# User:JaviPrieto/Integrals

$\int_a^b \! f(x)\,dx \,$
$\int_a^b \! f(x)\,dx = F(b) - F(a)\,$

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 $\dot{x}$ or $x'\,\!$, which Newton used to indicate differentiation, and the box notation was difficult for printers to reproduce, so these notations were not widely adopted.

$\int_a^b f(x)\,dx .$
$\int_0^1 \sqrt x \, dx \,\!.$
$\textstyle \sqrt {\frac {1} {5}} \left ( \frac {1} {5} - 0 \right ) + \sqrt {\frac {2} {5}} \left ( \frac {2} {5} - \frac {1} {5} \right ) + \cdots + \sqrt {\frac {5} {5}} \left ( \frac {5} {5} - \frac {4} {5} \right ) \approx 0.7497.\,\!$
$\int_0^1 \sqrt x \,dx = \int_0^1 x^{\frac{1}{2}} \,dx = F(1)- F(0) = {\textstyle \frac 2 3}.$
$\int f(x) \, dx \,\!$
$\int_A f(x) \, d\mu \,\!$
$\int_{A} d\omega = \int_{\part A} \omega , \,\!$
$a = x_0 \le t_1 \le x_1 \le t_2 \le x_2 \le \cdots \le x_{n-1} \le t_n \le x_n = b . \,\!$
$\sum_{i=1}^{n} f(t_i) \Delta_i ;$
$\left| S - \sum_{i=1}^{n} f(t_i)\Delta_i \right| < \epsilon.$
$\int 1_A d\mu = \mu(A)$.
\begin{align} \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{align}
$\int_E s \, d\mu = \sum_{i=1}^{n} a_i \, \mu(A_i \cap E) .$
$\int_E f \, d\mu = \sup\left\{\int_E s \, d\mu\, \colon 0 \leq s\leq f\text{ and } s\text{ is a simple function}\right\};$
\begin{align} 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{align}
$\int_E |f| \, d\mu < \infty , \,\!$
$\int_E f \, d\mu = \int_E f^+ \, d\mu - \int_E f^- \, d\mu . \,\!$
$f \mapsto \int_a^b f \; dx$
$\int_a^b (\alpha f + \beta g)(x) \, dx = \alpha \int_a^b f(x) \,dx + \beta \int_a^b g(x) \, dx. \,$
$f\mapsto \int_E f d\mu$
$\int_E (\alpha f + \beta g) \, d\mu = \alpha \int_E f \, d\mu + \beta \int_E g \, d\mu.$
$f\mapsto\int_E f d\mu, \,$
$m(b - a) \leq \int_a^b f(x) \, dx \leq M(b - a).$
$\int_a^b f(x) \, dx \leq \int_a^b g(x) \, dx.$
$\int_a^b f(x) \, dx < \int_a^b g(x) \, dx.$
$\int_c^d f(x) \, dx \leq \int_a^b f(x) \, dx.$
$(fg)(x)= f(x) g(x), \; f^2 (x) = (f(x))^2, \; |f| (x) = |f(x)|.\,$
$\left| \int_a^b f(x) \, dx \right| \leq \int_a^b | f(x) | \, dx.$
$\left( \int_a^b (fg)(x) \, dx \right)^2 \leq \left( \int_a^b f(x)^2 \, dx \right) \left( \int_a^b g(x)^2 \, dx \right).$
$\left|\int f(x)g(x)\,dx\right| \leq\left(\int \left|f(x)\right|^p\,dx \right)^{1/p} \left(\int\left|g(x)\right|^q\,dx\right)^{1/q}.$
$\left(\int \left|f(x)+g(x)\right|^p\,dx \right)^{1/p} \leq\left(\int \left|f(x)\right|^p\,dx \right)^{1/p} +\left(\int \left|g(x)\right|^p\,dx \right)^{1/p}.$
$\int_a^b f(x) \, dx$
$\int_a^b f(x) \, dx = - \int_b^a f(x) \, dx.$
$\int_a^a f(x) \, dx = 0.$
$\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx.$
\begin{align} \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{align}
$\int_M \omega = - \int_{M'} \omega \,.$

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 $\mu,$ and integrates over a subset A, without any notion of orientation; one writes $\textstyle{\int_A f\,d\mu = \int_{[a,b]} f\,d\mu}$ 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.

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

Consider, for example, the function $\tfrac{1}{(x+1)\sqrt{x}}$ 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 $\tfrac{\pi}{6}$. 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, $\tfrac{\pi}{2} - 2\arctan \tfrac{1}{\sqrt{t}}$. This has a finite limit as t goes to infinity, namely $\tfrac{\pi}{2}$. Similarly, the integral from 1/3 to 1 allows a Riemann sum as well, coincidentally again producing $\tfrac{\pi}{6}$. Replacing 1/3 by an arbitrary positive value s (with s < 1) is equally safe, giving $-\tfrac{\pi}{2} + 2\arctan\tfrac{1}{\sqrt{s}}$. This, too, has a finite limit as s goes to zero, namely $\tfrac{\pi}{2}$. Combining the limits of the two fragments, the result of this improper integral is

\begin{align} \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{align}

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 $\tfrac{1}{x}$ does not converge; and over the unbounded interval 1 to 8 the integral of $\tfrac{1}{\sqrt{x}}$ does not converge.

The improper integral
$\int_{-1}^{1} \frac{dx}{\sqrt[3]{x^2}} = 6$
is unbounded internally, but both left and right limits exist.
\begin{align} \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{align}
$\int_{-1}^{1} \frac{dx}{x} \,\!$
$\int_E f(x) \, dx.$
$\iint_D 5 \ dx\, dy$
$\int_4^9 \int_2^7 \ 5 \ dx\, dy$
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 $F(b) - F(a)$ method or otherwise, the result is again integrated with respect to the other variable. The result will equate to the volume under the surface.
$\iiint_\mathrm{cuboid} 1 \, dx\, dy\, dz$
$W=\vec F\cdot\vec s.$

For an object moving along a path in a vector field $\vec F$ 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 $\vec s$ to $\vec s + d\vec s$. This gives the line integral

$W=\int_C \vec F\cdot d\vec s.$
$\int_S {\mathbf v}\cdot \,d{\mathbf {S}}.$
$\int_S f\,dx^1 \cdots dx^m.$
$dx^a \wedge dx^a = 0 \,\!$
$d\omega = \sum_{i=1}^n \frac{\partial f}{\partial x_i} dx^i \wedge dx^a.$
$\int_{\Omega} d\omega = \int_{\partial\Omega} \omega \,\!$

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—$\textstyle\int_a^b f(x)\,dx = F(b)-F(a).$

$\int_{-2}^{2} \tfrac15 \left( \tfrac{1}{100}(322 + 3 x (98 + x (37 + x))) - 24 \frac{x}{1+x^2} \right) dx ,$

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

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