# User:JaviPrieto/Integrals

${\displaystyle \int _{a}^{b}\!f(x)\,dx\,}$
${\displaystyle \int _{a}^{b}\!f(x)\,dx=F(b)-F(a)\,}$

Isaac Newton used a small vertical bar above a variable to indicate integration, or placed the variable inside a box. The vertical bar was easily confused with ${\displaystyle {\dot {x}}}$ or ${\displaystyle x'\,\!}$, which Newton used to indicate differentiation, and the box notation was difficult for printers to reproduce, so these notations were not widely adopted.

${\displaystyle \int _{a}^{b}f(x)\,dx.}$
${\displaystyle \int _{0}^{1}{\sqrt {x}}\,dx\,\!.}$
${\displaystyle \textstyle {\sqrt {\frac {1}{5}}}\left({\frac {1}{5}}-0\right)+{\sqrt {\frac {2}{5}}}\left({\frac {2}{5}}-{\frac {1}{5}}\right)+\cdots +{\sqrt {\frac {5}{5}}}\left({\frac {5}{5}}-{\frac {4}{5}}\right)\approx 0.7497.\,\!}$
${\displaystyle \int _{0}^{1}{\sqrt {x}}\,dx=\int _{0}^{1}x^{\frac {1}{2}}\,dx=F(1)-F(0)={\textstyle {\frac {2}{3}}}.}$
${\displaystyle \int f(x)\,dx\,\!}$
${\displaystyle \int _{A}f(x)\,d\mu \,\!}$
${\displaystyle \int _{A}d\omega =\int _{\partial A}\omega ,\,\!}$
${\displaystyle a=x_{0}\leq t_{1}\leq x_{1}\leq t_{2}\leq x_{2}\leq \cdots \leq x_{n-1}\leq t_{n}\leq x_{n}=b.\,\!}$
${\displaystyle \sum _{i=1}^{n}f(t_{i})\Delta _{i};}$
${\displaystyle \left|S-\sum _{i=1}^{n}f(t_{i})\Delta _{i}\right|<\epsilon .}$
${\displaystyle \int 1_{A}d\mu =\mu (A)}$.
{\displaystyle {\begin{aligned}\int s\,d\mu &{}=\int \left(\sum _{i=1}^{n}a_{i}1_{A_{i}}\right)d\mu \\&{}=\sum _{i=1}^{n}a_{i}\int 1_{A_{i}}\,d\mu \\&{}=\sum _{i=1}^{n}a_{i}\,\mu (A_{i})\end{aligned}}}
${\displaystyle \int _{E}s\,d\mu =\sum _{i=1}^{n}a_{i}\,\mu (A_{i}\cap E).}$
${\displaystyle \int _{E}f\,d\mu =\sup \left\{\int _{E}s\,d\mu \,\colon 0\leq s\leq f{\text{ and }}s{\text{ is a simple function}}\right\};}$
{\displaystyle {\begin{aligned}f^{+}(x)&{}={\begin{cases}f(x),&{\text{if }}f(x)>0\\0,&{\text{otherwise}}\end{cases}}\\f^{-}(x)&{}={\begin{cases}-f(x),&{\text{if }}f(x)<0\\0,&{\text{otherwise}}\end{cases}}\end{aligned}}}
${\displaystyle \int _{E}|f|\,d\mu <\infty ,\,\!}$
${\displaystyle \int _{E}f\,d\mu =\int _{E}f^{+}\,d\mu -\int _{E}f^{-}\,d\mu .\,\!}$
${\displaystyle f\mapsto \int _{a}^{b}f\;dx}$
${\displaystyle \int _{a}^{b}(\alpha f+\beta g)(x)\,dx=\alpha \int _{a}^{b}f(x)\,dx+\beta \int _{a}^{b}g(x)\,dx.\,}$
${\displaystyle f\mapsto \int _{E}fd\mu }$
${\displaystyle \int _{E}(\alpha f+\beta g)\,d\mu =\alpha \int _{E}f\,d\mu +\beta \int _{E}g\,d\mu .}$
${\displaystyle f\mapsto \int _{E}fd\mu ,\,}$
${\displaystyle m(b-a)\leq \int _{a}^{b}f(x)\,dx\leq M(b-a).}$
${\displaystyle \int _{a}^{b}f(x)\,dx\leq \int _{a}^{b}g(x)\,dx.}$
${\displaystyle \int _{a}^{b}f(x)\,dx<\int _{a}^{b}g(x)\,dx.}$
${\displaystyle \int _{c}^{d}f(x)\,dx\leq \int _{a}^{b}f(x)\,dx.}$
${\displaystyle (fg)(x)=f(x)g(x),\;f^{2}(x)=(f(x))^{2},\;|f|(x)=|f(x)|.\,}$
${\displaystyle \left|\int _{a}^{b}f(x)\,dx\right|\leq \int _{a}^{b}|f(x)|\,dx.}$
${\displaystyle \left(\int _{a}^{b}(fg)(x)\,dx\right)^{2}\leq \left(\int _{a}^{b}f(x)^{2}\,dx\right)\left(\int _{a}^{b}g(x)^{2}\,dx\right).}$
${\displaystyle \left|\int f(x)g(x)\,dx\right|\leq \left(\int \left|f(x)\right|^{p}\,dx\right)^{1/p}\left(\int \left|g(x)\right|^{q}\,dx\right)^{1/q}.}$
${\displaystyle \left(\int \left|f(x)+g(x)\right|^{p}\,dx\right)^{1/p}\leq \left(\int \left|f(x)\right|^{p}\,dx\right)^{1/p}+\left(\int \left|g(x)\right|^{p}\,dx\right)^{1/p}.}$
${\displaystyle \int _{a}^{b}f(x)\,dx}$
${\displaystyle \int _{a}^{b}f(x)\,dx=-\int _{b}^{a}f(x)\,dx.}$
${\displaystyle \int _{a}^{a}f(x)\,dx=0.}$
${\displaystyle \int _{a}^{b}f(x)\,dx=\int _{a}^{c}f(x)\,dx+\int _{c}^{b}f(x)\,dx.}$
{\displaystyle {\begin{aligned}\int _{a}^{c}f(x)\,dx&{}=\int _{a}^{b}f(x)\,dx-\int _{c}^{b}f(x)\,dx\\&{}=\int _{a}^{b}f(x)\,dx+\int _{b}^{c}f(x)\,dx\end{aligned}}}
${\displaystyle \int _{M}\omega =-\int _{M'}\omega \,.}$

These conventions correspond to interpreting the integrand as a differential form, integrated over a chain. In measure theory, by contrast, one interprets the integrand as a function f with respect to a measure ${\displaystyle \mu ,}$ and integrates over a subset A, without any notion of orientation; one writes ${\displaystyle \textstyle {\int _{A}f\,d\mu =\int _{[a,b]}f\,d\mu }}$ to indicate integration over a subset A. This is a minor distinction in one dimension, but becomes subtler on higher dimensional manifolds; see Differential form: Relation with measures for details.

${\displaystyle F(x)=\int _{a}^{x}f(t)\,dt.}$
${\displaystyle \int _{a}^{b}f(t)\,dt=F(b)-F(a).}$
The improper integral
${\displaystyle \int _{0}^{\infty }{\frac {dx}{(x+1){\sqrt {x}}}}=\pi }$
has unbounded intervals for both domain and range.
${\displaystyle \int _{a}^{\infty }f(x)dx=\lim _{b\to \infty }\int _{a}^{b}f(x)dx}$
${\displaystyle \int _{a}^{b}f(x)dx=\lim _{\epsilon \to 0}\int _{a+\epsilon }^{b}f(x)dx}$

Consider, for example, the function ${\displaystyle {\tfrac {1}{(x+1){\sqrt {x}}}}}$ integrated from 0 to 8 (shown right). At the lower bound, as x goes to 0 the function goes to 8, and the upper bound is itself 8, though the function goes to 0. Thus this is a doubly improper integral. Integrated, say, from 1 to 3, an ordinary Riemann sum suffices to produce a result of ${\displaystyle {\tfrac {\pi }{6}}}$. To integrate from 1 to 8, a Riemann sum is not possible. However, any finite upper bound, say t (with t > 1), gives a well-defined result, ${\displaystyle {\tfrac {\pi }{2}}-2\arctan {\tfrac {1}{\sqrt {t}}}}$. This has a finite limit as t goes to infinity, namely ${\displaystyle {\tfrac {\pi }{2}}}$. Similarly, the integral from 1/3 to 1 allows a Riemann sum as well, coincidentally again producing ${\displaystyle {\tfrac {\pi }{6}}}$. Replacing 1/3 by an arbitrary positive value s (with s < 1) is equally safe, giving ${\displaystyle -{\tfrac {\pi }{2}}+2\arctan {\tfrac {1}{\sqrt {s}}}}$. This, too, has a finite limit as s goes to zero, namely ${\displaystyle {\tfrac {\pi }{2}}}$. Combining the limits of the two fragments, the result of this improper integral is

{\displaystyle {\begin{aligned}\int _{0}^{\infty }{\frac {dx}{(x+1){\sqrt {x}}}}&{}=\lim _{s\to 0}\int _{s}^{1}{\frac {dx}{(x+1){\sqrt {x}}}}+\lim _{t\to \infty }\int _{1}^{t}{\frac {dx}{(x+1){\sqrt {x}}}}\\&{}=\lim _{s\to 0}\left(-{\frac {\pi }{2}}+2\arctan {\frac {1}{\sqrt {s}}}\right)+\lim _{t\to \infty }\left({\frac {\pi }{2}}-2\arctan {\frac {1}{\sqrt {t}}}\right)\\&{}={\frac {\pi }{2}}+{\frac {\pi }{2}}\\&{}=\pi .\end{aligned}}}

This process does not guarantee success; a limit may fail to exist, or may be unbounded. For example, over the bounded interval 0 to 1 the integral of ${\displaystyle {\tfrac {1}{x}}}$ does not converge; and over the unbounded interval 1 to 8 the integral of ${\displaystyle {\tfrac {1}{\sqrt {x}}}}$ does not converge.

The improper integral
${\displaystyle \int _{-1}^{1}{\frac {dx}{\sqrt[{3}]{x^{2}}}}=6}$
is unbounded internally, but both left and right limits exist.
{\displaystyle {\begin{aligned}\int _{-1}^{1}{\frac {dx}{\sqrt[{3}]{x^{2}}}}&{}=\lim _{s\to 0}\int _{-1}^{-s}{\frac {dx}{\sqrt[{3}]{x^{2}}}}+\lim _{t\to 0}\int _{t}^{1}{\frac {dx}{\sqrt[{3}]{x^{2}}}}\\&{}=\lim _{s\to 0}3(1-{\sqrt[{3}]{s}})+\lim _{t\to 0}3(1-{\sqrt[{3}]{t}})\\&{}=3+3\\&{}=6.\end{aligned}}}
${\displaystyle \int _{-1}^{1}{\frac {dx}{x}}\,\!}$
${\displaystyle \int _{E}f(x)\,dx.}$
${\displaystyle \iint _{D}5\ dx\,dy}$
${\displaystyle \int _{4}^{9}\int _{2}^{7}\ 5\ dx\,dy}$
From here, integration is conducted with respect to either x or y first; in this example, integration is first done with respect to x as the interval corresponding to x is the inner integral. Once the first integration is completed via the ${\displaystyle F(b)-F(a)}$ method or otherwise, the result is again integrated with respect to the other variable. The result will equate to the volume under the surface.
${\displaystyle \iiint _{\mathrm {cuboid} }1\,dx\,dy\,dz}$
${\displaystyle W={\vec {F}}\cdot {\vec {s}}.}$

For an object moving along a path in a vector field ${\displaystyle {\vec {F}}}$ such as an electric field or gravitational field, the total work done by the field on the object is obtained by summing up the differential work done in moving from ${\displaystyle {\vec {s}}}$ to ${\displaystyle {\vec {s}}+d{\vec {s}}}$. This gives the line integral

${\displaystyle W=\int _{C}{\vec {F}}\cdot d{\vec {s}}.}$
${\displaystyle \int _{S}{\mathbf {v} }\cdot \,d{\mathbf {S} }.}$
${\displaystyle \int _{S}f\,dx^{1}\cdots dx^{m}.}$
${\displaystyle dx^{a}\wedge dx^{a}=0\,\!}$
${\displaystyle d\omega =\sum _{i=1}^{n}{\frac {\partial f}{\partial x_{i}}}dx^{i}\wedge dx^{a}.}$
${\displaystyle \int _{\Omega }d\omega =\int _{\partial \Omega }\omega \,\!}$

The most basic technique for computing definite integrals of one real variable is based on the fundamental theorem of calculus. Let f(x) be the function of x to be integrated over a given interval [a, b]. Then, find an antiderivative of f; that is, a function F such that F' = f on the interval. By the fundamental theorem of calculus—provided the integrand and integral have no singularities on the path of integration—${\displaystyle \textstyle \int _{a}^{b}f(x)\,dx=F(b)-F(a).}$

${\displaystyle \int _{-2}^{2}{\tfrac {1}{5}}\left({\tfrac {1}{100}}(322+3x(98+x(37+x)))-24{\frac {x}{1+x^{2}}}\right)dx,}$

| colspan="4" | ${\displaystyle \textstyle \int _{-2.25}^{1.75}f(x)\,dx=4.1639019006585897075\ldots }$

${\displaystyle \int _{a}^{b}f(x)\,dx\approx {\frac {b-a}{6}}\left[f(a)+4f\left({\frac {a+b}{2}}\right)+f(b)\right],}$
${\displaystyle \left|-{\frac {(b-a)^{5}}{2880}}f^{(4)}(\xi )\right|.}$