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

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.
 * $$F(x) = \int_a^x f(t)\, dt.$$
 * $$\int_a^b f(t)\, dt = F(b) - F(a).$$


 * $$\int_{a}^{\infty} f(x)dx = \lim_{b \to \infty} \int_{a}^{b} f(x)dx$$
 * $$\int_{a}^{b} f(x)dx = \lim_{\epsilon \to 0} \int_{a+\epsilon}^{b} f(x)dx$$

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 &gt; 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 &lt; 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.
 * $$\begin{align} \int_{-1}^{1} \frac{dx}{\sqrt[3]{x^2}} &{} = \lim_{s \to 0} \int_{-1}^{-s} \frac{dx}{\sqrt[3]{x^2}}  + \lim_{t \to 0} \int_{t}^{1} \frac{dx}{\sqrt[3]{x^2}} \\  &{} = \lim_{s \to 0} 3(1-\sqrt[3]{s}) + \lim_{t \to 0} 3(1-\sqrt[3]{t}) \\  &{} = 3 + 3 \\  &{} = 6.\end{align}$$
 * $$ \int_{-1}^{1} \frac{dx}{x} \,\!$$
 * $$\int_E f(x) \, dx.$$
 * $$\iint_D 5 \ dx\, dy$$
 * $$\int_4^9 \int_2^7 \ 5 \ dx\, dy$$
 * 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&mdash;provided the integrand and integral have no singularities on the path of integration&mdash;$$\textstyle\int_a^b f(x)\,dx = F(b)-F(a).$$
 * $$ \int_{-2}^{2} \tfrac15 \left( \tfrac{1}{100}(322 + 3 x (98 + x (37 + x))) - 24 \frac{x}{1+x^2} \right) dx, $$


 * colspan="4" | $$\textstyle \int_{-2.25}^{1.75} f(x)\,dx = 4.1639019006585897075\ldots$$
 * $$ \int_a^b f(x) \, dx \approx \frac{b-a}{6}\left[f(a) + 4f\left(\frac{a+b}{2}\right)+f(b)\right],$$
 * $$ \left|-\frac{(b-a)^5}{2880} f^{(4)}(\xi)\right|.$$