Anti Derivatives

Introduction to anti derivatives:

The process of solving for anti-derivatives is called anti-differentiation. It is also called as indefinite integration. The opposite function of differentiation is called anti-derivative, which is the process of finding a derivative. Anti-derivatives are related to definite integrals through the fundamental theorem of calculus.

Anti-differentiation is the process of finding the set of all anti - derivatives of a given function. The symbol `int ` denotes the operation of anti-differentiation.

` int` f(x) dx = F(x) + c 

where,         F'(x) = `d/dx` (F(x)) = f(x)

Formulas for Anti-derivatives:

1. `int` dx = x + c

2. `int ` a f(x) dx =  a `int` f(x) dx

3. ` int` x n dx = `(x^n+1) / (n+1)` + c

4. `int` `(1/x)` dx = log x + c

5. `int`[f(x) ± F(x)] dx =`int` f(x) dx ± `int`F(x) dx

6.  `int` e x dx = e x + c

7.` int ` cos x .dx = sin x. + c

8. `int` sin x.dx = - cos x + c

9. `int` sec2x. dx = tan x + c

10.  `int` cosec x. cot x. dx = -cosec x + c

11. `int` sec x . tan x .dx = sec x + c

12.  `int` cosec2 x .dx = -cot  x + c

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Anti-derivatives Problems:

problem 1:

If f(x) = 4x^3 - 6x^4 + x^2, find the anti-derivative of f(x).

Solution:

Given, anti-derivative of f(x) = `int` f(x) dx

=`int` (4x^3 - 6x^4 + x^2) dx.

`int`(4x^3 - 6x^4 + x^2) dx =`int` 4x^3dx -`int` 6x^4 dx + `int`x^2 dx.

= 4`int`x^3dx - 6`int`x^4 dx +`int`x^2 dx.

= 4`(x^4/4)` - 6`(x^5/5)` + `(x^3/3)` + c.

= `(4/4) ` x^4 - `(6/5)` x5 +  `(x^3/3)` + c.

`int` (4x^3 - 6x^4 + x^2) dx =  x^4 - `(6/5)` x5 +  `(x^3/3)` + c.

Answer:

`int` (4x^3 - 6x^4 + x^2) dx =  x^4 - `(6/5)` x5 +  `(x^3/3)` + c.

problem 2:

Find the anti-derivative of the quadratic expression:`int` (x^2- x - 4)1/4(2x -1) dx

Solution:

Let u = x^2 -  x - 4

Therefore, `(du)/dx` = 2x - 1       or    `(dx) / (du)` = `1/ (2x - 1)`

So, (2x - 1) dx = du 

Now, `int` (x^2 - x - 4)1/4(2x - 1)dx =`int`(x^2- x - 4)1/4 du                                   

=`int` u1/4du

= `(4/5) ` u5/4

= `(4/5) ` (x^2- x - 4)5/4 + c

Answer:

`int` (x^2- x - 4)1/4 (2x - 1) dx   = `(4/5) ` (x^2- x - 4)5/4 + c

problem 3:

Find the anti-derivative of sin2 x

Solution:

`int`sin2 x  dx =  `int` `(1/2)`  (1- cos 2x) dx                                             ; we know, sin2x = `(1- cos 2x) / 2`

= `int``(1/2)` 1 dx -  `int` `(1/2)`cos 2x dx

=  `(1/2)`  `int`dx - `(1/2)` `int` cos 2x dx                               ;  `int` cos 2x dx = `(1/2)` sin 2x + c

= `(1/2)` x -  `(1/2)` `(1/2)` sin 2x + c

Answer:

`int`sin2 x  dx = ` (1/2)` x - `(1/4)` sin 2x + c