设f(x)在[0,1]上有二阶连续导数，证明：∫^(0,1)f(x)dx=1⼀2 (f(0)+f(1))- 1⼀2 ∫^(0,1)x(1-x)f"(x)dx

f(0)=f(x)+f'(x)(0-x)+0.5f''(a)(0-x)^2
f(1)=f(x)+f'(x)(1-x)+0.5f''(b)(1-x)^2
两式相减，移项，颤则悉取绝对值得|f'盯渣(x)|=|f(1)-f(0)+0.5f''(a)x^2-0.5f''(b)(1-x)^2|<=
|f(1)-f(0)|+0.5a(x^2+(1-x)^2)<=|f(1)-f(0)|+0.5a，最后不等式是因为二次函数x^2+(1-x)^2在【0
1】上茄乎的最大值是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'滚枯(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
∫大枯洞^(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/败橘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.