设f^{\prime\prime}(x)在[1,3]上连续，且f(2)=0. 证明：存在\xi\in[1,3]，使得f^{\prime\prime}(\xi)=3\int_1^3 f(x)dx.
错误思路：对f(x)在x=2处泰勒展开，然后两边都有取值范围限制……
实际上，我们注意到：
\left[\int_{1}^{x}f(t)dt\right]^\prime=f(x)
正解：记g(x)=\int_1^xf(t)dt，g(1)=0，g^\prime(x)=f(x)，g^\prime(2)=0.
g(x)在x=2处的泰勒展开：
g(x)=g(2)+g^\prime(2)(x-2)+\dfrac{1}{2}g^{\prime\prime}(2)(x-2)^2+\dfrac{1}{6}g^{(3)}(\mu)(x-2)^3,\mu\text{在2和}x\text{之间}
故g(3)=g(2)+\dfrac{1}{2}g^{(2)}(2)+\dfrac{1}{6}g^{(3)}(\mu_1),\mu_1\in(2,3)\\
g(1)=g(2)+\dfrac{1}{2}g^{(2)}(2)-\dfrac{1}{6}g^{(3)}(\mu_2),\mu_2\in(1,2)
即\int_1^3f(x)dx=g(3)=\dfrac{f^{\prime\prime}(\mu_1)+f^{\prime\prime}(\mu_2)}{6}
由于f^{\prime\prime}(x)在[1,3]连续，记f^{\prime\prime}(x)的最大值和最小值分别为M和m. 则\dfrac{f^{\prime\prime}(\mu_1)+f^{\prime\prime}(\mu_2)}{2}\in[m,M]
由介值定理，\exist \xi\in[1,3]，使f^{\prime\prime}(\xi)=\dfrac{f^{\prime\prime}(\mu_1)+f^{\prime\prime}(\mu_2)}{2}，进而f^{\prime\prime}(\xi)=3\int_1^3f(x)dx.
证毕.
