- 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.

\[ \int f(x) \, dx \]

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.

$\frac {d}{dx} (x^2 + 1) =2x$

$\frac {d}{dx} (x^2 + 5) =2x$

$\frac {d}{dx} (x^2 + 5/2) =2x$

So

$ \int (2x) \, dx = x^2 + C$

Here $C \in R$

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

$\frac {d}{dx} \int f(x) dx= f(x)$

and $ \int f'(x) dx= f(x) + C$

Let

$\frac {d}{dx} F(x) = f(x)$

Then

$\int f(x) \; dx = F(x) + C$

Now

$\frac {d}{dx} \int f(x) \; dx = \frac {d}{dx} F(x) =f(x)$

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

$\int [f(x) + g(x)]\; dx= \int f(x) \; dx + \int g(x) \; dx$

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

$\int [f(x) - g(x)]\; dx= \int f(x) \; dx - \int g(x) \; dx$

$\int k f(x) \; dx= k\int f(x) \; dx $

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 $

$\int (cx) \; dx = \frac {cx^2}{2} + C$ ( Where c is a constant)

$\int (x^n) \; dx = \frac {x^{n+1}}{n+1}$

$\int (e^x) \; dx = e^x + C$

$\int (\frac {1}{x}) \; dx = ln |x| + c$

$\int (a^x) \; dx = \frac {a^x}{ log a} + C$

$\int (log_{a} x) \; dx=\frac {1}{x ln a} + C$

$\int (\sin x) \; dx= - \cos x + C$

$\int ( \sec^2 x) \; dx = \tan x + C$

$\int (\csc^2 x) \; dx = -\cot x + C$

$\int ( \sec (x) \tan (x) ) \; dx =\sec x + C$

$\int (\csc (x) \cot (x)) \; dx = -\csc x + C$

$ \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$

$\int (\frac {1}{\sqrt {1-x^2}}) \; dx = - \cos ^{-1}x +C$

$\int ( \frac {1}{1 + x^2}) \; dx =\tan ^{-1}x + C$

$\int ( \frac {1}{1 + x^2}) \; dx = -\cot ^{-1}x + C$

$\int (\frac {1}{|x|\sqrt {x^-1}}) \; dx = -sec^{-1} x + C $

$\int (\frac {1}{|x|\sqrt {x^-1}}) \; dx = -cosec^{-1} x + C $

**Notes**-
**NCERT Solutions & Assignments**

Class 12 Maths Class 12 Physics Class 12 Chemistry Class 12 Biology