# Path Integration

Path integration is one of the method of definite integrals. the path integral has the upper and lower limits of the given function. In definite integral, `int_a^b`f(x) dx . where, a, b and x is the complex numbers and path of integration from a to b. In the path integral is   `int` f(x) dx . where,  f(x) is continuous function. The path integral with an interval a to b `int_a^b`f(x) dx = F(b) - F(a). In this article we shall discuss about path integration

## General Integration Formulas:

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

2.  `int` ex dx = ex + c

3.  ` int ` log x dx = x (log x - 1) + c

4.  ` int ` sec2x.dx = tan x + c

5. ` int ` cosec2x.dx = - cot x + c

6.`int` ax dx = ax log x + c

7. ` int` sin x.dx = - cos x + c

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

These formulas are helps to solve the path integration.

## Path Integration - Problems:

Path integration - problem 1:

Let   z(t) = (3t -2, t+1) ,   `1 < t<=2` .  Let the density of the wire at the point   f(x, y) = x + y . Calculate the mass of the wire by using path integration.

Solution:

The given function is z ' (t) = (3 , 1 )

So,.       || z ' (t) || = `sqrt(3^2 + 1^2)` .

= `sqrt10` .

f (z(t))  = (3t - 2) + (t + 1)  = 4t - 1 .

Mass of the wire is `int_ z` f ds .

The path integral is  `int_ z` f ds  = ` int_a^b f (z(t)) ||z '(t)|| dt` .

=` int_1^2 (4t - 1) sqrt10 dt` .

= `sqrt10 [int_1^2 4t dt - int_1^2 dt]`

= `sqrt 10 [[(4 t^2)/2]_1^2 - [t]_1^2] ` .

= `sqrt10 [[(2 t^2)]_1^2 - [t]_1^2] ` .

= `sqrt10 [[2 ((2^2) - (1^2))] - [2 - 1]]` ..

=  `sqrt10 [[2(3)] - [1]]` ..

= `5 sqrt10.` .

if  z(t) given in cm  and f is grams/cm means the mass of the wire is `5 sqrt10.`  grams.

Path integration - problem 2:

Let   g(t) = (3t -2, t +1) ,   `0 < t<=1` .  Let the density of the wire at the point   f(x, y) = x + y . Calculate the mass of the wire by using path integration.

Solution:

The given function is g ' (t) = (3 , 1 )

So,.       || g ' (t) || = `sqrt(3^2 + 1^2)` .

= `sqrt10` .

f (g(t))  = (3t - 2) + (t + 1)  = 4t - 1 .

Mass of the wire is `int_ g` f ds .

The path integral is  `int_ g` f ds  = ` int_a^b f (g(t)) ||g '(t)|| dt` .

=` int_0^1 (4t - 1) sqrt10 dt` .

= `sqrt10 [int_0^1 4t dt - int_0^1 dt]`

= `sqrt 10 [[(4 t^2)/2]_0^1 - [t]_0^1] ` .

= `sqrt10 [[(2 t^2)]_0^1 - [t]_0^1] ` .

= `sqrt10 [[2 ((1^2) - (0^2))] - [1 - 0]]` ..

=  `sqrt10 [[2(1)] - [1]]` ..

= ` sqrt10.` .

if  z(t) given in cm  and f is grams/cm means the mass of the wire is ` sqrt10.`  grams.