User:JaviPrieto/Integrals

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\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|.