nyuuzyou/wikis huggingface dataset:
Integration by Partial Fractions/Examples/Arbitrary Example 2

== Example of Use of Integration by Partial Fractions ==
$\ds \int \dfrac {x + 3} {x^2 + 3 x + 2} = 2 \ln \size {x + 1} - \ln \size {x + 2} + C$
== Proof ==
== Sources ==
* 
* 
Category:Examples of Use of Integration by Partial Fractions
Integration by Parts

== Theorem ==
Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$.
Let $f$ and $g$ have primitives $F$ and $G$ respectively on $\closedint a b$.
Then:
=== Primitive ===
=== Definite Integral ===
This technique is called integration by parts.
== Also presented as ==
== Proof Technique ==
== Examples ==
== Sources ==
* 
* 
Category:Integration by Parts
Category:Integral Calculus
Category:Named Theorems
Category:Proof Techniques
Integration by Parts/Also presented as

== Integration by Parts: Also presented as ==
Integration by Parts is often seen presented in this sort of form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
or:
:$\ds \int u \rd v = u v - \int v \rd u$
where it is understood that $u$ and $v$ are functions of the independent variable.
== Sources ==
* : $\text {II}$. Calculus: Integration: Integration by Parts
* : Chapter $10$: Integration: $10.4$. Standard integrals: General Rules: $\text {VI}$.
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integration by Parts: $3.3.12$
* : Part $\text {III}$: Derivatives and Integrals: Chapter $18$: Integration in Elementary Terms: Theorem $1$ (Integration by Parts)
* : $\S 14$: General Rules of Integration: $14.4$
* : Front endpapers: A Brief Table of Integrals: $1$.
* : Appendix $\text I$: Table of Indefinite Integrals $4$.
* 
* : Appendix: Table $2$: Integrals
* 
* : Appendix: Table $2$: Integrals
* : $\S 16$: Indefinite Integrals: General Rules of Integration: $16.4.$
Category:Integration by Parts
Integration by Parts/Corollary

== Corollary to Integration by Parts ==
Let $u$ and $v$ be real functions which are integrable on their domain.
Then:
:$\ds \int u v \rd x = \paren {\int u \rd x} v - \int \paren {\int u \rd x} \dfrac {\d v} {\d x} \rd x$
== Proof ==
From Integration by Parts:
:$(1): \quad \ds \int u \dfrac {\d v} {\d x} \rd x = u v - \int v \dfrac {\d u} {\d x} \rd x$
In $(1)$, let us make the identifications:
This gives:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integration by Parts: $3.3.13$
Category:Integration by Parts
Integration by Parts/Definite Integral

== Theorem ==
Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$.
Let $f$ and $g$ have primitives $F$ and $G$ respectively on $\closedint a b$.
Then:
:$\ds \int_a^b \map f t \map G t \rd t = \bigintlimits {\map F t \map G t} a b - \int_a^b \map F t \map g t \rd t$
== Proof ==
By Product Rule for Derivatives:
:$\map D {F G} = f G + F g$
Thus $F G$ is a primitive of $f G + F g$ on $\closedint a b$.
Hence, by the Fundamental Theorem of Calculus:
:$\ds \int_a^b \paren {\map f t \map G t + \map F t \map g t} \rd t = \bigintlimits {\map F t \map G t} a b$
The result follows.
== Proof Technique ==
== Sources ==
* : Part $\text {III}$: Derivatives and Integrals: Chapter $18$: Integration in Elementary Terms: Theorem $1$ (Integration by Parts)
* : $6$. Integral Calculus: Appendix: Rules and Techniques of Integration: $2.1$ Integration by parts
* : $\S 13.21$
Category:Integration by Parts
Integration by Parts/Primitive

== Theorem ==
Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$.
Let $f$ and $g$ have primitives $F$ and $G$ respectively on $\closedint a b$.
Then:
:$\ds \int \map f t \map G t \rd t = \map F t \map G t - \int \map F t \map g t \rd t$
on $\closedint a b$.
=== Corollary ===
== Proof $1$ ==
== Proof $2$ ==
== Also presented as ==
== Proof Technique ==
== Sources ==
* : Appendix $\text I$: Table of Indefinite Integrals $3$.
* 
Category:Integration by Parts
Integration by Parts/Proof Technique

== Integration by Parts: Proof Technique ==
The technique of solving an integral in the form $\ds \int \map f t \map G t \rd t$ in this manner is called integration by parts.
Its validity as a solution technique stems from the fact that it may be possible to choose $f$ and $G$ such that $G$ is easier to differentiate than to integrate.
Thus the plan is to reduce the integral to one such that $\ds \int \map F t \map g t \rd t$ is easier to solve than $\ds \int \map f t \map G t \rd t$.
It may be, of course, that one or more further applications of this technique are needed before the solution can be extracted.
== Sources ==
* 
Category:Integration by Parts
Integration by Parts/Riemann-Stieltjes Integral

== Theorem ==
Let $f, \alpha$ be a real functions that are bounded on $\closedint a b$.
Suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint a b$.
Then, $\alpha$ is Riemann-Stieltjes integrable with respect to $f$ on $\closedint a b$ and:
:$\ds \int_a^b f \rd \alpha + \int_a^b \alpha \rd f = \map f b \map \alpha b - \map f a \map \alpha a$
== Proof ==
Let $\epsilon > 0$ be arbitrary.
By definition of the Riemann-Stieltjes integral, let $P_\epsilon$ be a subdivision of $\closedint a b$ such that:
:For every $P$ finer than $P_\epsilon$, $\size {\map S {P, f, \alpha} - \int_a^b f \rd \alpha} < \epsilon$
Let $P = \set {x_0, \dotsc, x_n}$ be an arbitrary subdivision finer than $P_\epsilon$.
Let $t_k \in \closedint {x_{k - 1} } {x_k}$ for all $k \in \set {1, \dotsc, n}$.
Then:
We also have:
Combining the above:
Now, consider $y_\ell \in \R$ defined for all $\ell \in \set {0, 1, \dotsc, 2 n - 1, 2 n}$ as:
:$y_{2 k} = x_k$
:$y_{2 k - 1} = t_k$
Since $y_{2 k - 2} = x_{k - 1} \le t_k = y_{2 k - 1}$ and $t_{2 k - 1} = t_k \le x_k = y_{2 k}$, we have:
:$y_0 \le y_1 \le \dotso \le y_{2 n - 1} \le y_{2 n}$
As $y_0 = x_0 = a$ and $y_{2 n} = x_n = b$, it follows that:
:$P' := \set {y_0, y_1, \dotsc, y_{2 n - 1}, y_{2 n}}$
is a subdivision of $\closedint a b$.
Furthermore, since $P_\epsilon \subseteq P \subseteq P'$, it follows that $P'$ is finer than $P_\epsilon$.
Let $t'_\ell \in \closedint {y_{\ell - 1}} {y_\ell}$ be defined for all $\ell \in \set {1, \dotsc, 2 n}$ as:
:$t'_{2 k} := x_k = y_{2 k} \in \closedint {y_{2 k - 1}} {y_{2 k}}$
:$t'_{2 k - 1} := x_{k - 1} = y_{2 k - 2} \in \closedint {y_{2 k - 2}} {y_{2 k - 1}}$
Then:
Therefore:
Since $P$ finer than $P_\epsilon$ and $\epsilon > 0$ were arbitrary, by definition of the Riemann-Stieltjes integral:
:$\int_a^b \alpha \rd f = \map f b \map \alpha b - \map f a \map \alpha a - \int_a^b f \rd \alpha$
The result follows.
== Sources ==
* : Chapter $7$ The Riemann-Stieltjes Integral: $\S 7.5$: Theorem $7.6$
Category:Riemann-Stieltjes Integral
Integration by Substitution

== Theorem ==
Let $\phi$ be a real function which has a derivative on the closed interval $\closedint a b$.
Let $I$ be an open interval which contains the image of $\closedint a b$ under $\phi$.
Let $f$ be a real function which is continuous on $I$.
=== Primitive ===
=== Definite Integral ===
The technique of solving an integral in this manner is called integration by substitution.
== Also known as ==
== Proof Technique ==
== Examples ==
== Also see ==
* Weierstrass Substitution
== Sources ==
* 
* 
* 
* 
* 
* 
Category:Integration by Substitution
Category:Integral Calculus
Category:Named Theorems
Category:Proof Techniques
Integration by Substitution/Also known as

== Integration by Substitution: Also known as ==
Because the most usual substitution variable used is $u$, this method is often referred to Integration by Substitution as $u$-substitution in the source works for introductory-level calculus courses.
Some sources refer to this technique as Change of Variable, but that has a number of different meanings depending upon context.
== Sources ==
* 
* 
* 
* 
* 
* 
Category:Integration by Substitution
Integration by Substitution/Corollary

== Corollary to Integration by Substitution ==
Let $f : \R \to \R$ be a real function.
Let $f$ be integrable.
Let $a$, $b$, and $c$ be real numbers.
Then:
:$\ds \int_{a - c}^{b - c} \map f t \rd t = \int_a^b \map f {t - c} \rd t$
== Proof ==
Let $\map \phi u = u - c$.
By Sum Rule for Derivatives, Derivative of Identity Function, and Derivative of Constant, we have:
:$\map {\phi'} u = 1$
By Integration by Substitution:
Category:Integration by Substitution
Integration by Substitution/Definite Integral

== Theorem ==
Let $\phi$ be a real function which has a derivative on the closed interval $\closedint a b$.
Let $I$ be an open interval which contains the image of $\closedint a b$ under $\phi$.
Let $f$ be a real function which is continuous on $I$.
If $\map \phi a \le \map \phi b$, then the definite integral of $f$ from $a$ to $b$ can be evaluated by:
:$\ds \int_{\map \phi a}^{\map \phi b} \map f t \rd t = \int_a^b \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$
where $t = \map \phi u$.
If $\map \phi a > \map \phi b$, then the definite integral of $f$ from $a$ to $b$ can be evaluated by:
:$\ds - \int_{\map \phi b}^{\map \phi a} \map f t \rd t = \int_a^b \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$
The technique of solving an integral in this manner is called integration by substitution.
=== Corollary ===
== Proof ==
Let $F$ be an antiderivative of $f$.
We have:
Hence $\map F {\map \phi u}$ is an antiderivative of $\map f {\map \phi u} \dfrac \d {\d u} \map \phi u$.
Thus:
If $\map \phi a \le \map \phi b$, we also have:
which was to be proved. 
If instead $\map \phi a > \map \phi b$, we have:
which was to be proved.
== Also known as ==
== Proof Technique ==
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 8$. Change of Variable
* : $6$. Integral Calculus: Appendix: Rules and Techniques of Integration: $2.2$ Integration by substitution
* : $\S 13.22$
* 
* 
* : $\S 4.5, \S 8.4$
Category:Integration by Substitution
Category:Definite Integrals
Integration by Substitution/Examples

== Examples of Use of Integration by Substitution ==
=== Primitive of $\paren {2 x + 3} \sqrt {x^2 + 3 x + 2}$ ===
=== Primitive of $\frac {\cos x} {\paren {1 + \sin x}^2}$ ===
=== Primitive of $\sqrt {1 - x}$ ===
Category:Integral Substitutions
Integration by Substitution/Examples/(2x + 3) by Root of x^2 + 3x + 2

== Example of Use of Integration by Substitution ==
:$\ds \int \paren {2 x + 3} \sqrt {x^2 + 3 x + 2} \rd x = \dfrac 2 3 {\paren {\sqrt {x^2 + 3 x + 2} }^3} + C$
== Proof ==
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $6$.
Category:Integral Substitutions
Integration by Substitution/Examples/Cosine over Square of 1 + Sine

== Example of Use of Integration by Substitution ==
:$\ds \int \dfrac {\cos x} {\paren {1 + \sin x}^2} \rd x = -\dfrac 1 {1 + \sin x} + C$
== Proof ==
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $7$.
Category:Integral Substitutions
Integration by Substitution/Examples/Primitive of Root of 1 minus x

== Example of Use of Integration by Substitution ==
:$\ds \int \sqrt {1 - x} \rd x = -\dfrac 2 3 \paren {1 - x}^{3 / 2} + C$
== Proof ==
== Sources ==
* 
* 
Category:Integral Substitutions
Integration by Substitution/Primitive

== Theorem ==
Let $\phi$ be a real function which has a derivative on the closed interval $\closedint a b$.
Let $I$ be an open interval which contains the image of $\closedint a b$ under $\phi$.
Let $f$ be a real function which is continuous on $I$.
The primitive of $f$ can be evaluated by:
:$\ds \int \map f x \rd x = \int \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$
where $x = \map \phi u$.
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Also known as ==
== Proof Technique ==
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 8$. Change of Variable
* : Chapter $10$: Integration: $10.4$. Standard integrals: General Rules: $\text {V}$.
* 
* 
* : $\S 4.5, \S 8.4$
Category:Integration by Substitution
Integration by Substitution/Primitive/Also presented as

== Integration by Substitution: Also presented as == 
The rule of Integration by Substitution can also be seen in the form:
:$\ds \int \map f x = \int \map \phi u \dfrac {\d x} {\d u} \rd u$
and:
:$\ds \int \map F {\map \phi x} \rd x = \int \dfrac {\map F u} {\map {f'} x} \rd u$
where $x = \map \phi u$.
== Sources ==
* : Chapter $10$: Integration: $10.4$. Standard integrals: General Rules: $\text {V}$.
Category:Integration by Substitution
Integration by Substitution/Primitive/Proof 1

== Theorem ==
== Proof ==
Let $\map F x = \ds \int \map f x \rd x$.
Thus by definition $\map F x$ is a primitive of $\map f x$.
So $\map F x$ is an antiderivative of $\map f {\map \phi u} \dfrac \d {\d u} \map \phi u$.
Therefore:
where $x = \map \phi u$.
Category:Integration by Substitution
Integration by Substitution/Proof Technique

== Integration by Substitution: Proof Technique ==
The usefulness of the technique of Integration by Substitution stems from the fact that it may be possible to choose $\phi$ such that $\map f {\map \phi u} \dfrac \d {\d u} \map \phi u$ (despite its seeming complexity in this context) may be easier to integrate.
If $\phi$ is a trigonometric function, the use of trigonometric identities to simplify the integrand is called integration by trigonometric substitution (or simply trig substitution).
Care must be taken to address the domain and image of $\phi$.
This consideration frequently arises when inverse trigonometric functions are involved.
== Also known as ==
== Sources ==
* : $\S 4.5, \S 8.4$
Category:Integration by Substitution
Integration by Substitution/Riemann-Stieltjes Integral

== Theorem ==
Let $g$ be a real function that is continuous and strictly monotone on $\closedint a b$.
Let $\Bbb I = g \closedint a b$ be the image of $g$ under $\closedint a b$.
Let $f, \alpha$ be real functions that are bounded on $\Bbb I$.
Further suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\Bbb I$.
Let $h, \beta : \Bbb I \to \R$ be defined as:
:$\map h x = \map f {\map g x}$
:$\map \beta x = \map \alpha {\map g x}$
Then, $h$ is Riemann-Stieltjes integrable with respect to $\beta$ on $\Bbb I$ and:
:$\ds \int_a^b h \rd \beta = \int_{\map g a}^{\map g b} f \rd \alpha$
== Proof ==
We have the following special cases:
=== Increasing ===
=== Decreasing ===
By definition of strictly monotone, $g$ is either strictly increasing or strictly decreasing.
First, suppose $g$ is strictly increasing.
By the Intermediate Value Theorem, it is clear that:
:$\Bbb I = \closedint {\map g a} {\map g b}$
Then, the result follows from the special case Increasing above.
Now, suppose $g$ is strictly decreasing.
Once again, by the Intermediate Value Theorem, it is clear that:
:$\Bbb I = \closedint {\map g b} {\map g a}$
Then:
In each case, the result holds, so the theorem follows from Proof by Cases.
== Sources ==
* : Chapter $7$ The Riemann-Stieltjes Integral: $\S 7.6$: Theorem $7.7$
Category:Riemann-Stieltjes Integral
Integration by Substitution/Riemann-Stieltjes Integral/Decreasing

== Theorem ==
Let $g$ be a real function that is continuous and strictly decreasing on $\closedint a b$.
Let $f, \alpha$ be real functions that are bounded on $\closedint {\map g b} {\map g a}$.
Further suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint {\map g b} {\map g a}$.
Let $h, \beta : \closedint a b \to \R$ be defined as:
:$\map h x = \map f {\map g x}$
:$\map \beta x = \map \alpha {\map g x}$
Then, $h$ is Riemann-Stieltjes integrable with respect to $\beta$ on $\closedint a b$ and:
:$\ds \int_a^b h \rd \beta = - \int_{\map g b}^{\map g a} f \rd \alpha$
== Proof ==
For any $x \in \closedint a b$, $\map g x > \map g a$ implies $x < a$ by definition of strictly decreasing, so:
:$\forall x \in \closedint a b: \map g x \le \map g a$
Likewise, $\map g x > \map g b$ implies $x < b$, so:
:$\forall x \in \closedint a b: \map g x \ge \map g b$
Together:
:$\Img g \subseteq \closedint {\map g b} {\map g a}$
But by the Intermediate Value Theorem:
:$\closedint {\map g b} {\map g a} \subseteq \Img g$
Therefore:
:$\Img g = \closedint {\map g b} {\map g a}$
Then, by Inverse of Strictly Monotone Function, $g$ has an inverse $g^{-1} : \closedint {\map g b} {\map g a} \to \closedint a b$.
Furthermore, $g^{-1}$ is strictly decreasing.
Let $\epsilon > 0$ be arbitrary.
By definition of the Riemann-Stieltjes integral, there exists a subdivision $P'_\epsilon$ of $\closedint {\map g b} {\map g a}$ such that:
:For each $P'$ finer than $P'_\epsilon$, $\ds \size {\map S {P', f, \alpha} - \int_{\map g b}^{\map g a} f \rd \alpha} < \epsilon$.
Suppose $P'_\epsilon = \set {y_0, y_1, \dotsc, y_{n - 1}, y_n}$.
Then, define $P_\epsilon = g^{-1} \sqbrk {P'_\epsilon}$.
Since $g^{-1}$ is strictly decreasing:
:$a = \map {g^{-1}} {\map g a} = \map {g^{-1}} {y_n} < \map {g^{-1}} {y_{n - 1}} < \dotso < \map {g^{-1}} {y_1} < \map {g^{-1}} {y_0} = \map {g^{-1}} {\map g b} = b$
we have that $P_\epsilon$ is a subdivision of $\closedint a b$.
Now, let $P = \set {x_0, \dotsc, x_m}$ be a subdivision of $\closedint a b$ that is finer than $P_\epsilon$.
For each $k \in \set {0, 1, \dotsc, m - 1, m}$, define:
:$x'_k = x_{m - k}$
Define $P' = \set {\map g {x'_0}, \map g {x'_1}, \dotsc, \map g {x'_{m - 1}}, \map g {x'_m}$.
Since $g$ is strictly decreasing, $P'$ is a subdivision of $\closedint {\map g b} {\map g a}$.
Furthermore, for every $k$:
:$y_k = \map g {\map {g^{-1}} {y_k}} \in g \sqbrk {P_\epsilon} \subseteq g \sqbrk P = P'$
Thus, $P'$ is finer than $P'_\epsilon$.
Hence:
Therefore:
Since $\epsilon > 0$ was arbitrary, by definition of the Riemann-Stieltjes integral:
:$\ds \int_a^b h \rd \beta = - \int_{\map g a}^{\map g b} f \rd \alpha$
== Sources ==
* : Chapter $7$ The Riemann-Stieltjes Integral: $\S 7.6$: Theorem $7.7$
Category:Riemann-Stieltjes Integral
Integration by Substitution/Riemann-Stieltjes Integral/Increasing

== Theorem ==
Let $g$ be a real function that is continuous and strictly increasing on $\closedint a b$.
Let $f, \alpha$ be real functions that are bounded on $\closedint {\map g a} {\map g b}$.
Further suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint {\map g a} {\map g b}$.
Let $h, \beta : \closedint a b \to \R$ be defined as:
:$\map h x = \map f {\map g x}$
:$\map \beta x = \map \alpha {\map g x}$
Then, $h$ is Riemann-Stieltjes integrable with respect to $\beta$ on $\closedint a b$ and:
:$\ds \int_a^b h \rd \beta = \int_{\map g a}^{\map g b} f \rd \alpha$
== Proof ==
For any $x \in \closedint a b$, $\map g x < \map g a$ implies $x < a$ by definition of strictly increasing, so:
:$\forall x \in \closedint a b: \map g x \ge \map g a$
Likewise, $\map g x > \map g b$ implies $x > b$, so:
:$\forall x \in \closedint a b: \map g x \le \map g b$
Together:
:$\Img g \subseteq \closedint {\map g a} {\map g b}$
But by the Intermediate Value Theorem:
:$\closedint {\map g a} {\map g b} \subseteq \Img g$
Therefore:
:$\Img g = \closedint {\map g a} {\map g b}$
Then, by Inverse of Strictly Monotone Function, $g$ has an inverse $g^{-1} : \closedint {\map g a} {\map g b} \to \closedint a b$.
Furthermore, $g^{-1}$ is strictly increasing.
Let $\epsilon > 0$ be arbitrary.
By definition of the Riemann-Stieltjes integral, there exists a subdivision $P'_\epsilon$ of $\closedint {\map g a} {\map g b}$ such that:
:For each $P'$ finer than $P'_\epsilon$, $\ds \size {\map S {P', f, \alpha} - \int_{\map g a}^{\map g b} f \rd \alpha} < \epsilon$.
Suppose $P'_\epsilon = \set {y_0, y_1, \dotsc, y_{n - 1}, y_n}$.
Then, define $P_\epsilon = g^{-1} \sqbrk {P'_\epsilon}$.
Since $g^{-1}$ is strictly increasing:
:$a = \map {g^{-1}} {\map g a} = \map {g^{-1}} {y_0} < \map {g^{-1}} {y_1} < \dotso < \map {g^{-1}} {y_{n - 1}} < \map {g^{-1}} {y_n} = \map {g^{-1}} {\map g b} = b$
we have that $P_\epsilon$ is a subdivision of $\closedint a b$.
Now, let $P = \set {x_0, \dotsc, x_m}$ be a subdivision of $\closedint a b$ that is finer than $P_\epsilon$.
Define $P' = g \sqbrk P$.
As above, since $g$ is strictly increasing, $P'$ is a subdivision of $\closedint {\map g a} {\map g b}$.
Furthermore, for every $k$:
:$y_k = \map g {\map {g^{-1}} {y_k}} \in g \sqbrk {P_\epsilon} \subseteq g \sqbrk P = P'$
Thus, $P'$ is finer than $P'_\epsilon$.
Hence:
Therefore:
Since $\epsilon > 0$ was arbitrary, by definition of the Riemann-Stieltjes integral:
:$\ds \int_a^b h \rd \beta = \int_{\map g a}^{\map g b} f \rd \alpha$
== Sources ==
* : Chapter $7$ The Riemann-Stieltjes Integral: $\S 7.6$: Theorem $7.7$
Category:Riemann-Stieltjes Integral
Integration on Polynomials is Linear Operator

== Theorem ==
Let $\map P \R$ be the vector space of all polynomial functions on the real number line $\R$.
Let $S$ be the mapping defined as:
:$\ds \forall p \in \map P \R: \forall x \in \R: \map S {\map p x} = \int_0^x \map p t \rd t$
Then $S$ is a linear operator on $\map P \R$.
== Proof ==
Let $\map f x, \map g x$ be real functions which are integrable on $\R$.
Let $\closedint a b$ be a closed interval of $\R$.
Then from Linear Combination of Definite Integrals, $\lambda f + \mu g$ is integrable on $\closedint a b$ and:
:$\ds \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t$
From Real Polynomial Function is Continuous:
:$\forall p \in \map P \R: p$ is continuous on $\R$.
From Continuous Real Function is Darboux Integrable:
:$\forall p \in \map P \R: p$ is integrable on $\R$.
The result follows from Condition for Linear Transformation.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.6$
Category:Linear Operators
Category:Integral Calculus
Category:Real Polynomial Functions
Interconnection between Vacuum Permittivity and Vacuum Permeability

== Motivation behind Vacuum Permittivity and Vacuum Permeability ==
Right from the start we see the equation $c^2 = \dfrac 1 {\varepsilon_0 \mu_0}$ and we understand immediately that these parameters, vacuum permittivity and vacuum permeability, are tightly wound up in the structure of the electromagnetic spectrum.
One of them defines the strength of the electric attraction.
The other defines the strength of the magnetic force.
Hence it is apparent that the units of measurement we use are to a certain extent arbitrary, as the concepts being discussed are more abstract than tangible physical measurements.
But in order to keep the relationships within the grasp of the tangible, we need to make sure the units we use are accessible enough to interface neatly with the existing measurement systems we already have.
Prevalent is the SI system, in which we can set up units of considerable convenience.
We notice that if we define $\mu_0$ as $4 \pi \times 10^{-7}$ in henries per metre, which evaluates to $\dfrac {\mathrm {kg} \times \mathrm m} {\mathrm A^2 \times \mathrm s^4}$, then many of the equations work out with manageable numbers all approximately the same size.
From this convenient number, we define the ampere, and work out what conditions we require in order to produce it.
We use the Biot-Savart Law to work out what to expect, and then measure the force and work out the rest of the numbers from there.
Hence having chosen $\mu_0$ in that way, we must then define $\varepsilon_0$ as $\dfrac 1 {\mu_0 c^2}$.
Or instead, we could declare what value $\varepsilon_0$ should be based on accurately and reliably reproducible standards, that is $\dfrac {e^2} {2 \alpha h c}$, and then force $\mu_0$ to be $\dfrac 1 {\varepsilon_0 c^2}$.
The latter is the direction it currently lies, after the $2019$ standards redefinition.
We discuss the CGS system in passing, in order to complain about its confusingness and fiddly-smallness.
The author of this page thanks everybody in his educational system that he was never asked to learn it in the first place, coming straight in as they did with amps, coulombs, volts, webers, teslas, farads and henries.
Category:Vacuum Permittivity
Category:Vacuum Permeability
Category:Value of Vacuum Permittivity
Category:Value of Vacuum Permeability