# Integral Calculas

## Integral Calculas

• In the differential calculus, we are given a function and we have to find the derivative or differential of this function.
• Integral calculus is the inverse of differential calculus
• Here we are to find a function whose differential is given.
• Integration is a process which is the inverse of differentiation
• While differential calculus concerns rates of change and slopes of curves (represented by derivatives), integral calculus deals with areas under curves and the accumulation of quantities.

## Indefinite Integral or Antiderivative

The process of finding the function when its derivative is given is called the indefinite integral. The notation used is:
$\int f(x) \, dx$
let $\frac {d}{dx} F(x) =f(x)$
Then we can write
$\int f(x) \, dx = F(x) + C$
C is called constant of integration. All these integrals differ by a constant
The result of the indefinite integral is always a family of functions since a constant can be added to any function without changing its derivative.

Example
$\frac {d}{dx} (x^2 + 1) =2x$
$\frac {d}{dx} (x^2 + 5) =2x$
$\frac {d}{dx} (x^2 + 5/2) =2x$
So
$\int (2x) \, dx = x^2 + C$
Here $C \in R$

## Geometrical interpretation of indefinite integral

Indefinite integral is always a family of curves .The different values of C will correspond to different members of this family and these members can be obtained by shifting any one of the curves parallel to itself.
Lets take the previous example
$\int (2x) \, dx = x^2 + C$
The different curves will be
$y=x^2 +1$
$y=x^2 -1$
$y=x^2 +2$
$y=x^2 + 3$
These are all parabola as shown below. Each of which can be obtained by the shifting any one of the curves parallel to itself

Further if we find the tangents of all these curves at a point x=a on the curve, we have that
$\frac {dy}{dx} _{x=a} = 2a$
So all the tangents will be parallel at the point x=a
Thus
$\int f(x) \, dx = F(x) + C=F_c(x)$
implies that the tangents to all the curves $y = F_C (x)$, $C in R$, at the points of intersection of the curves by the line x = a, ($a \in R$), are parallel

## Properties of indefinite integral

Property I: The process of differentiation and integration are inverses of each other
$\frac {d}{dx} \int f(x) dx= f(x)$
and $\int f'(x) dx= f(x) + C$
Proof
Let
$\frac {d}{dx} F(x) = f(x)$
Then
$\int f(x) \; dx = F(x) + C$
Now
$\frac {d}{dx} \int f(x) \; dx = \frac {d}{dx} F(x) =f(x)$

Property II: Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent
Proof
Let f and g be two functions such that
$\frac {d}{dx} \int f(x) \; dx=\frac {d}{dx} \int g(x) \; dx$
$\frac {d}{dx} [\int f(x) dx -\int g(x) dx] =0$
or
$\int f(x) dx -\int g(x) dx =C$
$\int f(x) dx= \int g(x) dx +C$
So families are cuves ($\int f(x) dx + C_1$) and ($\int g(x) dx + C_2$) are equivalent
Hence in this $\int f(x) dx$ and $\int g(x) dx$ are equivalent

Property III
$\int [f(x) + g(x)]\; dx= \int f(x) \; dx + \int g(x) \; dx$
Proof
LHS
$\frac {d}{dx} \int [f(x) + g(x)] dx= f(x) + g(x)$
RHS
$\frac {d}{dx} [\int f(x) dx + \int g(x) dx]=\frac {d}{dx} \int f(x) dx + \frac {d}{dx} \int g(x) dx =f(x) + g(x)$
From Property II, then
$\int [f(x) + g(x)] \; dx= \int f(x) \; dx + \int g(x) \; dx$

Similarly , we can prove
$\int [f(x) - g(x)]\; dx= \int f(x) \; dx - \int g(x) \; dx$

Property IV
$\int k f(x) \; dx= k\int f(x) \; dx$
Proof
LHS
$\frac {d}{dx} \int k f(x) dx= k f(x)$
RHS
$\frac {d}{dx} k\int f(x) \; dx = k \frac {d}{dx} \int f(x) \; dx = k f(x)$
From Property II, then
$\int k f(x) \; dx= k\int f(x) \; dx$

## Basic Integration formulas

$\int (c) \; dx = x + C$  ( Where c is a constant)
$\int (cx) \; dx = \frac {cx^2}{2} + C$ ( Where c is a constant)
$\int (x^n) \; dx = \frac {x^{n+1}}{n+1}$
$\int (e^x) \; dx = e^x + C$
$\int (\frac {1}{x}) \; dx = ln |x| + c$
$\int (a^x) \; dx = \frac {a^x}{ log a} + C$
$\int (log_{a} x) \; dx=\frac {1}{x ln a} + C$

## Integration formulas for Trigonometric Functions

$\int (\cos x) \; dx= \sin x + C$
$\int (\sin x) \; dx= - \cos x + C$
$\int ( \sec^2 x) \; dx = \tan x + C$
$\int (\csc^2 x) \; dx = -\cot x + C$
$\int ( \sec (x) \tan (x) ) \; dx =\sec x + C$
$\int (\csc (x) \cot (x)) \; dx = -\csc x + C$
$\int (\tan x) \; dx = ln |\sec x| + C$
$\int (\cot x) \; dx = ln |\sin x| + C$
$\int (\sec x) \; dx = ln |\sec x + \tan x| + C$
$\int (\csc x) \; dx = ln |\csc x - \cot x| + C$

## Integration formulas Related to Inverse Trigonometric Functions

$\int ( \frac {1}{\sqrt {1-x^2} } ) \; dx= \sin^{-1}x + C$
$\int (\frac {1}{\sqrt {1-x^2}}) \; dx = - \cos ^{-1}x +C$
$\int ( \frac {1}{1 + x^2}) \; dx =\tan ^{-1}x + C$
$\int ( \frac {1}{1 + x^2}) \; dx = -\cot ^{-1}x + C$
$\int (\frac {1}{|x|\sqrt {x^-1}}) \; dx = -sec^{-1} x + C$
$\int (\frac {1}{|x|\sqrt {x^-1}}) \; dx = -cosec^{-1} x + C$

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