A-A+
设f(x)在[0 1]上有连续的二阶导数 且f(0)=f(1)证明:存在ξ属于(0 1)使得
问题详情
设f(x)在[0，1]上有连续的二阶导数，且f'(0)=f'(1)证明:存在ξ属于(0,1)使得∫(0->1)f(x)dx=[f(0)+f(1)]/2
参考答案
分部积分,
∫(0->1)f(x)dx=∫(0->1)d[f(x)(x-1/2)]-∫(0->1)f'(x)(x-1/2)dx
=[f(0)+f(1)]/2-∫(0->1)f'(x)(x-1/2)dx
=[f(0)+f(1)]/2-∫(0->1)d[f'(x)(x^2/2-x/2+1/4)]+∫(0->1)f''(x)(x^2/2-x/2+1/4)dx
=[f(0)+f(1)]/2-0+∫(0->1)f''(x)(x^2/2-x/2+1/4)dx.
由于∫(0->1)(x^2/2-x/2+1/4)dx=1/6,
并且x^2/2-x/2+1/4>=0,0
