# Continuous Integration

By bringing together minuscule data integrals assign a number to a function in such a manner that describes volume, area and displacement. The two major operations of calculus are integration and its opposite that is differentiation.

Given a function f of a real variable x and an interval [a, b] of the real line, the definite integral

The connection between differential and **integral calculus** is through the fundamental theorem of calculus. In particular, if f(x) is continuous in the interval a <= x <=b and G(x) is a function such that (dG)/(dx) = f(x) for all values of x in (a, b), then

Let f be continuous on an interval I. Choose a point a in I. Then the function F(x) defined by

\(\large F(x)=\int_{a}^{x}f(t)\;dt\)

Let c be in I, and let x be infinitely close to c and between the endpoints of I. By the Addition Property,

\(\large \int_{a}^{c}f(t)\;dt=\int_{a}^{x}f(t)\;dt+\int_{x}^{c}f(t)\;dt,\)

\(\large \int_{a}^{c}f(t)\;dt-\int_{a}^{x}f(t)\;dt+\int_{x}^{c}f(t)\;dt,\)

\(\large f(c)-F(x)=\int_{x}^{c}f(t)\;dt.\)

Example to find out is continuous integration:

In what is continuous integration, let *f(y) *= *in y, u*(α) = *a* and *v*(α)=α

In this case, the function doesn’t depend on α Consequently, it has to be substitute with u, v, and f.

\(\frac{(d)}{(d \alpha)}\int_a^{\alpha}\;In,y,dy= \frac{(f\alpha)(d, a)}{(d\;\alpha)}\)