(|\int f| = |\sum_k y_k (x_{k+1} - x_k)| \le \sum_k |y_k|(x_{k+1} - x_k) = \int |f|\). The second inequality follows from \(|y_k| \le \|f\|_\infty\) and \(\sum (x_{k+1} - x_k) = b - a\).
Definition — Integral of a piecewise continuous function
Let \(f \in \mathcal{CM}([a, b], K)\) and let \((\varphi_n)\) be a sequence of step functions on \([a, b]\) converging uniformly to \(f\) (such a sequence exists by Theorem T2.1). Then the sequence \(\left( \int_{[a, b]} \varphi_n \right)_{n \in \mathbb{N}}\) converges in \(K\), and its limit does not depend on the choice of \((\varphi_n)\). This common limit is the integral of \(f\) on \([a, b]\), denoted $$ \textcolor{colordef}{\int_{[a, b]} f := \lim_{n \to +\infty} \int_{[a, b]} \varphi_n}. $$ For a step function, this definition coincides with the previous one (D3.1). Proposition — Properties of the integral
Let \(f, g \in \mathcal{CM}([a, b], K)\) with \(a \le b\) and \(\lambda, \mu \in K\). - (i) \textcolor{colorprop}{Linearity}: \(\int_{[a, b]} (\lambda f + \mu g) = \lambda \int_{[a, b]} f + \mu \int_{[a, b]} g\).
- (ii) \textcolor{colorprop}{Triangle inequality}: \(\left| \int_{[a, b]} f \right| \le \int_{[a, b]} |f| \le (b - a) \|f\|_\infty\).
- (iii) \textcolor{colorprop}{Chasles}: For every \(c \in [a, b]\), \(\int_{[a, b]} f = \int_{[a, c]} f + \int_{[c, b]} f\).
- (iv) \textcolor{colorprop}{Real/imaginary part}: \(\int_{[a, b]} f = \int_{[a, b]} \operatorname{Re}(f) + i \int_{[a, b]} \operatorname{Im}(f)\). In particular, if \(f\) is real-valued, \(\int_{[a, b]} f \in \mathbb{R}\).
- (v) \textcolor{colorprop}{Modification on a finite set}: If \(f\) and \(g\) agree on \([a, b]\) except on a finite set, then \(\int_{[a, b]} f = \int_{[a, b]} g\).
For each property, we prove the result first for step functions, then pass to the limit using a sequence \((\varphi_n)\) uniformly approaching \(f\) (and \((\psi_n)\) for \(g\)). Let \((\varphi_n)\), \((\psi_n)\) be step functions with \(\varphi_n \to f\), \(\psi_n \to g\) uniformly.
- (i) For step functions, linearity is P3.1(i). For piecewise continuous: \(\lambda \varphi_n + \mu \psi_n\) is step and \(\to \lambda f + \mu g\) uniformly, so \(\int (\lambda \varphi_n + \mu \psi_n) = \lambda \int \varphi_n + \mu \int \psi_n\) tends to \(\lambda \int f + \mu \int g\), and by definition of \(\int (\lambda f + \mu g)\) also tends to \(\int (\lambda f + \mu g)\).
- (ii) For step functions, this is P3.1(ii). For piecewise continuous: \(|\int \varphi_n| \le \int |\varphi_n| \le (b - a) \|\varphi_n\|_\infty\). Pass to the limit: \(|\int f| \le \int |f|\) (since \(|\varphi_n|\) is step and \(\to |f|\) uniformly, by reverse triangle), and \(\int |f| \le (b - a) \|f\|_\infty\) since \(\|\varphi_n\|_\infty \to \|f\|_\infty\).
- (iii) For step functions, \(f\) admits an adapted subdivision containing \(c\) (refine if needed), and the Chasles relation is a direct splitting of the sum at index \(q\) with \(x_q = c\). For piecewise continuous: restrict \(\varphi_n\) to \([a, c]\) and \([c, b]\); each restriction is step on its segment and uniformly approximates \(f|_{[a, c]}\) and \(f|_{[c, b]}\) respectively. Pass to the limit.
- (iv) For step functions, \(\operatorname{Re}\) and \(\operatorname{Im}\) commute with the sum. For piecewise continuous: \(\operatorname{Re}(\varphi_n) \to \operatorname{Re}(f)\) uniformly (since \(|\operatorname{Re}(z)| \le |z|\)), same for \(\operatorname{Im}\).
- (v) If \(f, g \in \mathcal{CM}([a, b], K)\) agree on \([a, b] \setminus \{a_1, \dots, a_q\}\) for a finite set \(\{a_1, \dots, a_q\}\), then \(g - f\) is step on \([a, b]\) with value \(0\) on the open complement of the \(a_j\)'s, so \(\int (g - f) = 0\) (the sum has only zero contributions). By linearity, \(\int g = \int f\).
Proposition — Real-valued case
Let \(f, g \in \mathcal{CM}([a, b], \mathbb{R})\) with \(a \le b\). - (i) \textcolor{colorprop}{Positivity}: \(f \ge 0 \Rightarrow \int_{[a, b]} f \ge 0\).
- (ii) \textcolor{colorprop}{Monotonicity}: \(f \le g \Rightarrow \int_{[a, b]} f \le \int_{[a, b]} g\).
- (iii) \textcolor{colorprop}{Strict positivity}: If \(f \ge 0\) on \([a, b]\), \(f\) is continuous at some \(x_0 \in [a, b]\) with \(f(x_0) > 0\), and \(a < b\), then \(\int_{[a, b]} f > 0\).
- (iv) \textcolor{colorprop}{Nullity for continuous nonnegative functions}: If \(f \in C([a, b], \mathbb{R}_+)\), \(a < b\), and \(\int_{[a, b]} f = 0\), then \(f \equiv 0\) on \([a, b]\).
- (i) For \(f \ge 0\) step, the sum \(\sum y_k (x_{k+1} - x_k)\) has nonnegative terms. For piecewise continuous \(f \ge 0\), approximate by step \(\varphi_n\) from the construction of T2.1: the values of \(\varphi_n\) are values of \(f\) or of its continuous one-sided extensions, hence nonnegative since \(f \ge 0\). Therefore \(\int \varphi_n \ge 0\) for every \(n\), and the limit \(\int f \ge 0\).
- (ii) Apply (i) to \(g - f \ge 0\), then use linearity.
- (iii) Minoration argument. Since \(f\) is continuous at \(x_0\) and \(f(x_0) > 0\), there is \(\eta > 0\) and a closed interval \(J = [x_0 - \delta, x_0 + \delta] \cap [a, b]\) around \(x_0\) with \(\delta > 0\) and \(f \ge \eta\) on \(J\). Let \(\varphi = \eta \cdot 1_J\) (a step function on \([a, b]\)). Then \(0 \le \varphi \le f\) on \([a, b]\), so by monotonicity (ii): $$ \int_{[a, b]} f \ge \int_{[a, b]} \varphi = \eta \cdot |J| > 0. $$
- (iv) Apply (iii) by contraposition: if \(f\) is not identically \(0\), there is some \(x_0 \in [a, b]\) with \(f(x_0) > 0\); since \(f\) is continuous on \([a, b]\), (iii) applies and \(\int f > 0\), contradicting \(\int f = 0\).
Definition — Mean value
The mean value of \(f \in \mathcal{CM}([a, b], K)\) on \([a, b]\) (with \(a < b\)) is $$ \textcolor{colordef}{\bar f := \frac{1}{b - a} \int_{[a, b]} f}. $$ Proposition — Bounds for the mean value
Let \(f \in \mathcal{CM}([a, b], \mathbb{R})\) with \(a < b\). If \(m \le f(x) \le M\) for every \(x \in [a, b]\), then $$ \textcolor{colorprop}{m (b - a) \le \int_a^b f(t)\,dt \le M (b - a)}, \qquad \text{hence} \qquad m \le \bar f \le M. $$ If moreover \(f\) is continuous on \([a, b]\), then there exists \(c \in [a, b]\) such that \(\bar f = f(c)\).
The inequality \(m \le f \le M\) on \([a, b]\) gives, by monotonicity of the integral (P3.3(ii)): $$ \int_a^b m \, dt \le \int_a^b f(t)\,dt \le \int_a^b M\,dt, \qquad \text{i.e.} \qquad m(b - a) \le \int_a^b f \le M(b - a). $$ Dividing by \(b - a > 0\) gives \(m \le \bar f \le M\).
If \(f\) is continuous on \([a, b]\), the extreme value theorem (from Limits and continuity) gives \(m_0, M_0 \in [a, b]\) with \(f(m_0) = \min f\) and \(f(M_0) = \max f\). Then \(f(m_0) \le \bar f \le f(M_0)\), and by the intermediate value theorem applied to \(f\) between \(m_0\) and \(M_0\), there exists \(c\) between \(m_0\) and \(M_0\) (hence in \([a, b]\)) with \(f(c) = \bar f\).
If \(f\) is continuous on \([a, b]\), the extreme value theorem (from Limits and continuity) gives \(m_0, M_0 \in [a, b]\) with \(f(m_0) = \min f\) and \(f(M_0) = \max f\). Then \(f(m_0) \le \bar f \le f(M_0)\), and by the intermediate value theorem applied to \(f\) between \(m_0\) and \(M_0\), there exists \(c\) between \(m_0\) and \(M_0\) (hence in \([a, b]\)) with \(f(c) = \bar f\).
Definition — Extension of the notation to \(b \le a\)
For \(f \in \mathcal{CM}(I, K)\) defined on a wider interval \(I\) and \(a, b \in I\) with \(b < a\), we extend the notation by $$ \textcolor{colordef}{\int_a^b f(t)\,dt := -\int_b^a f(t)\,dt}. $$ The case \(b = a\) gives \(\int_a^a f = 0\). Linearity and Chasles still hold for the extended notation; in (ii) of P3.2, the bound becomes \(\left| \int_a^b f \right| \le |b - a| \|f\|_\infty\) (with \(|b - a|\) in place of \(b - a\)). Proposition — Integrals of even / odd / periodic continuous functions
Let \(a \ge 0\) and \(f \in C([-a, a], K)\). - If \(f\) is \textcolor{colorprop}{even}: \(\int_{-a}^a f(x)\,dx = 2 \int_0^a f(x)\,dx\).
- If \(f\) is \textcolor{colorprop}{odd}: \(\int_{-a}^a f(x)\,dx = 0\).
Recalled from Primitives. These properties were stated and used at the lycée-style level in the previous chapter. They are restated here for reference; their proof uses the change-of-variable theorem T5.3 (in the section Link with primitives), which is stated for continuous functions --- hence the continuity hypothesis here.
Method — Compute an integral using symmetry
For an integral \(\int_a^b f\): - Check whether the integration domain is symmetric (\([-A, A]\)) and \(f\) is even or odd: use P3.5 (parity).
- Check whether \(f\) is \(T\)-periodic and the domain has length \(T\): shift to \([0, T]\) via P3.5 (periodicity).
- Otherwise, split via Chasles into subintervals where \(f\) has simpler form.
Example
Compute \(\int_{-\pi}^\pi \sin t \, dt\).
The integrand \(\sin\) is odd and the domain \([-\pi, \pi]\) is symmetric, so \(\int_{-\pi}^\pi \sin t \, dt = 0\) by P3.5 (odd case).
Example
Compute \(\int_0^{2\pi} \cos^2 t \, dt\) using periodicity and the half-angle formula.
The function \(\cos^2\) is \(\pi\)-periodic (since \(\cos^2(t + \pi) = (-\cos t)^2 = \cos^2 t\)), so $$ \int_0^{2\pi} \cos^2 t \, dt = 2 \int_0^\pi \cos^2 t \, dt. $$ Using \(\cos^2 t = (1 + \cos 2t)/2\): $$ \begin{aligned} \int_0^\pi \cos^2 t \, dt &= \int_0^\pi \frac{1 + \cos 2t}{2} \, dt && \text{(half-angle formula)} \\
&= \frac{\pi}{2} + \frac{1}{2} \int_0^\pi \cos 2t \, dt && \text{(linearity)} \\
&= \frac{\pi}{2} + 0 = \frac{\pi}{2} && \text{(since \(\int_0^\pi \cos 2t \, dt = [\sin 2t / 2]_0^\pi = 0\)).} \end{aligned} $$ Hence \(\int_0^{2\pi} \cos^2 t \, dt = \pi\).
Skills to practice
- Computing integrals using parity\(\virgule\) periodicity\(\virgule\) Chasles
- Establishing inequalities via positivity and monotonicity
- Computing the mean value of a piecewise continuous function
IV
Riemann sums
Discrete \(\to\) continuous
The Riemann sums approximate \(\int_a^b f\) by a left- (or right-) endpoint rectangular sum on a uniform partition of \([a, b]\) in \(n\) pieces. Geometrically, each rectangle has base \((b - a)/n\) and height \(f(a + k(b - a)/n)\). As \(n \to +\infty\), the sums converge to the integral. The programme requires the proof of this limit only in the Lipschitz case; the general piecewise-continuous case is admitted.
Theorem — Riemann sums
For every \(f \in \mathcal{CM}([a, b], K)\) (with \(a < b\)): $$ \textcolor{colorprop}{\frac{b - a}{n} \sum_{k=0}^{n - 1} f\!\left( a + k \frac{b - a}{n} \right) \xrightarrow[n \to +\infty]{} \int_a^b f(t)\,dt} $$ and the right-endpoint variant \(\frac{b - a}{n} \sum_{k=1}^n f(a + k (b - a)/n)\) has the same limit. If moreover \(f\) is \(K_0\)-Lipschitz on \([a, b]\), the error is bounded by $$ \left| \int_a^b f - \frac{b - a}{n} \sum_{k=0}^{n - 1} f\!\left(a + k \frac{b - a}{n}\right) \right| \le \frac{K_0 (b - a)^2}{2 n}. $$
Set \(x_{n, k} = a + k (b - a)/n\) for \(k \in \{0, \dots, n\}\). The left-endpoint Riemann sum equals the integral of the step function \(\varphi_n\) defined by \(\varphi_n(x) = f(x_{n, k})\) on \([x_{n, k}, x_{n, k+1}[\) and \(\varphi_n(b) = f(b)\): $$ \int_{[a, b]} \varphi_n = \sum_{k=0}^{n - 1} f(x_{n, k}) \cdot \frac{b - a}{n}. $$ Compute the error using \(f\) Lipschitz: $$ \begin{aligned} \left| \int_{[a, b]} f - \int_{[a, b]} \varphi_n \right| &= \left| \int_{[a, b]} (f - \varphi_n) \right| && \text{(linearity)} \\
&\le \sum_{k=0}^{n - 1} \int_{x_{n, k}}^{x_{n, k+1}} |f(x) - f(x_{n, k})| \, dx && \text{(triangle inequality + Chasles)} \\
&\le \sum_{k=0}^{n - 1} \int_{x_{n, k}}^{x_{n, k+1}} K_0 (x - x_{n, k}) \, dx && \text{(Lipschitz bound)} \\
&= \sum_{k=0}^{n - 1} K_0 \cdot \frac{(b - a)^2}{2 n^2} && \text{(integrate \(K_0 (x - x_{n, k})\) on \([x_{n, k}, x_{n, k+1}]\))} \\
&= \frac{K_0 (b - a)^2}{2 n}. \end{aligned} $$ This tends to \(0\) as \(n \to +\infty\). The right-endpoint variant is proved in the same way, using the step function equal to \(f(x_{n, k+1})\) on \([x_{n, k}, x_{n, k+1}[\).
General piecewise continuous case --- admitted
The general piecewise-continuous case is admitted here, in accordance with the programme. A complete proof can be obtained by combining two facts: first, the result is checked directly for step functions; second, every piecewise continuous function is uniformly approximated by step functions, and the Riemann sums are stable under a uniform error.
Figure --- Riemann sum (left endpoints)
Left-endpoint Riemann sum on a uniform partition: each rectangle has base \((b - a)/n\) and height \(f(a + k(b - a)/n)\). 
Method — Recognize a sum as a Riemann sum
To compute the limit of a sum \(\sum_{k} u_{n, k}\) as \(n \to +\infty\): - Rewrite as \(\frac{1}{n} \sum_{k=1}^n f(k/n)\) for a suitable \(f : [0, 1] \to K\), when possible.
- The limit is then \(\int_0^1 f(x)\,dx\) by Theorem T4.1 on the interval \([0, 1]\).
- More generally, \(\frac{b - a}{n} \sum_{k=0}^{n - 1} f(a + k(b-a)/n) \to \int_a^b f(x)\,dx\).
Example
Compute \(\lim_{n \to +\infty} \sum_{k=1}^n \frac{1}{n + k}\).
Factor out \(1/n\): $$ \begin{aligned} \sum_{k=1}^n \frac{1}{n + k} &= \sum_{k=1}^n \frac{1}{n (1 + k/n)} \\
&= \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + k/n} && \text{(factor \(1/n\)).} \end{aligned} $$ This is the right-endpoint Riemann sum of \(f(x) = 1/(1 + x)\) on \([0, 1]\). By Theorem T4.1: $$ \lim_{n \to +\infty} \sum_{k=1}^n \frac{1}{n + k} = \int_0^1 \frac{dx}{1 + x} = [\ln(1 + x)]_0^1 = \ln 2. $$
Example
Compute \(\lim_{n \to +\infty} \frac{1}{n} \sum_{k=1}^n \sin\!\left( \frac{k \pi}{n} \right)\).
Recognize the right-endpoint Riemann sum of \(f(x) = \sin(\pi x)\) on \([0, 1]\): $$ \frac{1}{n} \sum_{k=1}^n \sin\!\left(\frac{k \pi}{n}\right) = \frac{1}{n} \sum_{k=1}^n f(k/n) \xrightarrow[n \to +\infty]{} \int_0^1 \sin(\pi x) \, dx = \left[ -\frac{\cos(\pi x)}{\pi} \right]_0^1 = \frac{2}{\pi}. $$
Skills to practice
- Recognizing a sum as a Riemann sum
- Computing limits by integral identification
- Using Riemann sums to bound a sequence
V
Link with primitives
The fundamental bridge
The Fundamental Theorem of Calculus (FTC) bridges two notions that look unrelated at first sight: integration (algebraic area under a curve) and differentiation (slope of a tangent). For a continuous \(f\) on an interval \(I\), the function \(F(x) = \int_a^x f(t)\,dt\) is differentiable with \(F'(x) = f(x)\) --- every continuous function has a primitive, and the integral of \(f\) between \(a\) and \(b\) is the difference of any primitive at the two endpoints. The FTC also powers two practical techniques --- integration by parts and change of variable --- proved here at the rigorous theorem level (informal versions were introduced in Primitives).
Definition — Primitive
Let \(f : I \to K\) be a function on an interval \(I\). A primitive of \(f\) on \(I\) is a function \(F : I \to K\) that is differentiable on \(I\) with \(F'(x) = f(x)\) for every \(x \in I\). Proposition — Primitives are unique up to an additive constant
If \(F_1, F_2 : I \to K\) are two primitives of the same function \(f : I \to K\) on an interval \(I\), then \(F_1 - F_2\) is a constant on \(I\). In particular, primitives of \(f\) are exactly the functions \(F + \lambda\) for \(\lambda \in K\), where \(F\) is any one primitive.
Set \(G = F_1 - F_2\). Then \(G' = F_1' - F_2' = f - f = 0\) on \(I\). For \(K = \mathbb{C}\), argue componentwise: \((\operatorname{Re} G)' = \operatorname{Re}(G') = 0\) and \((\operatorname{Im} G)' = \operatorname{Im}(G') = 0\), both real-valued differentiable on an interval. By the real mean value theorem applied separately to \(\operatorname{Re} G\) and \(\operatorname{Im} G\) (from Differentiability, P5.1), each is constant, so \(G\) is constant. For \(K = \mathbb{R}\), this reduces to the single real argument.
Theorem — Fundamental theorem of calculus
Let \(f : I \to K\) be continuous on an interval \(I\) and \(a \in I\). Then: - (i) The function \(F : x \mapsto \int_a^x f(t)\,dt\) is a \textcolor{colorprop}{primitive of \(f\)} on \(I\).
- (ii) For every primitive \(G\) of \(f\) on \(I\) and every \(b \in I\): \(\int_a^b f(t)\,dt = G(b) - G(a)\).
- (i) Fix \(x \in I\) and show \(F\) is differentiable at \(x\) with \(F'(x) = f(x)\). Let \(\varepsilon > 0\). By continuity of \(f\) at \(x\), there is \(\alpha > 0\) such that \(|f(t) - f(x)| < \varepsilon\) whenever \(t \in I\) and \(|t - x| < \alpha\). For \(h \in \mathbb{R}^*\) with \(x + h \in I\) and \(|h| < \alpha\), using Chasles: $$ F(x + h) - F(x) = \int_x^{x + h} f(t)\,dt, $$ and so (assuming WLOG \(h > 0\); the \(h < 0\) case is symmetric): $$ \begin{aligned} \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| && \text{(write \(f(x) = \frac{1}{h}\int_x^{x+h} f(x)\,dt\))} \\ &\le \frac{1}{h} \int_x^{x + h} |f(t) - f(x)|\,dt && \text{(triangle inequality)} \\ &\le \frac{1}{h} \int_x^{x + h} \varepsilon\,dt = \varepsilon && \text{(by continuity bound on \([x, x+h]\)).} \end{aligned} $$ Hence \((F(x + h) - F(x))/h \to f(x)\) as \(h \to 0\), so \(F\) is differentiable at \(x\) with \(F'(x) = f(x)\).
- (ii) From (i), \(F\) is a primitive of \(f\). By P5.1 (uniqueness up to an additive constant), \(G = F + \lambda\) for some \(\lambda \in K\). Then \(G(b) - G(a) = (F(b) + \lambda) - (F(a) + \lambda) = F(b) - F(a) = \int_a^b f - \int_a^a f = \int_a^b f\).
Method — Differentiate an integral with a moving upper bound
For \(f\) continuous on an interval \(I\) and \(u : J \to I\) differentiable on an interval \(J\), define \(H(x) = \int_a^{u(x)} f(t)\,dt\) on \(J\). Then \(H\) is differentiable on \(J\) with $$ \textcolor{colorprop}{H'(x) = f(u(x)) \cdot u'(x)}. $$ Proof: \(H = F \circ u\) with \(F(y) = \int_a^y f\), and apply the chain rule.Caveat. The variable \(x\) must appear only in the bounds; if \(x\) appears inside the integrand too, the rule does not apply directly --- separate the bound dependency from the integrand dependency first.
Example
Compute the derivative of \(H(x) = \int_0^{x^2} e^{-t^2}\,dt\) on \(\mathbb{R}\).
The integrand \(f(t) = e^{-t^2}\) is continuous on \(\mathbb{R}\), and \(u(x) = x^2\) is \(C^\infty\) on \(\mathbb{R}\). By the moving-bound rule: $$ H'(x) = f(u(x)) \cdot u'(x) = e^{-(x^2)^2} \cdot 2x = 2x \, e^{-x^4}. $$
Theorem — Integration by parts (IBP)
Let \(u, v \in C^1(I, K)\) and \(a, b \in I\). Then $$ \textcolor{colorprop}{\int_a^b u'(t) v(t)\,dt = \left[ u(t) v(t) \right]_a^b - \int_a^b u(t) v'(t)\,dt}. $$
Since \(u, v \in C^1(I, K)\), the product \(uv\) is \(C^1\) with \((uv)' = u'v + uv'\). The function \((uv)'\) is continuous, so by FTC (ii): $$ \begin{aligned} [u(t) v(t)]_a^b &= \int_a^b (uv)'(t)\,dt && \text{(FTC)} \\
&= \int_a^b u'(t) v(t)\,dt + \int_a^b u(t) v'(t)\,dt && \text{(linearity + product rule).} \end{aligned} $$ Rearranging gives the IBP formula.
Method — Apply integration by parts
For an integral \(\int_a^b u'(t) v(t)\,dt\): - Identify \(u'\) (the factor easy to primitivise) and \(v\) (the factor easy to differentiate).
- Compute \(u\) (primitive of \(u'\)) and \(v'\).
- Apply the formula and integrate the second piece \(\int u v'\) --- it should be simpler than the original.
Example
Compute \(\int_0^1 t e^t\,dt\) by IBP.
Set \(u'(t) = e^t\), \(v(t) = t\). Then \(u(t) = e^t\), \(v'(t) = 1\). Apply IBP: $$ \begin{aligned} \int_0^1 t e^t\,dt &= [t e^t]_0^1 - \int_0^1 e^t\,dt && \text{(IBP, T5.2)} \\
&= (1 \cdot e^1 - 0) - [e^t]_0^1 && \text{(evaluate bracket)} \\
&= e - (e - 1) = 1. \end{aligned} $$
Theorem — Change of variable
Let \(\varphi \in C^1(I, \mathbb{R})\) taking values in \(J\), \(f \in C(J, K)\), and \(a, b \in I\). Then $$ \textcolor{colorprop}{\int_a^b f(\varphi(t)) \, \varphi'(t)\,dt = \int_{\varphi(a)}^{\varphi(b)} f(x)\,dx}. $$
Since \(f\) is continuous on \(J\), it has a primitive \(F\) on \(J\) by FTC. The composite \(F \circ \varphi\) is \(C^1\) on \(I\) (composition of \(C^1\) and \(C^1\) with chain rule), with \((F \circ \varphi)'(t) = F'(\varphi(t)) \cdot \varphi'(t) = f(\varphi(t)) \cdot \varphi'(t)\). The function \((F \circ \varphi)'\) is continuous, so by FTC (ii): $$ \begin{aligned} \int_a^b f(\varphi(t)) \varphi'(t)\,dt &= \int_a^b (F \circ \varphi)'(t)\,dt && \text{(chain rule on \(F \circ \varphi\))} \\
&= (F \circ \varphi)(b) - (F \circ \varphi)(a) && \text{(FTC on \(F \circ \varphi\))} \\
&= F(\varphi(b)) - F(\varphi(a)) \\
&= \int_{\varphi(a)}^{\varphi(b)} f(x)\,dx && \text{(FTC on \(F\)).} \end{aligned} $$
Method — Apply change of variable
For an integral \(\int_a^b g(t)\,dt\) that contains an explicit composition \(g(t) = f(\varphi(t)) \varphi'(t)\): - Identify \(x = \varphi(t)\); compute \(dx = \varphi'(t)\,dt\).
- Transport the bounds: \(t = a \to x = \varphi(a)\), \(t = b \to x = \varphi(b)\).
- Rewrite as \(\int_{\varphi(a)}^{\varphi(b)} f(x)\,dx\) and integrate.
Example
Compute \(\int_0^1 \frac{2t}{1 + t^2}\,dt\) by change of variable.
Set \(\varphi(t) = 1 + t^2\), \(\varphi'(t) = 2t\), and \(f(x) = 1/x\). Then \(f(\varphi(t)) \varphi'(t) = 2t / (1 + t^2)\). The bounds transport: \(\varphi(0) = 1\), \(\varphi(1) = 2\). By the change-of-variable theorem: $$ \int_0^1 \frac{2t}{1 + t^2}\,dt = \int_1^2 \frac{dx}{x} = [\ln x]_1^2 = \ln 2. $$
Skills to practice
- Differentiating an integral with a moving upper bound
- Applying integration by parts
- Applying change of variable
VI
Global Taylor formulas
Local vs global
The Taylor-Young formula from Differentiability is local: near a single point \(a\), a sufficiently differentiable function satisfies \(f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x - a)^k + o((x - a)^n)\) as \(x \to a\) --- it controls \(f\) only in the immediate vicinity of \(a\). The global formula of this section, the Taylor-Lagrange inequality, is of a different nature: it bounds the remainder uniformly over an entire segment, valid for every \(b\) in the interval. To prove it we first establish the Taylor formula with integral remainder, an exact expression of the remainder as an integral; from it the inequality follows by the triangle inequality on integrals.
What is required here. The inequality is the result to know and use; the integral-remainder formula is presented as the tool that proves it. The distinction of nature between the local Taylor-Young formula and the global Taylor-Lagrange inequality is the point to retain.
What is required here. The inequality is the result to know and use; the integral-remainder formula is presented as the tool that proves it. The distinction of nature between the local Taylor-Young formula and the global Taylor-Lagrange inequality is the point to retain.
Theorem — Taylor with integral remainder
Let \(f \in C^{n+1}(I, K)\) and \(a, b \in I\). Then $$ \textcolor{colorprop}{f(b) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (b - a)^k + \int_a^b \frac{f^{(n+1)}(t)}{n!} (b - t)^n\,dt}. $$ The last term is called the integral remainder (« reste intégral »).
By induction on \(n\).
- Initialization (\(n = 0\)). For \(f \in C^1(I, K)\), the formula reduces to \(f(b) = f(a) + \int_a^b f'(t)\,dt\), which is exactly FTC (ii).
- Heredity. Assume the formula holds for some \(n \in \mathbb{N}\) and for every \(C^{n+1}\) function on \(I\). Take \(f \in C^{n+2}(I, K)\). Then \(f \in C^{n+1}\), so by induction hypothesis: $$ f(b) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (b - a)^k + \int_a^b \frac{f^{(n+1)}(t)}{n!} (b - t)^n\,dt. $$ Apply IBP to the integral with \(u'(t) = (b - t)^n / n!\) (primitive \(u(t) = -(b - t)^{n+1}/(n + 1)!\)) and \(v(t) = f^{(n+1)}(t)\) (derivative \(v'(t) = f^{(n+2)}(t)\), continuous since \(f \in C^{n+2}\)): $$ \begin{aligned} \int_a^b \frac{f^{(n+1)}(t)}{n!} (b - t)^n\,dt &= \left[ f^{(n+1)}(t) \cdot \left( -\frac{(b - t)^{n+1}}{(n+1)!} \right) \right]_a^b - \int_a^b f^{(n+2)}(t) \cdot \left( -\frac{(b - t)^{n+1}}{(n+1)!} \right)\,dt \\ &= \frac{f^{(n+1)}(a)}{(n+1)!} (b - a)^{n+1} + \int_a^b \frac{f^{(n+2)}(t)}{(n+1)!} (b - t)^{n+1}\,dt && \text{(IBP, T5.2).} \end{aligned} $$ Substitute back to get the formula at rank \(n + 1\).
Theorem — Taylor-Lagrange inequality
Let \(f \in C^{n+1}(I, K)\) and \(a, b \in I\). Let \(J\) be the segment with endpoints \(a\) and \(b\) (i.e., \(J = [\min(a, b), \max(a, b)]\)). Then $$ \textcolor{colorprop}{\left| f(b) - \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (b - a)^k \right| \le \frac{|b - a|^{n+1}}{(n + 1)!} \, \|f^{(n+1)}\|_{\infty, J}}. $$ The use of \(J\) (rather than \([a, b]\)) makes the statement valid for \(b < a\) as well.
We first assume \(a \le b\) (the \(b < a\) case is handled at the end). By Theorem T6.1, the left-hand side equals \(\left| \int_a^b \frac{f^{(n+1)}(t)}{n!} (b - t)^n\,dt \right|\). Bound using the triangle inequality on integrals and \(|f^{(n+1)}(t)| \le \|f^{(n+1)}\|_{\infty, J}\) on \(J = [a, b]\): $$ \begin{aligned} \left| \int_a^b \frac{f^{(n+1)}(t)}{n!} (b - t)^n\,dt \right| &\le \int_a^b \frac{|f^{(n+1)}(t)|}{n!} (b - t)^n\,dt && \text{(triangle inequality)} \\
&\le \frac{\|f^{(n+1)}\|_{\infty, J}}{n!} \int_a^b (b - t)^n\,dt && \text{(uniform bound on \(|f^{(n+1)}|\))} \\
&= \frac{\|f^{(n+1)}\|_{\infty, J}}{n!} \cdot \frac{(b - a)^{n+1}}{n + 1} && \text{(integrate \((b-t)^n\))} \\
&= \frac{(b - a)^{n+1}}{(n + 1)!} \, \|f^{(n+1)}\|_{\infty, J}. \end{aligned} $$ For \(b < a\), write the integral with the orientation reversed: \(\left| \int_a^b R(t)\,dt \right| = \left| \int_b^a -R(t)\,dt \right| \le \int_b^a |R(t)|\,dt\) with \(R(t) = f^{(n+1)}(t) (b - t)^n / n!\). Then integrate \(|b - t|^n\) over the segment \(J = [b, a]\) and the same bound applies, with \(|b - a|^{n+1}\) in place of \((b - a)^{n+1}\).
Method — Bound a function using Taylor-Lagrange
To bound the error in a Taylor polynomial approximation of \(f\) around \(a\), on a neighborhood \(J\) of \(a\): - Compute the Taylor polynomial \(P_n(b) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (b - a)^k\) to a chosen order \(n\).
- Compute or bound \(f^{(n+1)}\) on \(J\); let \(M = \|f^{(n+1)}\|_{\infty, J}\).
- By T6.2: \(|f(b) - P_n(b)| \le \frac{M \cdot |b - a|^{n+1}}{(n + 1)!}\) for every \(b \in J\).
Example
Show that for every \(x \ge 0\): $$ x - \frac{x^2}{2} \le \ln(1 + x) \le x - \frac{x^2}{2} + \frac{x^3}{3}. $$
Apply Taylor with integral remainder (T6.1) to \(f(t) = \ln(1 + t)\) at \(a = 0\). Compute \(f^{(k)}(t) = (-1)^{k-1} (k - 1)! / (1 + t)^k\) for \(k \ge 1\), hence \(f^{(k)}(0) = (-1)^{k-1} (k - 1)!\).
- Order 2. \(f(x) = x - x^2/2 + \int_0^x \frac{2}{(1 + t)^3} \cdot \frac{(x - t)^2}{2}\,dt\). The remainder is \(\int_0^x \frac{(x - t)^2}{(1 + t)^3}\,dt \ge 0\) since \(x \ge 0\) and the integrand is nonnegative. Hence \(\ln(1 + x) \ge x - x^2/2\).
- Order 3. \(f(x) = x - x^2/2 + x^3/3 - \int_0^x \frac{6}{(1 + t)^4} \cdot \frac{(x - t)^3}{6}\,dt\). The remainder \(-\int_0^x \frac{(x - t)^3}{(1 + t)^4}\,dt \le 0\) since \(x \ge 0\) and the integrand is nonnegative. Hence \(\ln(1 + x) \le x - x^2/2 + x^3/3\).
Example
Bound \(|\sin x - (x - x^3/6)|\) by Taylor-Lagrange inequality on \([-1, 1]\).
\(f(t) = \sin t\) is \(C^\infty\) on \(\mathbb{R}\) with \(|f^{(k)}(t)| \le 1\) for all \(t, k\). At \(a = 0\): the Taylor polynomial of order 3 is \(P_3(x) = x - x^3/6\) (the order-2 and order-4 terms vanish by parity). Apply T6.2 with \(n = 4\), on \(J = [-1, 1]\): $$ \left| \sin x - \left( x - \frac{x^3}{6} \right) \right| \le \frac{|x|^5}{5!} \, \|f^{(5)}\|_{\infty, J} \le \frac{|x|^5}{120} \le \frac{1}{120}, $$ since \(|f^{(5)}| = |\cos t| \le 1\) and \(|x|^5 \le 1\) on \(J\).
Skills to practice
- Writing Taylor with integral remainder
- Bounding a function via Taylor-Lagrange inequality
- Proving inequalities via Taylor
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