# Question #89633

Sep 11, 2016

Integration refers to the inverse process of differentiation. Allow me to illustrate it briefly in the section below.

#### Explanation:

So example, we have a single valued and differentiable function $f \left(x\right)$

The differential coefficient or simply the derivative of $f \left(x\right)$ is defined as,

${f}_{1} \left(x\right) = \frac{\mathrm{df}}{\mathrm{dx}}$ which can be defined from the first principle. Where the subscript 1 denotes that it is the first derivative of $f$.

Now, let $F \left(x\right) = {f}_{1} \left(x\right)$ denote the derivative of $f \left(x\right)$

We define the integral of $F \left(x\right)$ as,

$\int F \left(x\right) \mathrm{dx} = f \left(x\right) + C$ where $C$ is called the constant of integration.

$C$ is completely arbitrary and may be defined in some cases by the boundary conditions of the particular problem.

Since, the integral of $F \left(x\right)$ can have an indefinite number of forms just differing by the value of the constant $C$, this is called the indefinite integral of $F \left(x\right)$.

So far we are done with the basic definition of integration.

I would like to extend the discussion to the definition of the definite integral.

The definite integral of $F \left(x\right)$ from $a$ to $b$ is simple defined as,

${\int}_{a}^{b} F \left(x\right) \mathrm{dx} = {\left[f \left(x\right)\right]}_{a}^{b}$

Now, the fundamental theorem of integral calculus states that,

${\int}_{a}^{b} F \left(x\right) \mathrm{dx} = f \left(b\right) - f \left(a\right)$.

There are simple rules for determining the integral of a function. We you are dealing with the indefinite integral, just put a constant and if you're dealing with the definite integral, use the fundamental theorem of integral calculus.

A more rigorous and geometric interpretation of definite integral is provided in most elementary calculus textbooks. Please consult one if you like.