Example 4 we set nodes Xg = 1 / 4, Xj = 3 / 4 , tryto work out the numerical integration formula basedon the integration (' f(x)dx and also work out itstruncation error.Solution, because the integration formula is interpolationits integration coefficient can be seen as31-A = J.6(m)dx=J±-4 dx= J-=(4x - 3)dx2x。-x1Tx-xo[")(4x -1)dx :A = ’1(x)dx =)dx =-2Jo x, - XoJo2So integration formula is上页()+()J°f(x)dx ~下页返圆
上页 下页 返回 Example 4 we set nodes , try to work out the numerical integration formula based on the integration and also work out its truncation error. 0 1 x x = = 1/ 4, 3 / 4 1 0 f x dx ( ) Solution, because the integration formula is interpolation, its integration coefficient can be seen as 1 1 1 0 0 0 0 0 0 1 3 4 1 1 ( ) (4 3) 2 2 x A l x dx dx x dx x x − = = = − − = − 1 1 1 0 1 1 0 0 0 1 0 1 1 ( ) (4 1) 2 2 x x A l x dx dx x dx x x − = = = − = − So integration formula is 1 0 1 1 3 ( ) 2 4 4 f x dx f f +
If f"(x) is available in [O,1l, so the truncation error isR()= (t-[()+(]= (a)dx-1 ()dx=()(x-x-)asAmong them e (0,1) , and depended onX.P(x) is linear interpolation functionbased on the nodes of Xo and Xi上页下页返圆
上页 下页 返回 If is available in [0,1], so the truncation error is " f x( ) 1 0 1 1 1 " 1 0 0 0 1 1 3 ( ) ( ) 2 4 4 1 1 3 ( ) ( ) ( ) 2 4 4 R f f x dx f f f x dx P x dx f x x dx = − + = − = − − Among them , and depended on x. is linear interpolation function based on the nodes of and (0,1) 1P x( ) 0 x x1
s 3 Newton--Cotes FormulaloworderNewton-CotesformulaanditsremainIn Newton-Cotes formula, when n=1,2,4 they arethree most common and useful formulas, whichcalled as low-order formula1trapeziaformulaanditsremainder(-1)"-kC(n)II(t-j)dtJn·k!(n-k)!O≤j<nj+k上页下页返圆
上页 下页 返回 §3 Newton—Cotes Formula low order Newton-Cotes formula and its remainder In Newton-Cotes formula, when n=1,2,4 they are three most common and useful formulas, which called as low-order formula 1 trapezia formula and its remainder − − − − = n j k j n n k n k t j t n k n k C 0 0 ( ) ( )d !( )! ( 1)
set n = 1, so xo = a, xi = b, h = b- aCotes coefficients arecl=-f'(t-1)dt =1c() ='tdt =2Integration formula isI(f) =(b-a)Ec("f(x)3.5k=0b-a[f(x)+ f(x)]2.52b-a1.5[f(a)+ f(b)]namely li(f)=2上页0.51.5下页返圆
上页 下页 返回 Cotes coefficients are Integration formula is -0.5 0 0.5 1 1.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 (1) C0 t dt = − − 1 0 ( 1) 2 1 = (1) C1 tdt = 1 0 2 1 = ( ) 1 I f = = − 1 0 (1) ( ) ( ) k k xk b a C f [ ( ) ( )] 2 0 x1 f x f b a + − = [ ( ) ( )] 2 f a f b b a + − =
Above formula is trapezium formula or two-point formulab-aT=I(f)[f(a)+ f(b)]R,(x) :n+i(x)(n+1)!the reminder of trapezium formula isR(T) = R(I) = R(x)dxSecond mean value=T'"(2)theorem(x-a)(x-b)dxforintegrals2J"(n) ["(x-a)(x-b)dxne[a,b]2(b-a)3f"(n) (b-a)3f"(n)2612(b-a)3M2M, = max / f"(x) IIR(T)≤SO12xe[a,b]上页下页trapezium formula has 1algebraic precision返圆
上页 下页 返回 x a x b dx b f a − − = ( )( ) 2 ( ) x a x b dx f b a − − = ( )( ) 2 () [a,b] Second mean value theorem for integrals 6 ( ) 2 ( ) 3 f b − a = − ( ) ( 1)! ( ) ( ) 1 ( 1) x n f R x n n n + + + = ( ) 12 ( ) 3 f b a − = − 2 3 12 ( ) | ( )| M b a R T − max | ( )| [ , ] M2 f x x a b = trapezium formula has 1 algebraic precision so [ ( ) ( )] 2 f a f b b a + − T = I1 ( f ) = the reminder of trapezium formula is ( ) ( ) 1 R T = R I = b a R1 (x)dx Above formula is trapezium formula or two-point formula