高等数学吧 关注：471,941贴子：3,872,592
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如果有人需要，以下是我手打的第一小问的 markdown 格式（贴吧没有代码块真的难顶）
例：设 $f(x)$ 在 $[a,b]$ 上存在二阶导数，试证明：存在 $\xi, \eta \in (a,b)$，使
(1) $\int_a^bf(t)dt = f(\dfrac{a+b}{2})(b-a)+\dfrac{1}{24}f''(\xi)(b-a)^3$
(2) $\int_a^bf(t)dt = \dfrac{1}{2}(f(a)+f(b))(b-a)-\dfrac{1}{12}f''(\eta)(b-a)^3$
证明：
(1) 化成变限积分：令 $\varphi(x) = \int_{x_0}^xf(t)dt$，将 $\varphi(x)$ 在 $x =x_0$ 处展开成泰勒公式至 $n=2$，有
$\varphi(x) = \varphi(x_0)+\varphi'(x_0)(x-x_0)+\dfrac{1}{2}\varphi''(x_0)(x-x_0)^2+\dfrac{1}{3!}\varphi'''(\xi)(x-x_0)^3$
由 $\varphi(x_0) = 0, \varphi'(x_0) = f(x_0), \varphi''(x_0) = f'(x_0), \varphi'''(\xi) = f''(\xi)$，其中 $\xi \in (x_0,x)$ 或 $\xi \in (x,x_0)$
以 $\varphi(x) = \int_{x_0}^xf(t)dt$ 代入得
$\int_{x_0}^xf(t)dt = f(x_0)(x-x_0)+\dfrac{1}{2}f'(x_0)(x-x_0)^2+\dfrac{1}{6}f''(\xi)(x-x_0)^3$
若两点都是导数信息或函数信息，则对两点分别展开：对照欲证的式子，令 $x_0 = \dfrac{a+b}{2}$，再分别以 $x = a, x = b$ 代入，得：
$\int_{\frac{a+b}{2}}^af(t)dt = f(\dfrac{a+b}{2})(\dfrac{a-b}{2})+\dfrac{1}{2}f'(\dfrac{a+b}{2})(\dfrac{a-b}{2})^2+\dfrac{1}{6}f''(\xi_1)(\dfrac{a-b}{2})^3$
$\int_{\frac{a+b}{2}}^bf(t)dt = f(\dfrac{a+b}{2})(\dfrac{b-a}{2})+\dfrac{1}{2}f'(\dfrac{a+b}{2})(\dfrac{b-a}{2})^2+\dfrac{1}{6}f''(\xi_2)(\dfrac{b-a}{2})^3$
第二个式子 - 第一个式子，得
$\int_{a}^bf(t)dt = f(\dfrac{a+b}{2})(b-a)+\dfrac{1}{48}[f''(\xi_1)+f''(\xi_2)](b-a)^3$
显然，接下来要证明存在 $\xi_3 = \dfrac{1}{2}[f''(\xi_1)+f''(\xi_2)]$
题目没给二阶导连续，但是我不知道除了连续函数的介值定理之外还能怎么证
例：设 $f(x)$ 在 $[a,b]$ 上存在二阶导数，试证明：存在 $\xi, \eta \in (a,b)$，使
(1) $\int_a^bf(t)dt = f(\dfrac{a+b}{2})(b-a)+\dfrac{1}{24}f''(\xi)(b-a)^3$
(2) $\int_a^bf(t)dt = \dfrac{1}{2}(f(a)+f(b))(b-a)-\dfrac{1}{12}f''(\eta)(b-a)^3$
证明：
(1) 化成变限积分：令 $\varphi(x) = \int_{x_0}^xf(t)dt$，将 $\varphi(x)$ 在 $x =x_0$ 处展开成泰勒公式至 $n=2$，有
$\varphi(x) = \varphi(x_0)+\varphi'(x_0)(x-x_0)+\dfrac{1}{2}\varphi''(x_0)(x-x_0)^2+\dfrac{1}{3!}\varphi'''(\xi)(x-x_0)^3$
由 $\varphi(x_0) = 0, \varphi'(x_0) = f(x_0), \varphi''(x_0) = f'(x_0), \varphi'''(\xi) = f''(\xi)$，其中 $\xi \in (x_0,x)$ 或 $\xi \in (x,x_0)$
以 $\varphi(x) = \int_{x_0}^xf(t)dt$ 代入得
$\int_{x_0}^xf(t)dt = f(x_0)(x-x_0)+\dfrac{1}{2}f'(x_0)(x-x_0)^2+\dfrac{1}{6}f''(\xi)(x-x_0)^3$
若两点都是导数信息或函数信息，则对两点分别展开：对照欲证的式子，令 $x_0 = \dfrac{a+b}{2}$，再分别以 $x = a, x = b$ 代入，得：
$\int_{\frac{a+b}{2}}^af(t)dt = f(\dfrac{a+b}{2})(\dfrac{a-b}{2})+\dfrac{1}{2}f'(\dfrac{a+b}{2})(\dfrac{a-b}{2})^2+\dfrac{1}{6}f''(\xi_1)(\dfrac{a-b}{2})^3$
$\int_{\frac{a+b}{2}}^bf(t)dt = f(\dfrac{a+b}{2})(\dfrac{b-a}{2})+\dfrac{1}{2}f'(\dfrac{a+b}{2})(\dfrac{b-a}{2})^2+\dfrac{1}{6}f''(\xi_2)(\dfrac{b-a}{2})^3$
第二个式子 - 第一个式子，得
$\int_{a}^bf(t)dt = f(\dfrac{a+b}{2})(b-a)+\dfrac{1}{48}[f''(\xi_1)+f''(\xi_2)](b-a)^3$
显然，接下来要证明存在 $\xi_3 = \dfrac{1}{2}[f''(\xi_1)+f''(\xi_2)]$
题目没给二阶导连续，但是我不知道除了连续函数的介值定理之外还能怎么证
照例的 markdown 格式
书本上的第二小问的答案：
(2) 用常数 k 值法
令 $\dfrac{\int_a^bf(t)dt-\dfrac{1}{2}(f(a)+f(b))(b-a)}{(b-a)^3} = K$
作函数 $F(x) = \int_a^x{f(t)dt}-\dfrac{1}{2}(f(x)+f(a))(x-a)-K(x-a)^3$
有 $F(a) = F(b) = 0$，所以存在 $\eta_1 \in (a,b)$ 使 $F'(\eta_1) = 0$，即
$f(\eta_1)-\dfrac{1}{2}f'(\eta_1)(\eta_1-a)-\dfrac{1}{2}(f(\eta_1)+f(a))-3K(\eta_1-a)^2 = 0$
化简为
$f(\eta_1)-f(a)-f'(\eta_1)(\eta_1-a)-6K(\eta_1-a)^2 = 0$
又由泰勒公式有
$f(a) = f(\eta_1) + f'(\eta_1)(a-\eta_1) + \dfrac{1}{2}f''(\eta)(a-\eta_1)^2$
联立上述二式，得 $-6K = \dfrac{1}{2}f''(\eta)$
即 $f''(\eta) = -12K$
即 $f''(\eta) = -12\dfrac{\int_a^bf(t)dt-\dfrac{1}{2}(f(a)+f(b))(b-a)}{(b-a)^3}$
原式得证
这个书本的第二小问的答案倒是很靠谱
或许可以参考这个答案做第一小问
(1) 用常数 k 值法
令 $\dfrac{\int_a^bf(t)dt-\dfrac{1}{2}(f(\dfrac{a+b}{2}))(b-a)}{(b-a)^3} = K$
作函数 $F(x) = \int_a^x{f(t)dt}-\dfrac{1}{2}f(\dfrac{x+b}{2})(x-a)-K(x-a)^3$
有 $F(a) = F(b) = 0$，所以存在 $\xi_1 \in (a,b)$ 使 $F'(\xi_1) = 0$，即
$f(\xi_1) - \dfrac{1}{4}f'(\dfrac{\xi_1+b}{2})(\xi_1 - a) - \dfrac{1}{2}f(\dfrac{\xi_1+b}{2}) - 3K(\xi_1-a)^2 = 0$
又由泰勒公式有
$f(\xi_1) = f(\dfrac{\xi_1+b}{2}) + f'(\dfrac{\xi_1+b}{2})(\dfrac{\xi_1-b}{2}) + \dfrac{1}{2}f''(\xi_2)(\dfrac{\xi_1-b}{2})^2$
联立上述二式，得
$f(\dfrac{\xi_1+b}{2}) + f'(\dfrac{\xi_1+b}{2})(\dfrac{\xi_1-b}{2}) + \dfrac{1}{2}f''(\xi_2)(\dfrac{\xi_1-b}{2})^2 - \dfrac{1}{4}f'(\dfrac{\xi_1+b}{2})(\xi_1 - a) - \dfrac{1}{2}f(\dfrac{\xi_1+b}{2}) - 3K(\xi_1-a)^2 = 0$
化简为
$\dfrac{1}{2}f(\dfrac{\xi_1+b}{2}) + \dfrac{1}{4}f'(\dfrac{\xi_1+b}{2})(\xi_1+a-2b)+\dfrac{1}{2}f''(\xi_2)(\dfrac{\xi_1-b}{2})^2 - 3K(\xi_1-a)^2 = 0$
感觉也做不下去了……
书本上的第二小问的答案：
(2) 用常数 k 值法
令 $\dfrac{\int_a^bf(t)dt-\dfrac{1}{2}(f(a)+f(b))(b-a)}{(b-a)^3} = K$
作函数 $F(x) = \int_a^x{f(t)dt}-\dfrac{1}{2}(f(x)+f(a))(x-a)-K(x-a)^3$
有 $F(a) = F(b) = 0$，所以存在 $\eta_1 \in (a,b)$ 使 $F'(\eta_1) = 0$，即
$f(\eta_1)-\dfrac{1}{2}f'(\eta_1)(\eta_1-a)-\dfrac{1}{2}(f(\eta_1)+f(a))-3K(\eta_1-a)^2 = 0$
化简为
$f(\eta_1)-f(a)-f'(\eta_1)(\eta_1-a)-6K(\eta_1-a)^2 = 0$
又由泰勒公式有
$f(a) = f(\eta_1) + f'(\eta_1)(a-\eta_1) + \dfrac{1}{2}f''(\eta)(a-\eta_1)^2$
联立上述二式，得 $-6K = \dfrac{1}{2}f''(\eta)$
即 $f''(\eta) = -12K$
即 $f''(\eta) = -12\dfrac{\int_a^bf(t)dt-\dfrac{1}{2}(f(a)+f(b))(b-a)}{(b-a)^3}$
原式得证
这个书本的第二小问的答案倒是很靠谱
或许可以参考这个答案做第一小问
(1) 用常数 k 值法
令 $\dfrac{\int_a^bf(t)dt-\dfrac{1}{2}(f(\dfrac{a+b}{2}))(b-a)}{(b-a)^3} = K$
作函数 $F(x) = \int_a^x{f(t)dt}-\dfrac{1}{2}f(\dfrac{x+b}{2})(x-a)-K(x-a)^3$
有 $F(a) = F(b) = 0$，所以存在 $\xi_1 \in (a,b)$ 使 $F'(\xi_1) = 0$，即
$f(\xi_1) - \dfrac{1}{4}f'(\dfrac{\xi_1+b}{2})(\xi_1 - a) - \dfrac{1}{2}f(\dfrac{\xi_1+b}{2}) - 3K(\xi_1-a)^2 = 0$
又由泰勒公式有
$f(\xi_1) = f(\dfrac{\xi_1+b}{2}) + f'(\dfrac{\xi_1+b}{2})(\dfrac{\xi_1-b}{2}) + \dfrac{1}{2}f''(\xi_2)(\dfrac{\xi_1-b}{2})^2$
联立上述二式，得
$f(\dfrac{\xi_1+b}{2}) + f'(\dfrac{\xi_1+b}{2})(\dfrac{\xi_1-b}{2}) + \dfrac{1}{2}f''(\xi_2)(\dfrac{\xi_1-b}{2})^2 - \dfrac{1}{4}f'(\dfrac{\xi_1+b}{2})(\xi_1 - a) - \dfrac{1}{2}f(\dfrac{\xi_1+b}{2}) - 3K(\xi_1-a)^2 = 0$
化简为
$\dfrac{1}{2}f(\dfrac{\xi_1+b}{2}) + \dfrac{1}{4}f'(\dfrac{\xi_1+b}{2})(\xi_1+a-2b)+\dfrac{1}{2}f''(\xi_2)(\dfrac{\xi_1-b}{2})^2 - 3K(\xi_1-a)^2 = 0$
感觉也做不下去了……
