数值分析第二章答案(3)

b

f (x) S (x) a

2

ba

2

b

b

2

dx

2

ba

f (x) a

f (x) a

2

b

dx S (x) dx 2 f (x)S (x)dxdx S (x) dx 2 S (x) f (x) S (x) dx

a

a

b

b

2

b

从而有

a f (x)

ba

dx S (x) dx

a

2

2

f (x) S (x)

dx 2 S (x) f (x) S (x) dx

a

b

(2) S (x) f (x) S (x) dx

a

b

S (x)d f (x) S (x)

a

b

bb

S (x) f (x) S (x) f (x) S (x) d[S (x)]

aa

S (b) f (b) S (b) S (a) f (a) S (a) S (x) f (x) S (x) dx

ab

xk xk 1xk 1

S (b) f (b) S (b) S (a) f (a) S (a) S () f (x) S (x) dx

xk

2k 0 S (b) f (b) S (b) S (a) f (a) S (a) S (

k 0n 1

n 1

xk 1xk xk 1

) f(x) S(x)

xk2

S (b) f (b) S (b) S (a) f (a) S (a)