# Numerical Integrations

**Trapezoidal method rule**

According to trapezoidal rule for evaluating a definite integral,

=**Simpsons 1/3**^{rd}rule

**=****Simpsons 3/8th rule**

**=**[(*y*_{0}+*y*) + 3(_{n}*y*_{1}+*y*_{2}+*y*_{4}+ ... +*y*_{n}_{â€“1})+ 2(*y*_{3}+*y*_{6}+*y*_{9}+ ... +*y*_{n}_{â€“3})]

# Solution of Differential Equations

**Runge-Kutta method**

The rule for finding the increment*k*of*y*corresponding to the increment*h*of*x*.

=*f*(*x*,*y*) ;*y*(*x*_{0}) =*y*_{0}

Calculation is done as follows :

*k*1 =*h f*(*x*0,*y*0),*k*2 =

*k*3 = ,*k*4 =*hf*(*x*0 +*h*,*y*0 +*k*3)

*k*= (*k*1 + 2*k*2 + 2*k*3 +*k*4) = (*k*1 +*k*4) + (*k*2 +*k*3)

The approximate value,*y*1 =*y*0 +*k***Taylorâ€™s series method**

Given =*y*â€² =*f*(*x*,*y*) and*f*(*x*_{0}) =*y*_{0}

By taylorâ€™s theorem the series about a point*x*=*x*0

*y*=*y*0 + (*x*â€“*x*0) (*y*â€²)0 +

From this equation we can find*y*1 of*y*for*x*=*x*1 and then*y*â€²*, y*â€³*, y*â€³â€² are found out.**Eulerâ€™s method**

Given =*y*â€² =*f*(*x*,*y*) and*y*(*x*_{0}) =*y*_{0}

when*h*â†’ 0, by Taylorâ€™s series

*y*(*x*+*b*) =*y*(*x*) +*h y*â€²(*x*) =*y*(*x*) +*h**f*(*x*,*y*)

Thus,*yn*+1 =*yn h f*(*xn*,*yn*)

where,*h*=*i.e.*(*xn = x*0 +*nh*)