Double Integration by Parts

Introduction for double integration by parts:
The integration process represents the inverse process of the differentiation in progress. If `dy/dx` =h(x) here y is the function then the double integrals `int` `int`h(x) dx dy= y. There are two types of integration namely indefinite integral (without  limits) and definite integral (with limits). If we consider u = f(x), v = h(x), then the product rule in its simplest form is: `intint u dv/dx dxdy = int [uv - int v (du)/dx dx]dy`

In this article we shall discuss about the double integration by parts with the examples for the solution.

Examples to Explain "double Integration by Parts "

Double integration by parts on this function  `int int 5x^3 ln x dxdy` ,solve for the answer after calculating the integration on the first step by the integration by parts.
Solution:

First we take inner integral on the functions   `int 5x^3 ln x dx`

It is solved under the integration by parts as the first step for the prosecution of the steps.

The logarithmic form is exist in the given integration function.

Take ln x as u and `x^3 dx` as dv

u =5 `ln x`                   dv =`x^3 dx`

`(du)/dx = 5/x`                   v = `int x^3 dx`    

du = `5dx/x`                  v = `x^4/4 `      

We have  `int u (dv)/dx dx = uv - int v du/dx dx`

0r `int u dv = uv - int v du`

`int 5x^3 ln x dx`  = 5 ln x . `x^4/4 - ` `int x^4/4 5dx/x`             

=  5 ln x . `x^4/4 - ` `5/4int x^3 dx` 

=  5 ln x . `x^4/4 - `   `5/4 (1/4 x^4) `           

= `5 ln x (x^4)/4``-(5x^4)/16`          

This is the result obtained under the integration by parts for the inner integral.

Operate it with the outer integral functions by,

`int int 5x^3 ln x dxdy` =  `int [5 ln x (x^4)/4-(5x^4)/16 ] dy`         

=  `int` 5 ln x . `x^4/4 dy ` ` - int (5x^4)/16 dy` 

= `5y ln x (x^4)/4``-(5yx^4)/16 + c `          

This is the result obtained under the double integration by parts.

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Problems to Explain "double Integration by Parts "

Double integration by parts on this function  `int int x log x dx dy` ,solve for the answer after calculating the integration on the first step by the integration by parts.
Solution:

First we take inner integral on the functions   `int x log x dx`

It is solved under the integration by parts as the first step for the prosecution of the steps.

`int x log xdx` = `int ( log x) (x dx)`

The logarithmic function on the x function with another function x is not integrated directly ,so we take it as

u = logx       and        dv = x dx

? du = 1/x dx              v = `x^2/2`           

? `int x log x dx` = (logx) `x^2/2`  -  `int (x^2/2) (1/x dx)`                        

= `x^2/2`  (logx) `-1/2` `int x dx`       

? `int x log x dx` =  `x^2/2`  (logx) - `1/4 x^2 `

This is the result obtained under the integration by parts for the inner integral.

Operate it with the outer integral functions by,

`int int x log x dx dy`  =    `int [x^2/2(log x) - 1/4 x^2 ] dy`           

=    `int [x^2/2(log x)]dy - int [1/4 x^2 ] dy`

=       `x^2y/2(log x) - 1/4 x^2y dy + c`              

This is the result obtained under the double integration by parts.