PROGRAM 9 SimpsonOne Third Rule Taking n 2

PROGRAM 9 : Simpson’One Third Rule : Taking n = 2 in general quadrature formula in eqn, (1) , the highest non-zero difference is second order difference and hence the next higher order terms are neglected. Hence from eqn. (1) ∫x 0+2 h y dx = h [ (2 y +(4/2) ∆y + {(8/3)-(4/2)} (∆2 y )/2!] x 0 0 = h [2 y 0 + 2 (y 1 -y 0 ) + (1/3) (y 2 -2 y 1+y 0)] =(h/3) [y 0 +4 y 1 + y 2] Similarly for the next intervals, x 0+4 h y dx = (h/3) [ y + 4 y + y ] x 0+2 h ∫ 2 3 4. . x 0+nh y dx = (h/3) [ y x 0+(n-2) h ∫ n-2 +4 yn-1 + yn ] where n is multiple of 2. On adding all these terms, I = a∫b y dx = (h/3) [ ( y 0+yn) + 4( y 1 + y 3 +…. +yn-1) +2 (y 2 + y 4+…+ yn-2)]

The value of definite (finite) integral by Simpson’s one third Rule is given by I = a∫b y dx = (h/3) [ ( y 0+yn) + 4( y 1 + y 3 +…. +yn-1) +2 (y 2 + y 4+…+ yn-2)] where y 0 = F(xo) = F (a) y 1 = F(xo + H) = F (a + H) y 2= F(xo + 2 H) = F (a +2 H). . yn = F(xo + n. H) = F (a +n. H ) =F (b)

C C C PROGRAM TO CALCULATE VALUE OF DEFINITE INTEGRAL USING SIMPSON’S 1/3 RULE FOR F(X) = X **2 WRITE(*, *)’ENTER LOWER AND UPPER LIMITS’ READ(*, *) A, B WRITE(*, *)’ENTER NUMBER OF SEGMENTS(EVEN NO. )’ READ(*, *) N H=(B-A)/N SUM = F(A) + F(B) + 4. 0* F(A + H) DO 10 I = 3, N-1, 2 SUM = SUM + 4. 0* F(A + I*H) + 2. 0* F(A + (I-1)*H ) 10 CONTINUE RESULT = H*SUM/3. 0 `

WRITE(*, *) ’INTEGRAL FROM ‘, A, ‘TO’, B WRITE(*, *) ’WHEN H = ‘, H, ‘IS’, RESULT STOP END