Question

设f(x)在[0,1]上有二阶连续导数,证明:∫ (-1,2)f(x)dx=1/2[f(1)+f(2)]-1/2∫(1,

Answer 1

用分部积分法.

∫^(0,1)x(1-x)f"(x)dx (u= x(1-x) v'= f''(x) u' =1-2x v= f'(x)

=[x(1-x) f'(x) ] (0,1) - ∫^(0,1)(1-2x)f'(x)dx 再设u1= 1-2x v1 = f'(x) (u1)' =-2 (v1)'= f(x)

= 0 - (1- 2x) f(x) (0,1) - 2 ∫^(0,1)f(x)dx

=f(1) +f(0) -2 ∫^(0,1)fx)dx

移项,整理即得:：∫^(0,1)f(x)dx=1/2 (f(0)+f(1))- 1/2 ∫^(0,1)x(1-x)f"(x)

其中:[x(1-x) f'(x) ] (0,1) 表示:函数[x(1-x) f'(x) ] 在x=1的值减去它在 x=0的值.另处类似.