avewright/proofwiki-math huggingface dataset:
Theorem: Modulus of Gamma Function of One Half plus Imaginary Number

Let $t \in \R$ be a real number.
Then: 
:$\cmod {\map \Gamma {\dfrac 1 2 + i t} } = \sqrt {\pi \map \sech {\pi t} }$
where:
:$\Gamma$ is the Gamma function
:$\sech$ is the hyperbolic secant function.

Proof:
{{begin-eqn}}
{{eqn	| l = \cmod {\map \Gamma {\frac 1 2 + i t} }^2
	| r = \map \Gamma {\frac 1 2 + i t} \overline {\map \Gamma {\frac 1 2 + i t} }
	| c = Modulus in Terms of Conjugate
}}
{{eqn	| r = \map \Gamma {\frac 1 2 + i t} \map \Gamma {\frac 1 2 - i t}
	| c = Complex Conjugate of Gamma Function
}}
{{eqn	| r = \map \Gamma {\frac 1 2 + i t} \map \Gamma {1 - \paren {\frac 1 2 + i t} }
	| c = applying some algebra
}}
{{eqn	| r = \pi \map \csc {\pi \paren {\frac 1 2 + i t} }
	| c = Euler's Reflection Formula
}}
{{eqn	| r = \pi \map \sec {\pi i t}
	| c = Sine of Complement equals Cosine
}}
{{eqn	| r = \pi \map \sech {\pi t}
	| c = Hyperbolic Cosine in terms of Cosine
}}
{{end-eqn}}
As $\cmod z \ge 0$ for all complex numbers $z$, we can take the non-negative square root of both sides and write:
:$\cmod {\map \Gamma {\dfrac 1 2 + i t} } = \sqrt {\pi \map \sech {\pi t} }$
{{qed}}
Category:Gamma Function
Category:Hyperbolic Secant Function
2livusfxejek9s219i2q3olje9w7s1s
Theorem: Integral to Infinity of Sine p x over x

:$\ds \int_0^\infty \frac {\sin p x} x \rd x = \begin {cases} \dfrac \pi 2 & : p > 0 \\ \\ 0 & : p = 0 \\ \\ -\dfrac \pi 2 & : p < 0 \end {cases}$

Proof:
Let $p > 0$.
We have: 
{{begin-eqn}}
{{eqn	| l = \int_0^\infty \frac {\sin p x} x \rd x
	| r = \frac 1 p \int_0^\infty \frac {\sin t} { \frac 1 p t} \rd t
	| c = substituting $t = p x$
}}
{{eqn	| r = \int_0^\infty \frac {\sin t} t \rd t
}}
{{eqn	| r = \frac \pi 2
	| c = Dirichlet Integral
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn	| l = \int_0^\infty \frac {\sin \left({- p x}\right)} x \rd x
	| r = -\int_0^\infty \frac {\sin p x} x \rd x
	| c = Sine Function is Odd
}}
{{eqn	| r = -\frac \pi 2
	| c = per above computation
}}
{{end-eqn}}
For $p = 0$, we have: 
{{begin-eqn}}
{{eqn	| l = \int_0^\infty \frac {\sin 0 x} x \rd x 
	| r = \int_0^\infty \frac 0 x \rd x
	| c = Sine of Zero is Zero
}}
{{eqn	| r = 0
}}
{{end-eqn}}
Hence the result. 
{{qed}}
Theorem: Integral to Infinity of Sine p x over x

:$\ds \int_0^\infty \frac {\sin p x} x \rd x = \begin {cases} \dfrac \pi 2 & : p > 0 \\ \\ 0 & : p = 0 \\ \\ -\dfrac \pi 2 & : p < 0 \end {cases}$

Proof:
{{ProofWanted|Use Primitive of Sine of a x over x, and also the analytic solution as found in {{BookLink|Integration|R.P. Gillespie|ed = 2nd|edpage = Second Edition}} }}
Theorem: Radical of Sum of Ideals

Let $A$ be a commutative ring with unity.
Let $\mathfrak a, \mathfrak b \subseteq A$ be ideals.
Then for the radical of their sum we have:
:$\map \Rad {\mathfrak a + \mathfrak b} = \map \Rad {\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} }$

Proof:
From Ideal of Ring is Contained in Radical:
:$\mathfrak a \subseteq \map \Rad {\mathfrak a}$
From Sum of Larger Ideals is Larger:
:$\mathfrak a + \mathfrak b \subseteq \map \Rad {\mathfrak a} + \map \Rad {\mathfrak b}$
From Radical of Ideal Preserves Inclusion:
:$\map \Rad {\mathfrak a + \mathfrak b} \subseteq \map \Rad {\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} }$
{{qed|lemma}}
Let $x \in \map \Rad {\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} }$.
Then there exists $n \in \N$ such that the power $x^n \in \map \Rad {\mathfrak a} + \map \Rad {\mathfrak b}$.
By definition of sum of ideals, there exist $a \in \map \Rad {\mathfrak a}$ and $b \in \map \Rad {\mathfrak b}$ with $x^n = a + b$.
Let $p, q \in \N$ with $a^p \in \mathfrak a$ and $b^q \in \mathfrak b$.
By the Binomial Theorem:
:$\paren {a + b}^{p + q - 1} \in \mathfrak a + \mathfrak b$
{{explain|a link needed here}}
Thus:
:$x^{n \cdot \paren {p + q - 1} } \in \mathfrak a + \mathfrak b$
That is:
:$x \in \map \Rad {\mathfrak a + \mathfrak b}$
Hence by definition of subset:
:$\map \Rad {\mathfrak a + \mathfrak b} \supseteq \map \Rad {\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} }$
{{qed|lemma}}
The result follows by definition of set equality.
{{qed}}
Category:Radical of Ideals
b65nfestpyxd3urnryclg75vnyy6s31
Theorem: Ideal of Ring is Contained in Radical

Let $A$ be a commutative ring with unity.
Let $\mathfrak a \subseteq A$ be an ideal.
Then $\mathfrak a$ is contained in its radical:
:$\mathfrak a \subseteq \operatorname{Rad} \left({\mathfrak a}\right)$

Proof:
Let $a \in \mathfrak a$.
By definition of power:
: $a^1 = a$
Thus:
: $a \in \operatorname{Rad} \left({\mathfrak a}\right)$
{{qed}}
Category:Radical of Ideals
893sgr06yi4pmfepysm4qb70v66010g
Theorem: Radical of Ideal Preserves Inclusion

Let $A$ be a commutative ring with unity.
Let $\mathfrak a \subseteq \mathfrak b \subseteq A$ be an ideals.
Then we have an inclusion of their radicals:
:$\operatorname{Rad} \left({\mathfrak a}\right) \subseteq \operatorname{Rad} \left({\mathfrak b}\right)$

Proof:
Let $x \in \operatorname{Rad} \left({\mathfrak a}\right)$.
We show that $x \in \operatorname{Rad} \left({\mathfrak b}\right)$.
By definition of radical, there exists $n \in \N$ such that the power $x^n \in \mathfrak a$.
Because $\mathfrak a \subseteq \mathfrak b$, also $x^n \in \mathfrak b$.
Thus $x \in \operatorname{Rad} \left({\mathfrak b}\right)$.
{{qed}}
Category:Radical of Ideals
k3nevcqozzdlobgpyk4rf16r31o9smw
Theorem: Ideals with Coprime Radicals are Coprime

Let $A$ be a commutative ring with unity.
Let $\mathfrak a, \mathfrak b \subseteq A$ be ideals.
Let their radicals be coprime:
:$\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} = \ideal 1$
Then $\mathfrak a$ and $\mathfrak b$ are coprime:
:$\mathfrak a + \mathfrak b = \ideal 1$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \map \Rad {\mathfrak a + \mathfrak b}
      | r = \map \Rad {\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} }
      | c = Radical of Sum of Ideals
}}
{{eqn | r = \map \Rad {\ideal 1}
      | c = $\map \Rad {\mathfrak a}$ and $\map \Rad {\mathfrak b}$ are coprime
}}
{{eqn | r = \ideal 1
      | c = Radical of Unit Ideal
}}
{{end-eqn}}
By Unit Ideal iff Radical is Unit Ideal:
:$\mathfrak a + \mathfrak b = \ideal 1$
That is, $\mathfrak a$ and $\mathfrak b$ are coprime.
{{qed}}
Category:Radical of Ideals
Category:Coprime Ideals
7ya59d37ix7vckmcp54whscnbfpubl0
Theorem: Radical of Unit Ideal

Let $A$ be a commutative ring with unity.
Let $\ideal 1$ be its unit ideal.
Then its radical equals $\ideal 1$:
:$\map \Rad {\ideal 1} = \ideal 1$.

Proof:
By definition of ideal:
:$\map \Rad {\ideal 1} \subseteq A$
By Ideal of Ring is Contained in Radical:
:$\ideal 1 = A \subseteq \Rad {\ideal 1}$.
By definition of set equality:
:$\map \Rad {\ideal 1} = \ideal 1$
{{qed}}
Category:Radical of Ideals
hqnjo3hlzuix5y1p4fx9jk3cm5rh1u1
Theorem: Radical of Power of Prime Ideal

Let $A$ be a commutative ring with unity.
Let $\mathfrak p \subseteq A$ be a prime ideal.
Let $n > 0$ be a natural number.
Then the radical of the $n$th power of $\mathfrak p$ equals $\mathfrak p$:
:$\map {\operatorname{Rad} } {\mathfrak p^n} = \mathfrak p$

Proof:
$(\subseteq):$
Let $a \in \map {\operatorname{Rad} } {\mathfrak p^n}$.
Then by definition, $a^k \in \mathfrak p^n$ for some integer $k$.
By Power of Ideal is Subset, $\mathfrak p^n \subseteq \mathfrak p$.
Hence, $a^k \in \mathfrak p$.
By Power in Prime Ideal, $a \in \mathfrak p$.
{{qed|lemma}}
$(\supseteq):$
Let $b \in \mathfrak p$.
Then by definition $b^n \in \mathfrak p^n$.
Hence $b \in \map {\operatorname{Rad} } {\mathfrak p^n}$.
{{qed}}
Category:Radical of Ideals
Category:Ideal Theory
Category:Commutative Algebra
Category:Ring Theory
38ehspfzboic7xwi8yj2pjwiznby4jk
Theorem: Nilpotent Ring Element plus Unity is Unit

Let $A$ be a ring with unity.
Let $1 \in A$ be its unity.
Let $a \in A$ be nilpotent.
Then $1 + a$ is a unit of $A$.

Proof:
Because $a$ is nilpotent, there exists a natural number $n > 0$ with $a^n = 0$.
By Sum of Geometric Sequence in Ring:
:$\paren {1 + a} \cdot \ds \sum_{k \mathop = 0}^{n - 1} \paren {-a}^k = 1 + \paren {-a}^n$
:$\paren {\ds \sum_{k \mathop = 0}^{n - 1} \paren {-a}^k} \cdot \paren {1 + a} = 1 + \paren {-a}^n$
where $\sum$ denotes summation.
By Negative of Nilpotent Ring Element:
:$\paren {-a}^n = 0$
Thus $1 + a$ is a unit.
{{qed}}
Theorem: Ideal is Contained in Contraction of Extension

Let $A$ and $B$ be commutative rings with unity.
Let $f : A \to B$ be a ring homomorphism.
Let $\mathfrak a \subseteq A$ be an ideal.
Then $\mathfrak a$ is contained in the contraction of its extension by $f$:
:$\mathfrak a \subseteq \mathfrak a^{ec}$

Proof:
By definition of extension and generated ideal:
:$f \sqbrk {\mathfrak a} \subseteq \mathfrak a^e$
By Subset of Domain is Subset of Preimage of Image:
:$\mathfrak a \subseteq f^{-1} \sqbrk {f \sqbrk {\mathfrak a}}$
By Preimage of Subset is Subset of Preimage:
:$\mathfrak a \subseteq f^{-1} \sqbrk {f \sqbrk {\mathfrak a }} \subseteq f^{-1} \sqbrk {\mathfrak a^e}$
{{qed}}
Category:Ideal Theory
Category:Commutative Algebra
Category:Ring Theory
c7f1dzls6siwagv658rbp3amqfu7l4l
Theorem: Equivalence of Definitions of Irreducible Polynomial over Field

Let $K$ be a field.
{{TFAE|def = Irreducible Polynomial}}

Proof:
Note that by Field is Integral Domain, $K$ is indeed an integral domain.
{{proof wanted}}
Category:Irreducible Polynomials
e32op3s3m1i7mw9bk3haxpgjj5vsf1o
Theorem: Linear Combination of Integrals/Indefinite

:$\ds \int \paren {\lambda \map f x + \mu \map g x} \rd x = \lambda \int \map f x \rd x + \mu \int \map g x \rd x$

Proof:
Let $F$ and $G$ be primitives of $f$ and $g$ respectively on $\closedint a b$.
By Linear Combination of Derivatives, $H = \lambda F + \mu G$ is a primitive of $\lambda f + \mu g$ on $\closedint a b$.
Hence:
{{begin-eqn}}
{{eqn | l = \int \paren {\lambda \map f t + \mu \map g t} \rd t
      | r = \lambda \map F t + \mu \map G t
      | c =
}}
{{eqn | r = \lambda \int \map f t \rd t + \mu \int \map g t \rd t
      | c =
}}
{{end-eqn}}
{{qed}}
Theorem: Linear Combination of Integrals/Definite

:$\ds \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t$

Proof:
Let $F$ and $G$ be primitives of $f$ and $g$ respectively on $\closedint a b$.
By Linear Combination of Derivatives, $H = \lambda F + \mu G$ is a primitive of $\lambda f + \mu g$ on $\closedint a b$.
Hence by the Fundamental Theorem of Calculus:
{{begin-eqn}}
{{eqn | l = \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t
      | r = \bigintlimits {\lambda \map F t + \mu \map G t} a b
      | c = 
}}
{{eqn | r = \lambda \bigintlimits {\map F t} a b + \mu \bigintlimits {\map G t} a b
      | c = 
}}
{{eqn | r = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Linear Combination of Integrals/Definite

:$\ds \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t$

Proof:
It is clear that for step functions $s$ and $t$:
{{handwaving|"It is clear that ..."}}
{{MissingLinks}}
:$\ds \int_a^b \lambda \map s x + \mu \map t x \rd x = \lambda \int_a^b \map s x \rd x + \mu \int_a^b \map t x \rd x$
Under any partition, the lower Darboux sums and upper Darboux sums of $f$ and $g$ are step functions, so the above formula relates the lower Darboux sums and upper Darboux sums of $f$ and $g$ to the lower Darboux sums and upper Darboux sums of the linear combinations of $f$ and $g$.
Because this identity is preserved for all possible partitions of $\closedint a b$, it is preserved for the supremum and infimum of all possible lower Darboux sums and upper Darboux sums, so the linear combinations of $f$ and $g$ are integrable.
{{begin-eqn}}
{{eqn | l = \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t
      | r = \sup \set {\sum_{\nu \mathop = 1}^n \map {m_\nu^{\paren {\lambda f + \mu g} } } {x_\nu - x_{\nu - 1} }: \forall \nu \in \closedint 1 n x_\nu > x_{\nu - 1} }
      | c = 
}}
{{eqn | r = \lambda \sup \set {\sum_{\nu \mathop = 1}^n \map {m_\nu^{\paren f} } {x_\nu - x_{\nu - 1} }: \forall \nu \in \closedint 1 n x_\nu > x_{\nu - 1} } + \mu \sup \set {\sum_{\nu \mathop = 1}^n \map {m_\nu^{\paren g} } {x_\nu - x_{\nu - 1} }: \forall \nu \in \closedint 1 n x_\nu > x_{\nu - 1} }
      | c = 
}}
{{eqn | r = \lambda \int_a^b \map f x \rd x + \mu \int_a^b \map g x \rd x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Correspondence between Abelian Groups and Z-Modules/Isomorphism of Categories

Let $\Z$ be the ring of integers.
Let $\mathbf{Ab}$ be the category of abelian groups.
Let $\mathbf{\mathbb Z-Mod}$ be the category of unitary $\Z$-modules.
Then the:
* forgetful functor $\mathbf{\mathbb Z-Mod} \to \mathbf{Ab}$
* associated Z-module functor $\mathbf{Ab} \to \mathbf{\mathbb Z-Mod}$
are strict inverse functors.
In particular, $\mathbf{Ab}$ and $\mathbf{\mathbb Z-Mod}$ are isomorphic.

Proof:
{{proof wanted}}
Category:Examples of Isomorphisms of Categories
Category:Category of Modules
Category:Category of Abelian Groups
tvzmplj60h6ilgkmjxv76mf8rkg4nml
Theorem: Correspondence between Abelian Groups and Z-Modules/Homomorphisms

Let $G, H$ be abelian groups.
Let $f : G \to H$ be a mapping.
{{TFAE}}
:$(1): \quad f$ is a group homomorphism.
:$(2): \quad f$ is a $\Z$-module homomorphism between the $\Z$-modules associated with $G$ and $H$.

Proof:
{{proof wanted}}
Category:Z-Module Associated with Abelian Group
Category:Abelian Groups
Category:Group Homomorphisms
Category:Linear Transformations
6tpem2e4d6yhy6iek9epmlpiw2q2d3t
Theorem: Definite Integral of Constant Multiple of Real Function

Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.
{{mistake|$f$ is only integrable here, but theorems used requiring that $f$ is continuous are used in both proofs}}
Let $c \in \R$ be a real number.
Then:
:$\ds \int_a^b c \map f x \rd x = c \int_a^b \map f x \rd x$

Proof:
Let $F$ be a primitive of $f$ on $\closedint a b$.
By Primitive of Constant Multiple of Function, $H = c F$ is a primitive of $c f$ on $\closedint a b$.
Hence by the Fundamental Theorem of Calculus:
{{begin-eqn}}
{{eqn | l = \int_a^b c \map f x \rd x
      | r = \bigintlimits {c \map F x} a b
      | c = 
}}
{{eqn | r = c \bigintlimits {\map F x} a b
      | c = 
}}
{{eqn | r = c \int_a^b \map f x \rd x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Definite Integral of Constant Multiple of Real Function

Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.
{{mistake|$f$ is only integrable here, but theorems used requiring that $f$ is continuous are used in both proofs}}
Let $c \in \R$ be a real number.
Then:
:$\ds \int_a^b c \map f x \rd x = c \int_a^b \map f x \rd x$

Proof:
Let $F$ be a primitive of $f$ on $\closedint a b$.
By Linear Combination of Definite Integrals:
:$\ds \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t$
for real functions $f$ and $g$ which are integrable on the closed interval $\closedint a b$, where $\lambda$ and $\mu$ be real numbers.
The result follows by setting $\lambda = c$ and $\mu = 0$.
{{qed}}
Theorem: Mapping to Singleton is Unique

Let $S$ be a set.
Let $T$ be a singleton.
Then there exists a unique mapping $S \to T$.

Proof:
Let $T = \set t$.
Let $f$ and $g$ both be mappings from $S$ to $T$.
From Mapping is Constant iff Image is Singleton:
:$\forall s \in S: \map f s = t$
and:
:$\forall s \in S: \map g s = t$
The result follows by Equality of Mappings.
{{qed}}
Theorem: Mean Value Theorem for Integrals/Generalization

Let $f$ and $g$ be continuous real functions on the closed interval $\closedint a b$ such that:
:$\forall x \in \closedint a b: \map g x \ge 0$
Then there exists a real number $k \in \closedint a b$ such that:
:$\ds \int_a^b \map f x \map g x \rd x = \map f k \int_a^b \map g x \rd x$

Proof:
Let:
:$\ds \int_a^b \map g x \rd x = 0$
We are given that:
:$\forall x \in \closedint a b: \map g x \ge 0$
Hence by Continuous Non-Negative Real Function with Zero Integral is Zero Function:
:$\forall x \in \closedint a b: \map g x = 0$
Hence:
:$\ds \int_a^b \map f x \cdot 0 \rd x = \map f k \cdot 0$
and so the result holds for any choice of $k$.
{{qed|lemma}}
Let:
:$\ds \int_a^b \map g x \rd x \ne 0$
From Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint a b$.
By the Extreme Value Theorem, there exist $m, M \in \closedint a b$ such that:
:$\ds \map f m = \min_{x \mathop \in \closedint a b} \map f x$
:$\ds \map f M = \max_{x \mathop \in \closedint a b} \map f x$
Then the following inequality holds for all $x$ in $\closedint a b$:
:$\map f m \le \map f x \le \map f M$
Multiplying by $\map g x$, and using that:
:$\forall x \in \closedint a b: \map g x \ge 0$
we get:
:$\map f m \map g x \le \map f x \map g x \le \map f M \map g x$
Integrating from $a$ to $b$ gives:
:$\ds \int_a^b \map f m \map g x \rd x \le \int_a^b \map f x \map g x \rd x \le \int_a^b \map f M \map g x \rd x$
By Linear Combination of Definite Integrals:
:$\ds \map f m \int_a^b \map g x \rd x \le \int_a^b \map f x \map g x \rd x \le \map f M \int_a^b \map g x \rd x$
Dividing by $\ds \int_a^b \map g x \rd x$ gives:
:$\ds \map f m \le \dfrac {\int_a^b \map f x \map g x \rd x} {\int_a^b \map g x \rd x} \le \map f M $
By the Intermediate Value Theorem, there exists some $k \in \openint a b$ such that:
{{begin-eqn}}
{{eqn | l = \dfrac {\int_a^b \map f x \map g x \rd x} {\int_a^b  \map g x \rd x}
      | r = \map f k
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int_a^b \map f x \map g x \rd x
      | r = \map f k \int_a^b \map g x \rd x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Rectangular Formula for Definite Integrals

Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.
Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ form a normal subdivision of $\closedint a b$:
:$\forall r \in \set {1, 2, \ldots, n}: x_r - x_{r - 1} = \dfrac {b - a} n$
Then the definite integral of $f$ {{WRT|Integration}} $x$ from $a$ to $b$ can be approximated as:
:$\ds \int_a^b \map f x \rd x \approx \sum_{r \mathop = 0}^{n - 1} h \map f {x_r}$
where $h = \dfrac {b - a} n$.

Proof:
{{ProofWanted|This will probably boil down to a graphical approach based on the structure of a Darboux integral. We need to explain rigorously what an "approximation" means, and we may also want to quantify the error.}}
Theorem: Trapezium Rule for Definite Integrals

Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.
Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ form a normal subdivision of $\closedint a b$:
:$\forall i \in \set {1, 2, \ldots, n}: x_i - x_{i - 1} = \dfrac {b - a} n$
Then the definite integral of $f$ {{WRT|Integration}} $x$ from $a$ to $b$ can be approximated  as:
:$\ds \int_a^b \map f x \rd x \approx \dfrac h 2 \paren {\map f {x_0} + \map f {x_n} + \sum_{i \mathop = 1}^{n - 1} 2 \map f {x_i} }$
where $h = \dfrac {b - a} n$.

Proof:
The geometric interpretation of a definite integral states that the area between the $4$ lines $x = a$, $x = b$, $y = 0$ and $y = \map f x$ is equal to $\ds \int_a^b \map f x \rd x$.
We approximate this area by dividing it into trapezia:
:460px
Consider the trapezium $T_i$ whose vertices are $\tuple {x_i, \map f {x_i}, x_{i + 1}, \map f {x_{i + 1} } }$ for some $i \in \set {0, 1, \ldots, n}$.
The area of $T_i$ is given by Area of Trapezium as:
:$\map \Area {T_i} = \dfrac {\map f {x_i} + \map f {x_{i + 1} } } 2 h$
Now let us consider the summation of all such areas:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^{n - 1} \map \Area {T_i}
      | r = \sum_{i \mathop = 0}^{n - 1} \dfrac {\map f {x_i} + \map f {x_{i + 1} } } 2 h
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 0}^{n - 1} \dfrac h 2 \paren {\map f {x_i} + \map f {x_{i + 1} } }
      | c = 
}}
{{eqn | r = \dfrac h 2 \paren {\map f {x_0} + \map f {x_n} + \sum_{i \mathop = 1}^{n - 1} 2 \map f {x_i} }
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Simpson's Rule

Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.
:$\ds \int_a^b \map f x \rd x \approx \dfrac {b - a} 6 \paren {\map f a + 4 \map f {\dfrac {a + b} 2} + \map f b}$
Hence the area under the curve is approximated by the area under the quadratic polynomial passing through $\tuple {x, \map f x}$ for the $3$ values $x = a$, $x = \dfrac {a + b} 2$ and $x = b$.
=== Repeated Simpson's Rule ===
{{:Simpson's Rule/Repeated}}

Proof:
{{ProofWanted|Graphical approach based on approximating the area under the curve as a series of parabolas.}}
Theorem: Definite Integral to Infinity of Reciprocal of x Squared plus a Squared

:$\ds \int_0^\infty \dfrac {\d x} {x^2 + a^2} = \frac \pi {2 a}$
for $a \ne 0$.

Proof:
{{begin-eqn}}
{{eqn | l = \int_0^\infty \dfrac {\d x} {x^2 + a^2}
      | r = \int_0^{\mathop \to +\infty} \dfrac {\d x} {x^2 + a^2}
      | c = 
}}
{{eqn | r = \lim_{\gamma \mathop \to +\infty} \int_0^\gamma \dfrac {\d x} {x^2 + a^2}
      | c = {{Defof|Improper Integral on Closed Interval Unbounded Above}}
}}
{{eqn | r = \lim_{\gamma \mathop \to +\infty} \intlimits {\frac 1 a \arctan \frac x a} 0 \gamma
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | r = \frac 1 a \lim_{\gamma \mathop \to +\infty} \paren {\arctan \frac \gamma a - \arctan 0}
      | c = 
}}
{{eqn | r = \frac 1 a \lim_{\gamma \mathop \to +\infty} \arctan \frac \gamma a
      | c = Arctangent of Zero is Zero
}}
{{eqn | r = \frac 1 a \frac \pi 2
      | c = Arctangent Tends to Half Pi as Argument Tends to Infinity
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Definite Integral to Infinity of Reciprocal of x Squared plus a Squared

:$\ds \int_0^\infty \dfrac {\d x} {x^2 + a^2} = \frac \pi {2 a}$
for $a \ne 0$.

Proof:
{{begin-eqn}}
{{eqn	| l = \int_0^\infty \dfrac {\d x} {x^2 + a^2}
	| r = \frac \pi { 2 a^{2 - 1} } \csc \left({\frac \pi 2}\right)
	| c = Definite Integral to Infinity of $\dfrac 1 {1 + x^n}$: Corollary
}}
{{eqn	| r = \frac \pi {2 a} 
	| c = Sine of Right Angle
}}
{{end-eqn}}
{{qed}}