构造表达式,分析表达式的性质

例题1:证明:$\int_{\rm{0}}^{\frac{\pi }{{\rm{2}}}} {\frac{{\sin x}}{{1 + {x^2}}}dx} < \int_{\rm{0}}^{\frac{\pi }{{\rm{2}}}} {\frac{{\cos x}}{{1 + {x^2}}}dx} $

解题思路点:

  1. 构造函数:$f(x{\rm{) = }}\frac{{\sin x - \cos x}}{{1 + {x^2}}}$
  2. 分析函数,由上面的构造函数容易得到$f(x)$关于$x = \frac{\pi }{4}$,将原积分式变形为:
$$ \begin{array}{l} \int_{\rm{0}}^{\frac{\pi }{4}} {\frac{{\sin x - \cos x}}{{1 + {x^2}}}dx} - \int_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + {x^2}}}dx} \\ = \int_{\rm{0}}^{\frac{\pi }{4}} {\frac{{\sin x - \cos x}}{{1 + {x^2}}}dx} + \int_{\rm{0}}^{\frac{\pi }{4}} {\frac{{\sin t - \cos t}}{{1 + {{(\frac{\pi }{2} - x)}^2}}}dt} < 0 \end{array} $$
得证。
**例题2:**设函数$f(x)$在$[0,1]$上可导,且$f(1)-f(0)=1$,证明:$\int_0^1 {{{(f'(x))}^2}dx} \ge 1$ **解题思路点:** 1. 题中积分式子包含平方,首先可以想到**柯西-施瓦兹**不等式:
$$ {(\int_a^b {f(x)g(x)dx} )^2} \le \int_a^b {{f^2}(x)dx} *\int_a^b {{g^2}(x)dx} $$
  1. 根据柯西-施瓦兹不等式构造:
$$ \int_0^1 {{{(f'(x))}^2}dx} \ge {(\int_0^1 {f'(x)} dx)^2} = 1 $$

得证。


例题3:设函数$f(x)$在$[a,b]$上可导,$f’(x)$在$[a,b]$上可积,且$f(a)+f(b)=0$,证明:$|f(x)| \le \frac{1}{2}\int_a^b {|f’(x)|dx} $

解题思路点:

  1. 分析题目右边所给的表达式其被积函数在$[a,b]$区间有正有负。构造表达式
$$ \begin{array}{l} \int_a^x {f'(x)dx} = f(x) - f(a)\\ \int_x^b {f'(x)dx} = f(b) - f(x) \end{array} $$
  1. 将上面构造的表达式进行初等变换:
$$ \begin{array}{l} 2f(x) = \int_a^x {f'(x)dx} - \int_x^b {f'(x)dx} \\ = > 2|f(x)| \le |\int_a^x {f'(x)dx} | + |\int_x^b {f'(x)dx} |\\ f(x) < \frac{1}{2}\int_a^b {|f'(x)|dx} \end{array} $$

得证。

拉格朗日中值定理的应用(二)

例题4:设函数$f(x)$在$[a,b]$上有连续的导数,且$f(a)=0$,证明:$\frac{{\rm{2}}}{{{{{\rm{(}}b - a)}^2}}}\int_a^b {|f(x)|dx} \le M$,其中$M = \mathop {\max }\limits_{a \le x \le b} |f'(x)|$

解题思路点:

  1. 因为$f(x) = \int_a^x {f'(x)dx} $,根据拉格朗日中值定理,
$$ |f(x)| = |\int_a^x {f'(x)dx} | \le \int_a^x {|f'(x)|dx} \le (x - a)M $$
  1. 根据上面的不等式可以得到:
    </div class=’full-math-div’> $$ \int_a^b {|f(x)|dx} \le \int_a^b {(x - a)Mdx} \le \frac{{{b^2} - 2ab}}{2}M \le \frac{{{b^2} - 2ab + {a^2}}}{2}M $$ </div>

得证。


例题5:设函数$f(X)$在$[a,b]$上有连续的导数,且$f(a)=f(b)=0$,求证$\int_a^b {|f(x)|dx} \le \frac{{{{(b - a)}^2}}}{4}M$,其中$M = \mathop {\max }\limits_{a \le x \le b} |f'(x)|$

解题思路点:

  1. 因为函数在$[a,b]$上具有连续的导数:

    $$ \begin{array}{l} f{\rm{(}}x) = f'(\xi )(x - a)\xi \in [a,b]\\ f(x) = f'(\mu )(x - b)\mu \in [a,b] \end{array} $$
  2. $f'(\xi ) \le M,f'(\mu ) \le M$,对于:
    $$ \begin{array}{l} \int_a^b {|f(x)|dx} = \int_a^{\frac{{a + b}}{2}} {|f(x)|dx} + \int_{\frac{{a + b}}{2}}^b {|f(x)|dx} \\ \le \int_a^{\frac{{a + b}}{2}} {(x - a)Mdx} + \int_{\frac{{a + b}}{2}}^b {(x - b)Mdx} \\ = \frac{{{{(a - b)}^2}}}{4}M \end{array} $$

得证。

泰勒展开式和拉格朗日中值定理综合应用

例题6:设函数 $f(x)$二阶可导,且$f^{(2)}(x) \geqslant 0$和$g(x)$为连续函数,对于任意的 $a>0$,证明:$\frac{1}{a}\int_0^a {f(g(x))dx} \ge f(\frac{1}{a}\int_0^a {g(x)dx} )$

解题思路点:

  1. 对函数进行泰勒展开:

    $$ \begin{array}{l} f(x) = f({x_0}) + f'({x_0})(x - {x_0}) + \frac{1}{2}f^{(2)}(\xi ){(x - {x_0})^2}\\ \ge f({x_0}) + f'({x_0})(x - {x_0}) \end{array} $$
  2. 令$x=g(t), {x_0}=\frac{1}{a}\int_0^a {g(x)dx} $,上式变形为:

$$ f(g(x)) \ge f(\frac{1}{a}\int_0^a {g(x)dx} ) + f'(\frac{1}{a}\int_0^a {g(x)dx} )(g(t) - \frac{1}{a}\int_0^a {g(x)dx} ) $$

对两边求积分:

$$ \int_0^a {f(g(x))dx} \ge af(\frac{1}{a}\int_0^a {g(x)dx} ) + f'(\frac{1}{a}\int_0^a {g(x)dx} )(\int_0^a {g(t) - \frac{1}{a}\int_0^a {g(x)dx} } ) $$

又因为:$\int_0^a {g(t) - \frac{1}{a}\int_0^a {g(x)dx} } = 0$

得证。


例题7:设函数$f(x)$在$[0,2]$有二阶连续的导数,且$f(1)=0$,证明:$\left| {\int_0^2 {f(x)dx} } \right| \le \frac{1}{3}M$,其中$M = \mathop {\max }\limits_{0 \le x \le 2} |f^{(2)}(x)|$。

解题思路点:

  1. 讲函数泰勒展开:
$$ \begin{array}{l} f(x) = f(1) + f'(1)(x - 1) + \frac{1}{2}f^{(2)}(\xi )(x - 1)\\ \le f'(1)(x - 1) + f^{(2)}(\xi ){(x - 1)^2}\\ f^{(2)}(\xi ) \le M \end{array} $$
  1. 对上面不等式两边求积分:
$$ \begin{array}{l} \int_0^2 {f(x)dx} = f'(1)\int_0^2 {x - 1dx} + \frac{1}{2}f^{(2)}(\xi )\int_0^2 {{{(x - 1)}^2}dx} \\ = \frac{1}{2}f^{(2)}(\xi )\int_0^2 {{{(x - 1)}^2}dx} \le \frac{1}{2}|f^{(2)}(\xi )|\int_0^2 {{{(x - 1)}^2}dx} \\ \le \frac{M}{2}\int_0^2 {{{(x - 1)}^2}dx} = \frac{1}{3}M \end{array} $$

得证。

例题8:设函数$f(x)$在$[0,1$有二阶连续的导数,且$f(0)=f(1)=0$,求证:$\int_0^1 {|f^{(2)}(x{\rm{)}}|dx} \ge 4\mathop {\max }\limits_{0 \le x \le 1} |f(x)|$

解题思路点:

  1. 由题目的已知条件可以得到存在$|f(\xi )| = \mathop {\max }\limits_{0 \le x \le 1} |f(x)|\xi \in {\rm{[0,1]}}$
  1. 函数在$[0,\xi]$和$[\xi, 1]$区间满足拉格朗日中值定理条件:
$$ \begin{array}{l} \xi f'(a) = f(\xi ),\\ (\xi - 1)f'(b) = f(\xi ) \end{array} $$
  1. 对题中所给表达式进行变形:
$$ \begin{array}{l} \int_0^1 {|f^{(2)}(x)|dx} \ge \int_a^b {|f^{(2)}(x)|dx} \ge |\int_a^b {f^{(2)}(x)dx} | = |f'(b) - f'(a)|\\ = \left| {\frac{{f(\xi )}}{{\xi - 1}} - \frac{{f(\xi )}}{\xi }} \right| = |f(\xi )|\frac{1}{{\xi (1 - \xi )}} \ge 4|f(\xi )| \end{array} $$

得证。


例题9:设函数$f(x)$在$[a,b]$有二阶连续的导数,且$f'(a)=f'(b)=0$,证明:存在$\xi \in (a,b) $,使得:

$$ \int_a^b {f(x)dx} = (b - a)\frac{{f(a) + f(b)}}{2} + \frac{{{{(b - a)}^3}}}{6}f^{(2)}(\xi ) $$

解题思路点:

  1. 令:

    $$ \begin{array}{l} \int_a^b {f(x)dx} = (b - a)f(a) + \frac{1}{6}f^{(2)}({\xi _2}){(b - a)^3}\\ \int_a^b {f(x)dx} = (b - a)f(b) + \frac{1}{6}f^{(2)}({\mu _2}){(b - a)^3}\\ {\xi _2},{\mu _2} \in [a,b] \end{array} $$
  2. 两式相加:

$$ 2\int_a^b {f(x)dx} = (b - a)(f(a) + f(b)) + \frac{{{{(b - a)}^3}}}{3}(f^{(2)}({\xi _2}) + f^{(2)}({\mu _2})) $$
  1. 因为$f^{(2)}(x)$在$[a,b]$连续,所以存在$f^{(2)}(\xi) = \frac{f^{(2)}(\xi _2)f^{(2)}(\mu _2)}{2}$。得证

三角函数的变形

例题10:证明:$\int_0^\pi {x{a^{\sin x}}dx} \int_0^{\frac{\pi }{2}} {{a^{ - \cos x}}dx} \ge \frac{{{\pi ^3}}}{4}$,$a>0$为常数。

解题思路点:

  1. 解决这题需要消去被积函数中的正弦函数指数和余弦函数指数。因此对第一个表达式作变换:
$$ \begin{gathered} \int_0^\pi {x{a^{\sin x}}dx} \xrightarrow{{x = t + \frac{\pi }{2}}}\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {(t + \frac{\pi }{2}){a^{\cos t}}dt} \hfill \\ = \frac{\pi }{2}\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{a^{\cos t}}dt} + \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {t{a^{\cos t}}dt} = \frac{\pi }{2}\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{a^{\cos t}}dt} \hfill \\ = \pi \int_0^{\frac{\pi }{2}} {{a^{\cos t}}dt} \hfill \\ \end{gathered} $$
  1. 利用柯西-施瓦兹[1]不等式消去三角函数:
$$ \pi \int_0^{\frac{\pi }{2}} {{a^{\cos t}}dt} \int_0^{\frac{\pi }{2}} {{a^{ - \cos x}}dx} \geqslant \pi {(\int_0^{\frac{\pi }{2}} {{a^{\frac{{ - \cos x}}{2} + \frac{{\cos x}}{2}}}dx} )^2} = \frac{{{\pi ^3}}}{4} $$

得证。

计算被积函数的原函数

例题11:求$I = \int {\frac{{x + \sin x\cos x}}{{{{(\cos x - x\sin x)}^2}}}dx} $

解:

$$ \begin{gathered} 原式 = \int {\frac{{x{{\sec }^2}x + \tan x}}{{{{(1 - x\tan x)}^2}}}dx} \hfill \\ = \int {\frac{1}{{{{(1 - x\tan x)}^2}}}d(x\tan x)} \hfill \\ = - \frac{1}{{1 - x\tan x}} + C \hfill \\ \end{gathered} $$

例题:12求:

$$ I = \int {\ln ({{(x + a)}^{x + a}}{{(x + b)}^{x + b}})\frac{1}{{(x + a)(x + b)}}dx} $$

解:

$$ \begin{gathered} 原式 = \int {(\frac{{\ln (x + a)}}{{x + b}} + \frac{{\ln (x + b)}}{{x + a}})dx} \hfill \\ = \int {\ln (x + a)d(x + b)} + \int {\frac{{\ln (x + b)}}{{x + a}}dx} \hfill \\ = \ln (x + a)\ln (x + b) - \int {\frac{{\ln (x + b)}}{{x + a}}dx} + \int {\frac{{\ln (x + b)}}{{x + a}}dx} \hfill \\ = \ln (x + a)\ln (x + b) + C \hfill \\ \end{gathered} $$

例题13:求:$I = \int {\frac{1}{{{\text{(2}}{x^2} + 1)\sqrt {1 + {x^2}} }}dx} $

解:

$$ \begin{gathered} x = \tan t \hfill \\ 原式 = \int {\frac{1}{{(2{{\tan }^2}t + 1)\sqrt {1 + {{\tan }^2}t} }}d\tan t} \hfill \\ = \int {\frac{{\cos x}}{{{{\sin }^2}t + 1}}dt} \hfill \\ = \int {\frac{1}{{{{\sin }^2}t + 1}}d\sin t} \hfill \\ = \arctan \frac{x}{{\sqrt {1 + {x^2}} }} + c \hfill \\ \end{gathered} $$

例题14:求:

$$ \int {\frac{1}{{\sqrt {(x - a)(b - x)} }}dx} (a < x < b) $$

解:

$$ \begin{gathered} \int {\frac{1}{{\sqrt {{a^2} - {{(x - b)}^2}} }}dx} = \arcsin \frac{{x + b}}{a} + c \hfill \\ = \int {\frac{1}{{\sqrt { - {x^2} + (a + b)x - ab} }}dx} \hfill \\ {\text{ = }}\int {\frac{{\text{1}}}{{\sqrt {{{(\frac{{b - a}}{2})}^2} - {{(x - \frac{{a + b}}{2})}^2}} }}dx} = \arcsin \frac{{x - \frac{{a + b}}{2}}}{{\frac{{b - a}}{2}}} + C \hfill \\ \end{gathered} $$

例题15:求:$I = \int {\frac{1}{{1 + {x^4}}}dx} $

$$ \begin{gathered} \frac{1}{{1 + {x^4}}} = \frac{1}{{({x^2} + \sqrt 2 x + 1)({x^2} - \sqrt 2 x + 1)}} \hfill \\ = \frac{1}{{2\sqrt 2 }}(\frac{{x + \sqrt 2 }}{{{x^2} + \sqrt 2 x + 1}} - \frac{{x - \sqrt 2 }}{{{x^2} - \sqrt 2 x + 1}}) \hfill \\ = \frac{1}{{4\sqrt 2 }}(\frac{{2x + \sqrt 2 }}{{{x^2} + \sqrt 2 x + 1}} + \frac{{\sqrt 2 }}{{{x^2} + \sqrt 2 x + 1}} - \frac{{2x - \sqrt 2 }}{{{x^2} - \sqrt 2 x + 1}} + \frac{{\sqrt 2 }}{{{x^2} - \sqrt 2 x + 1}}) \hfill \\ = \int {\frac{1}{{1 + {x^4}}}dx} = \frac{1}{{4\sqrt 2 }}\ln \frac{{{x^2} + \sqrt 2 x + 1}}{{{x^2} - \sqrt 2 x + 1}} + 2\arctan (\sqrt 2 x + 1) - 2\arctan (\sqrt 2 x - 1) \hfill \\ \end{gathered} $$

例题16:求:$I = \int {\frac{1}{{x + \sqrt {{x^2} - x + 1} }}dx} $

解:令$t = x + \sqrt {{x^2} - x + 1} $,则$x = \frac{{{t^2} - 1}}{{2t - 1}}$,$dx = \frac{{2({t^2} - t + 1)}}{{{{(2t - 1)}^2}}}dt$

$$ \begin{gathered} 原式 = \int {\frac{{2({t^2} - t + 1)}}{{t{{(2t - 1)}^2}}}} = \int {\frac{2}{t} - \frac{3}{{2t - 1}} + \frac{3}{{{{(2t - 1)}^2}}}dt} \hfill \\ = 2\ln |t| - \frac{3}{2}\ln |2t - 1| - \frac{3}{2}\frac{1}{{2t - 1}} + C \hfill \\ = 2\ln |x + \sqrt {{x^2} - x + 1} | - \frac{3}{2}\ln |2x + 2\sqrt {{x^2} - x + 1} - 1| - \frac{3}{2}\frac{1}{{2x + 2\sqrt {{x^2} - x + 1} - 1}} + C \hfill \\ \end{gathered} $$

参考资料