What is Continuous Integration

In what is continuous integration, Let f (x) be continuous and single valued for a <= x <= b. Divide the interval (a, b) of the x axis into n parts at the points (a -= X_0), X_1, X_2… ( X_n -= b) as in below figure.

Define Delta x_i = x_i - x_(i-1) and let xi_i be a value of x in the interval x_(i-1) < xi_i <= x_i. Form the sum of the areas of each approximately rectangular area to get

sum_(i=1)^(n) f(xi_i) Delta x_i

continuous integration

If we take the limit as n ->oo and Delta x_i -> 0, the approximation of the area under the curve as a rectangle becomes more accurate and we obtain the definition of the definite integral of f (x) between the limits a, b; thus

^lim_((n -> oo)_(Delta x_i -> 0)) sum_(i=1)^n f(xi_i) Delta x_i -= int_a^b f(x)dx

The points a, b are referred to as the lower and upper limits of integration, respectively. The definite integral

int_a^b f(x) dx

equals the area under the curve f(x) in the interval (a, b) for what is continuous integration.

Fundamental Theorem - what is continuous integration:

In what is continuous integration, 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

int_a^b f(x)dx = G(b) - G(a)

The definite integral acts like an antiderivative of f(x), or the inverse operation to differentiation. The concept of inverse differentiation can be made more explicit by introducing the notation

[G(x)]_(a)^(b) = G(b) - G(a)

and using (dG)/(dx) = f(x) in the definite integral above equation to give

int_a^b ((dG)/(dx))dx = [G(x)]_a^b

Example - what is continuous integration:

In what is continuous integration, let f(y) = ln y, u(alpha) = a, and v(alpha) = alpha. In this case, the function f does not depend on alpha. Consequently, we substitute u, v, and f into find.

(d)/(d alpha) int_a^(alpha) In y dy = f(alpha)(d a)/(d alpha) = In alpha

because (da)/(d alpha) = 0 and f(alpha) = ln alpha. This result is verified by first evaluating the integral

int_a^b In y dy = [y In y –y ]_a^alpha = alpha In alpha - alpha - (a In a - a)

Taking the derivative of above equation with respect to alpha gives

(d)/(d alpha)[alpha In alpha - alpha - (a In a –a)] =(d)/(d alpha) [alpha In alpha - alpha]

Answer is     = In alpha