# Fortran ~ Simpson's 1/3rd Rule

# Fortran ~ Simpson's 1/3rd Rule

- By putting n=2 into a basic equation by integration and taking the curve through (X0, Y0), (X1, Y1) and (X2, Y2) as a polynomial of degree 2 so that difference of order Three and greater than three become zero, than we get.
- F(X)=3.12*X**3+1.54*X**2-6.23
- This is known as Simpson's 1/3rd Rule.

Image Source ~ Crafted With ©Ishwaranand - 2020 ~ Image by ©Ishwaranand |

Remarks

- Simpson's 1/3rd rule requires the division of the interval [X0, Xn] into an even number od sub-intervals of width h.In this rule, the interpolating polynomial is of degree 2.
- Therefore this rule is also known as a parabolic rule.

**PROGRAM SIMPSON**

C

C PROGRAM TO DEMONSTRATE APPLICATION OF SIMPSONS 1/3rd RULE

C

C THIS PROGRAM NUMERICALLY INTEGRATES GIVEN EXPRESSION WITHIN

C SPECIFIED LIMITS OF INTEGRATION

C

F(X)=3.12*X**3+1.54*X**2-6.23

C

WRITE (*,*)'LIMITS OF INTEGRATION (A,B)'

READ(*,*)A,B

WRITE(*,*)'NUMBER OF STRIPES(Number must be Even)'

READ(*,*)NS

IF(NS/2*2.NE.NS)THEN

WRITE(*,*)'NUMBER OF STRIPES (must be Even)'

WRITE(*,*)'PROGRAM TERMINATED...'

STOP

ENDIF

C STRIP SIZE

H=(B-A)/NS

AREA=F(A)+F(B)

DO 100 X=A+H,B-H,2*H

AREA=AREA+4.0*F(X)

100 CONTINUE

DO 200 X=A+2*H,B-2*H,2*H

AREA=AREA+2.0*F(X)

200 CONTINUE

AREA=AREA*H/3.0

WRITE(*,*)'AREA UNDER THE CURVE=',AREA

STOP

END

#OUTPUT

LIMIT OF INTEGRATION (A, B)

1 2

NUMBER OF STRIPS (Number must be Even)

2

AREA UNDER THE CURVE = 54.546665

## No comments

Comment on Ishwaranand or Happy Day

Not, Add any types of links