📝 题目
3. 3.8 设 $f\left( x\right)$ 在 $\left\lbrack {a,b}\right\rbrack$ 上二阶连续可微,求证:
$$ f\left( x\right) - f\left( a\right) - {f}^{\prime }\left( a\right) \left( {x - a}\right) = {\int }_{a}^{x}{f}^{\prime \prime }\left( t\right) \left( {x - t}\right) \mathrm{d}t\;\left( {\forall x \in \left\lbrack {a,b}\right\rbrack }\right) . $$
💡 答案与解析
2. $$
3.3.7 原式 $\frac{u = 1 - \frac{t}{n}}{}{\int }_{0}^{1}\frac{1 - {u}^{n}}{1 - u}\mathrm{\;d}u = {\int }_{0}^{1}\left( {1 + u + {u}^{2} + \cdots + {u}^{n - 1}}\right) \mathrm{d}u$
$$ = 1 + \frac{1}{2} + \cdots + \frac{1}{n}\text{ . } $$
3.3.8 原式左边 $= {\int }_{a}^{x}\left\lbrack {{f}^{\prime }\left( t\right) - {f}^{\prime }\left( a\right) }\right\rbrack \mathrm{d}t = {\int }_{a}^{x}\left\lbrack {{f}^{\prime }\left( t\right) - {f}^{\prime }\left( a\right) }\right\rbrack \mathrm{d}\left( {t - x}\right)$ 分部积分 原式右边.
3.3.9 原式左边 $\overset{x = \frac{ab}{u}}{ = }{\int }_{a}^{b}f\left( \frac{ab}{u}\right) \frac{\ln \frac{ab}{u}}{u}\mathrm{\;d}u$
$= {\int }_{a}^{b}f\left( u\right) \frac{\ln \left( {ab}\right) }{u}\mathrm{\;d}u$ 一原式左边.
3.3.10 (1) 原式左边令 $x = \frac{{a}^{2}}{u}$ ;
(2)原式左边令 $u = {x}^{2}$ ；
(3)对 $f\left( x\right) \overset{\text{ 定义 }}{ = }g\left( {x + \frac{{a}^{2}}{x}}\right)$ 用第 (2) 小题结论.