高等数学·不定积分、定积分与反常积分

不定积分、定积分与反常积分

不定积分

一、不定积分概念

1.定义

$$\begin{array}{}\text{(1)}& & \mathrm{原}\mathrm{函}\mathrm{数}：\mathrm{设}\mathrm{对}\mathrm{于}\mathrm{区}\mathrm{间}I\mathrm{上}\mathrm{的}\mathrm{任}\mathrm{意}\mathrm{一}\mathrm{点}x\mathrm{均}\mathrm{有}{F}^{\prime }(x)=f(x),\mathrm{则}\mathrm{称}F(x)\mathrm{为}f(x)\mathrm{在}\mathrm{区}\mathrm{间}I\mathrm{上}\mathrm{的}\mathrm{一}\mathrm{个}\mathrm{原}\mathrm{函}\mathrm{数}\text{(2)}& & \mathrm{不}\mathrm{定}\mathrm{积}\mathrm{分}：\mathrm{设}\mathrm{函}\mathrm{数}f(x)\mathrm{于}\mathrm{区}\mathrm{间}I\mathrm{上}\mathrm{有}\mathrm{原}\mathrm{函}\mathrm{数},\mathrm{则}\mathrm{其}\mathrm{余}\mathrm{原}\mathrm{函}\mathrm{数}\mathrm{的}\mathrm{全}\mathrm{体}\mathrm{称}\mathrm{为}f(x)\mathrm{于}\mathrm{区}\mathrm{间}I\mathrm{上}\mathrm{的}\mathrm{不}\mathrm{定}\mathrm{积}\mathrm{分},\mathrm{记}\mathrm{为}\int f(x)dx\text{(3)}& & \mathrm{线}\mathrm{性}：\int [\alpha f(x)+\beta g(x)]dx=\alpha \int f(x)dx+\beta \int g(x)dx\end{array}$$

2.计算

$$\begin{array}{}\text{(4)}& & \mathrm{计}\mathrm{算}\mathrm{方}\mathrm{法}\{\begin{array}{ll}& 1.\mathrm{基}\mathrm{本}\mathrm{公}\mathrm{式}\\ & 2.\mathrm{线}\mathrm{性}\\ & 3.\mathrm{积}\mathrm{分}\mathrm{法}\{\begin{array}{ll}& 1.\mathrm{换}\mathrm{元}\mathrm{法}\\ & 2.\mathrm{分}\mathrm{部}\mathrm{积}\mathrm{分}\mathrm{法}\end{array}\end{array}\end{array}$$

(1)第一换元法(凑微分)

$$\begin{array}{}\text{(5)}& & \mathrm{设}{F}^{\prime }(u)=f(u),\mathrm{则}\int f(\mathrm{\Phi }(x)){\mathrm{\Phi }}^{\prime }(x)dx=\int f(\mathrm{\Phi }(x))d(\mathrm{\Phi }(x))=F(\mathrm{\Phi }(x))+C\text{(6)}& & \mathrm{注}\mathrm{解}：\mathrm{找}\mathrm{到}\mathrm{合}\mathrm{适}\mathrm{的}\mathrm{凑}\mathrm{微}\mathrm{分}{\mathrm{\Phi }}^{\prime }(x)dx=d(\mathrm{\Phi }(x))\end{array}$$

常见凑微分：

$$\begin{array}{}\text{(7)}& & 1.\int f(ax+b)dx=\frac{1}{a}\int f(ax+b)d(ax+b)(a\ne 0)\text{(8)}& & eg1.\int \sin (2x+3)dx=\frac{1}{2}\int \sin (2x+3)d(2x+3)=\frac{1}{2}\cos (2x+3)+C\text{(9)}& & 2.\int f(a{x}^n+b)x^{n-1}dx=\frac{1}{na}\int f(a{x}^n+b)d(a{x}^n+b)\text{(10)}& & eg2.\int \cos (2x^4+3)x^{3}dx=\frac{1}{4\ast 2}\int \cos (2x^4+3)d(2x^4+3)=\frac{1}{8}\cos (2x^4+3)+C\text{(11)}& & 3.\int f(a^x+c)a^{x}dx=\frac{1}{\ln a}\int f(a^x+c)d(a^x+c)\text{(12)}& & eg3.\int \sin (2^x+3)2^{x}dx=\frac{1}{\ln 2}\int \sin (2^x+3)d(2^x+3)=\frac{1}{\ln 2}\cos (2^x+3)\text{(13)}& & 4.\int f(\frac{1}{x})\frac{1}{x^2}dx=-\int f(\frac{1}{x})d(\frac{1}{x})\text{(14)}& & eg4.\int \ln (\frac{1}{x})\frac{1}{x^2}dx=-\int \ln (\frac{1}{x})d(\frac{1}{x})+C\text{(15)}& & 5.\int f(\ln |x|)\frac{1}{x}d(x)=\int f(\ln |x|)d(\ln |x|)\text{(16)}& & eg5.\int \sin (\ln |x|)\frac{1}{x}dx=\int \sin (\ln (|x|)d(\ln |x|)=\cos (\ln x)+C\text{(17)}& & 6.\int f(\sqrt{x})\frac{1}{\sqrt{x}}dx=2\int f(\sqrt{x})d(\sqrt{x})\text{(18)}& & 7.\int f(\sin x)\cos xdx=-\int (\sin x)d(\sin x)\text{(19)}& & 8.\int f(\cos x)\sin dx=\int f(\cos x)d(\cos x)\text{(20)}& & 9.\int f(\tan x){\sec }^{2}xdx=\int f(\tan x)d(\tan x)\text{(21)}& & 10.\int f(\cot x){\csc }^{2}xdx=-\int f(\cot x)d(\cot x)\text{(22)}& & 11.\int f(\arcsin x)\frac{1}{\sqrt{1-x^2}}dx=\int f(\arcsin x)d(\arcsin x)\text{(23)}& & 12.\int f(\arccos x)(-\frac{1}{\sqrt{1-x^2}})dx=\int f(\arccos x)d(\arccos x)\text{(24)}& & 13.\int f(\arctan x)\frac{1}{1+x^2}dx=\int f(\arctan x)d(\arctan x)\text{(25)}& & 14.\int f(\sqrt{x^2+a})\frac{x}{\sqrt{x^2+a}}dx=\int f(\sqrt{x^2+a})d(\sqrt{x^2+a})\text{(26)}& & \mathrm{注}\mathrm{解}：(\sqrt{x^2\pm a}{)}^{\prime }=\frac{x}{\sqrt{x^2+a}},(\sqrt{a^2-x^2}{)}^{\prime }=\frac{-x}{\sqrt{a^2-x^2}}\end{array}$$

(2)第二换元法

$$\begin{array}{}\text{(27)}& & \mathrm{设}{F}^{\prime }(u)=f(\mathrm{\Phi }(u)){\mathrm{\Phi }}^{\prime }(u),\mathrm{则}\text{(28)}& & \int f(x)dx\stackrel{x=\mathrm{\Phi }(u)}{=}\int f(\mathrm{\Phi }(u)){\mathrm{\Phi }}^{\prime }(u)du=F(u)+C=F({\mathrm{\Phi }}^{-1}(x))+C\text{(29)}& & \mathrm{注}\mathrm{解}：\mathrm{找}\mathrm{到}\mathrm{合}\mathrm{适}\mathrm{的}x=\mathrm{\Phi }(u)\end{array}$$

1)三角换元

$$\begin{array}{}\text{(30)}& & x=a\sin u,x=a\tan u,x=a\sec u\text{(31)}& & \sqrt{a^2-x^2}\stackrel{x=a\sin u}{=}a\cos u,u\in [-\frac{\pi }{2},\frac{\pi }{2}],x\in [-a,a]\text{(32)}& & \sqrt{a^2+x^2}\stackrel{x=a\tan u}{=}a\sec u,u\in (-\frac{\pi }{2},\frac{\pi }{2}),x\in (-\mathrm{\infty },\mathrm{\infty })\text{(33)}& & \sqrt{x^2-a^2}\stackrel{x=a\sec u}{=}a\tan u,u\in (\frac{\pi }{2},\pi ]\cup (0,\frac{\pi }{2}]\end{array}$$

2)倒变换

$$\begin{array}{}\text{(34)}& & x=\frac{1}{u}\mathrm{常}\mathrm{用}\mathrm{于}\mathrm{含}\frac{1}{x}\mathrm{的}\mathrm{函}\mathrm{数}\end{array}$$

3）指数(或对数)变换

$$\begin{array}{}\text{(35)}& & a^x=u\mathrm{或}x=\frac{\ln u}{\ln a}\mathrm{常}\mathrm{用}\mathrm{于}\mathrm{含}{a}^x\mathrm{的}\mathrm{函}\mathrm{数}\end{array}$$

4）用于有理化的变换

$$\begin{array}{}\text{(36)}& & \frac{1}{\sqrt{x}+\sqrt[3]{x}}\mathrm{用}x=u^6\text{(37)}& & \sqrt[n]{\frac{ax+b}{cx+d}}\mathrm{用}u=\sqrt[n]{\frac{ax+b}{cx+d}}\mathrm{或}x=-\frac{d{u}^n-b}{c{u}^n-a}\end{array}$$

(3)分部积分法

$$\begin{array}{}\text{(38)}& & \int u(x)v^{\prime }(x)dx=\int u(x)d(v(x))=u(x)v(x)-\int v(x)u^{\prime }(x)dx\text{(39)}& & \mathrm{注}\mathrm{解}：\mathrm{找}\mathrm{到}\mathrm{合}\mathrm{适}\mathrm{的}u(x),v(x)\end{array}$$

1)降幂法

$$\begin{array}{}\text{(40)}& & \int x^n{e}^{ax}dx,\int x^n\sin axdx,\int x^n\cos axdx\text{(41)}& & \mathrm{取}u(x)=x^n\end{array}$$

2)升幂法

$$\begin{array}{}\text{(42)}& & \int x^a\ln xdx,\int x^a\arcsin xdx,\int x^a\arccos xdx,\int x^a\arctan xdx\text{(43)}& & \mathrm{取}u(x)=\ln x\end{array}$$

3)循环法

$$\begin{array}{}\text{(44)}& & \int e^{ax}\sin axdx,\int e^{ax}\cos axdx\text{(45)}& & \mathrm{取}u(x)=e^{ax}\mathrm{或}\sin ax\end{array}$$

4)递推公式法

$$\begin{array}{}\text{(46)}& & \mathrm{与}n\mathrm{有}\mathrm{关}\mathrm{的}\mathrm{结}\mathrm{果}{I}_n，\mathrm{建}\mathrm{立}\mathrm{递}\mathrm{推}\mathrm{关}\mathrm{系}{I}_n=f(I_{n-1})\mathrm{或}f(I_{n-2})\end{array}$$

定积分

一、定积分概念

1.定义

$$\begin{array}{}\text{(47)}& & \mathrm{定}\mathrm{义}:\mathrm{设}\mathrm{函}\mathrm{数}f(x)\mathrm{在}\mathrm{区}\mathrm{间}[a,b]\mathrm{上}\mathrm{有}\mathrm{定}\mathrm{义}\mathrm{且}\mathrm{有}\mathrm{界}\text{(48)}& & (1)\mathrm{分}\mathrm{割}：\mathrm{将}[a,b]\mathrm{分}\mathrm{成}n\mathrm{个}[x_{i-1},x_i]\mathrm{小}\mathrm{区}\mathrm{间}\text{(49)}& & (2)\mathrm{求}\mathrm{和}：[x_{i-1},x_i]\mathrm{上}\mathrm{取}\mathrm{一}\mathrm{点}{\xi }_i,\sum _{i=1}^{n}f({\xi }_i)\mathrm{\Delta }{x}_i,\lambda =max\mathrm{\Delta }{x}_1,\mathrm{\Delta }{x}_2,...,\mathrm{\Delta }{x}_n\text{(50)}& & (3)\mathrm{取}\mathrm{极}\mathrm{限}：\mathrm{若}\underset{\lambda \to 0}{lim}\sum _{i=1}^{n}f({\xi }_i)\mathrm{\Delta }x\text{exist},\mathrm{且}\mathrm{极}\mathrm{值}\mathrm{不}\mathrm{依}\mathrm{赖}\mathrm{区}\mathrm{间}[a,b]\mathrm{分}\mathrm{发}\mathrm{以}\mathrm{及}\mathrm{点}{\xi }_i\mathrm{的}\mathrm{取}\mathrm{法},\mathrm{则}\mathrm{称}f(x)\mathrm{在}\mathrm{区}\mathrm{间}[a,b]\mathrm{上}\mathrm{可}\mathrm{积},\text{(51)}& & {\int }_a^{b}f(x)dx=\underset{\lambda \to 0}{lim}f(\xi )\mathrm{\Delta }{x}_i& \text{(52)}& & \mathrm{注}\mathrm{解}：\text{(53)}& & (1)\lambda \to 0\to \nleftarrow n\to \mathrm{\infty }\text{(54)}& & (2)\mathrm{定}\mathrm{积}\mathrm{分}\mathrm{表}\mathrm{示}\mathrm{一}\mathrm{个}\mathrm{值},\mathrm{与}\mathrm{积}\mathrm{分}\mathrm{区}\mathrm{间}[a,b]\mathrm{有}\mathrm{关},\mathrm{与}\mathrm{积}\mathrm{分}\mathrm{变}\mathrm{化}\mathrm{量}x\mathrm{无}\mathrm{关}\text{(55)}& & {\int }_a^{b}f(x)dx={\int }_a^{b}f(t)dt\text{(56)}& & (3)\mathrm{如}\mathrm{果}\mathrm{积}\mathrm{分}{\int }_0^{1}f(x)dx\text{exist},\mathrm{将}[0,1]n\mathrm{等}\mathrm{分}，\mathrm{此}\mathrm{时}\mathrm{\Delta }{x}_i=\frac{1}{n},\mathrm{取}{\xi }_i=\frac{i}{n},\text{(57)}& & {\int }_0^{1}f(x)dx=\underset{\lambda \to 0}{lim}\sum _{i=1}nf({\xi }_i)\mathrm{\Delta }{x}_i=\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^{n}f(\frac{i}{n})\end{array}$$

$$\begin{array}{}\text{(58)}& & {\int }_a^{b}f(x)dx=\underset{\lambda \to 0}{lim}\sum _{i=1}^{n}f({\xi }_i){\mathrm{\Delta }}_i=\{\begin{array}{ll}& \underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^{n}f(a+(i-1)\frac{b-a}{n})\frac{b-a}{n},\mathrm{左}\mathrm{侧}\\ & \underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^{n}f(a+i\frac{b-a}{n})\frac{b-a}{n},\mathrm{右}\mathrm{侧}\end{array}\text{(59)}& & \mathrm{中}\mathrm{点}：{\mathrm{\Phi }}_i=a+(i-1)\frac{b-a}{n}+\frac{b-a}{2n}\end{array}$$

定理：(线性)

$$\begin{array}{}\text{(60)}& & \int [\alpha f(x)+\beta g(x)]dx=\alpha \int f(x)dx+\beta \int g(x)dx\end{array}$$

注解：积分无小事

$$\begin{array}{}\text{(61)}& & \int e^{\pm x^2}dx,\int \frac{\sin x}{x}\mathrm{积}\mathrm{不}\mathrm{出}\mathrm{来}\text{(62)}& & F^{\prime }(x)=f(x),x\in I,\mathrm{连}\mathrm{续}\mathrm{函}\mathrm{数}\mathrm{一}\mathrm{定}\mathrm{存}\mathrm{在}\mathrm{原}\mathrm{函}\mathrm{数}，\mathrm{无}\mathrm{穷}\mathrm{多}\mathrm{个}\text{(63)}& & [F(x)+C{]}^{\prime }=f(x)\end{array}$$

2.定积分存在的充分条件

$$\begin{array}{}\text{(64)}& & \mathrm{若}f(x)\mathrm{在}[a,b]\mathrm{上}\mathrm{连}\mathrm{续},\mathrm{则}{\int }_a^{b}f(x)dx\mathrm{必}\mathrm{定}\mathrm{存}\mathrm{在}\text{(65)}& & \mathrm{若}f(x)\mathrm{在}[a,b]\mathrm{上}\mathrm{有}\mathrm{上}\mathrm{界},\mathrm{且}\mathrm{只}\mathrm{有}\mathrm{有}\mathrm{限}\mathrm{个}\mathrm{间}\mathrm{断}\mathrm{点},\mathrm{则}{\int }_a^{b}f(x)dx\mathrm{必}\mathrm{定}\mathrm{存}\mathrm{在}\text{(66)}& & \mathrm{若}f(x)\mathrm{在}[a,b]\mathrm{上}\mathrm{只}\mathrm{有}\mathrm{有}\mathrm{限}\mathrm{个}\mathrm{第}\mathrm{一}\mathrm{类}\mathrm{间}\mathrm{断}\mathrm{点},\mathrm{则}{\int }_a^{b}f(x)dx\mathrm{必}\mathrm{定}\mathrm{存}\mathrm{在}\end{array}$$

3.定积分的几何意义

$$\begin{array}{}\text{(67)}& & (1)f(x)⩾0,{\int }_a^{b}f(x)dx=S\end{array}$$

$$\begin{array}{}\text{(68)}& & (2)f(x)⩽0,{\int }_a^{b}f(x)dx=-S\end{array}$$

$$\begin{array}{}\text{(69)}& & (3)f(x)⩾0\cup f(x)⩽0,{\int }_a^{b}f(x)dx=S_1+S_3-S_2\end{array}$$

注解：

$$\begin{array}{}\text{(70)}& & （1）\mathrm{当}f(x)\ge 0\mathrm{时},\mathrm{定}\mathrm{积}\mathrm{分}\mathrm{的}\mathrm{几}\mathrm{何}\mathrm{意}\mathrm{义}\mathrm{是},\mathrm{以}\mathrm{区}\mathrm{间}[a,b]\mathrm{为}\mathrm{底},y=f(x)\mathrm{为}\mathrm{曲}\mathrm{边}\mathrm{的}\mathrm{曲}\mathrm{边}\mathrm{梯}\mathrm{形}\mathrm{面}\mathrm{积}\text{(71)}& & （2）\mathrm{定}\mathrm{积}\mathrm{分}\mathrm{是}\mathrm{一}\mathrm{个}\mathrm{常}\mathrm{数}，\mathrm{只}\mathrm{与}f\mathrm{和}\mathrm{区}\mathrm{间}[a,b]\mathrm{有}\mathrm{关},\mathrm{与}\mathrm{积}\mathrm{分}\mathrm{变}\mathrm{量}\mathrm{用}\mathrm{什}\mathrm{么}\mathrm{字}\mathrm{母}\mathrm{无}\mathrm{关}\text{(72)}& & {\int }_a^{b}f(x)dx={\int }_a^{b}f(t)dt\text{(73)}& & （3）{\int }_a^{b}dx=b-a\text{(74)}& & （4）{\int }_a^{a}f(x)=0,{\int }_a^{b}f(x)dx=-{\int }_b^{a}f(t)dt\end{array}$$

二、定积分的性质

1.不等式性质

$$\begin{array}{}\text{(75)}& & (1)\mathrm{保}\mathrm{序}\mathrm{性}：\mathrm{若}\mathrm{在}\mathrm{区}\mathrm{间}[a,b]\mathrm{上}f(x)⩽g(x),\mathrm{则}{\int }_a^{b}f(x)dx⩽{\int }_a^{b}g(x)dx\text{(76)}& & \mathrm{推}\mathrm{论}：\text{(77)}& & (1)f(x)\ge 0,\mathrm{\forall }x\in [a,b],\mathrm{则}{\int }_a^{b}f(x)dx\ge 0\text{(78)}& & (2)f(x)\ge 0,\mathrm{\forall }x\in [a,b],\mathrm{且}[c,d]\subset [a,b],\mathrm{则}{\int }_a^{b}f(x)dx\ge {\int }_c^{d}f(x)dx\text{(79)}& & (3)|{\int }_a^{b}f(x)dx|\le {\int }_a^b|f(x)|dx\text{(80)}& & -|f|\le f\le |f|\Rightarrow {\int }_a^b-|f|\le {\int }_a^{b}f\le {\int }_a^b|f|\Rightarrow |{\int }_a^{b}f|\le {\int }_a^b|f|\text{(81)}& & \mathrm{如}：x^2\le x^3,x\in [0,1],\mathrm{则}{\int }_0^1{x}^{3}dx\le {\int }_0^1{x}^{2}dx\end{array}$$

$$\begin{array}{}\text{(82)}& & (4)(\mathrm{估}\mathrm{值}\mathrm{不}\mathrm{等}\mathrm{式})\mathrm{若}M\mathrm{及}m\mathrm{分}\mathrm{别}\mathrm{是}f(x)\mathrm{在}[a,b]\mathrm{上}\mathrm{的}\mathrm{最}\mathrm{大}\mathrm{值}\mathrm{和}\mathrm{最}\mathrm{小}\mathrm{值},\text{(83)}& & \mathrm{则}m(b-a)⩽{\int }_a^{b}f(x)dx⩽M(b-a)\end{array}$$

$$\begin{array}{}\text{(84)}& & \mathrm{证}\mathrm{明}：M(b-a)=S_{AFDC}=S_1+S_2+S_3\text{(85)}& & m(b-a)=S_{EBDC}=S_3\text{(86)}& & {\int }_a^{b}f(x)dx=S_{ADBC}=S_2+S_3\text{(87)}& & S_3⩽S_2+S_3⩽S_1+S_2+S_3\text{(88)}& & \iff m(b-a)⩽{\int }_a^{b}f(x)dx⩽M(b-a)\end{array}$$

$$\begin{array}{}\text{(89)}& & (3)|{\int }_a^{b}f(x)dx|⩽{\int }_a^b|f(x)|dx\end{array}$$

2.中值定理

$$\begin{array}{}\text{(90)}& & (1)\mathrm{若}f(x)\mathrm{在}[a,b]\mathrm{上}\mathrm{连}\mathrm{续},\mathrm{则}{\int }_a^{b}f(x)dx=f(\xi )(b-a),(a<\xi <b)\text{(91)}& & \mathrm{称}\frac{1}{b-a}{\int }_a^{b}f(x)dx\mathrm{为}\mathrm{函}\mathrm{数}y=f(x)\mathrm{在}\mathrm{区}\mathrm{间}[a,b]\mathrm{上}\mathrm{的}\mathrm{平}\mathrm{均}\mathrm{值}\text{(92)}& & \mathrm{注}\mathrm{解}：F^{\prime }(x)=f(x),F(b)-F(a)={\int }_a^{b}f(x)dx,f(\xi )(b-a)=F^{\prime }(\xi )(b-a)\text{(93)}& & (2)\mathrm{若}f(x),g(x)\mathrm{在}[a,b]\mathrm{上}\mathrm{连}\mathrm{续}，g(x)\mathrm{不}\mathrm{变}\mathrm{号},\mathrm{则}{\int }_a^{b}f(x)g(x)dx=f(\xi ){\int }_a^{b}g(x)dx\end{array}$$

注解：

$$\begin{array}{}\text{(94)}& & {\int }_0^1\frac{x}{\sin x}dx\text{(95)}& & f(x)=\{\begin{array}{ll}& \frac{x}{\sin x},x\in [0,1]\\ & 1,x=0\end{array}\text{(96)}& & \mathrm{结}\mathrm{论}：\mathrm{有}\mathrm{限}\mathrm{处}\mathrm{点}\mathrm{的}\mathrm{函}\mathrm{数}\mathrm{不}\mathrm{影}\mathrm{响}\mathrm{定}\mathrm{积}\mathrm{分}\text{(97)}& & f(x)=\{\begin{array}{ll}& x+1,[1,2]\\ & x,[0,1]\end{array}\text{(98)}& & {\int }_0^{2}f(x)dx={\int }_0^{1}xdx+{\int }_1^2(x+1)dx\end{array}$$

$$\begin{array}{}\text{(99)}& & \mathrm{证}\mathrm{明}：\frac{1}{2}\le {\int }_0^{\frac{1}{2}}\frac{1}{\sqrt{1-x^n}}dx\le \frac{\pi }{6}\text{(100)}& & \mathrm{估}\mathrm{值}\mathrm{积}\mathrm{分}：x\in [0,\frac{1}{2}]\text{(101)}& & \end{array}$$

例题：

$$\begin{array}{}\text{(102)}& & 1.\mathrm{求}\mathrm{极}\mathrm{限}\underset{n\to \mathrm{\infty }}{lim}{\int }_0^1\frac{x^n{e}^x}{1+e^x}dx\text{(103)}& & \mathrm{根}\mathrm{据}\mathrm{积}\mathrm{分}\mathrm{容}\mathrm{易}\mathrm{知}\mathrm{道}0\le \frac{x^n{e}^x}{1+e^x}\le x^n,x\in [0,1],n\in N^{\ast }\text{(104)}& & \mathrm{用}\mathrm{积}\mathrm{分}\mathrm{的}\mathrm{保}\mathrm{号}\mathrm{性}\text{(105)}& & 0\le {\int }_0^1\frac{x^n{e}^x}{1+e^x}dx\le {\int }_0^1{x}^{n}dx=\frac{1}{n+1}\text{(106)}& & \mathrm{用}\mathrm{夹}\mathrm{逼}\mathrm{定}\mathrm{理}\text{(107)}& & \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n+1}=0\text{(108)}& & \underset{n\to \mathrm{\infty }}{lim}{\int }_0^1\frac{x^n{e}^x}{1+e^x}dx=0\end{array}$$

$$\begin{array}{}\text{(109)}& & 2.\mathrm{设}{I}_1={\int }_0^{\frac{4}{\pi }}\frac{\tan x}{x}dx,I_2={\int }_0^{\frac{4}{\pi }}\frac{x}{\tan x}dx\mathrm{则}\text{(110)}& & (A)I_1{I}_{2}1(B)1I_1{I}_2(C)I_2{I}_{1}1(D)1I_2{I}_1\text{(111)}& & \mathrm{解}：\mathrm{用}\mathrm{保}\mathrm{序}\mathrm{性}a<b,f(x)\le g(x),{\int }_a^{b}f(x)\le {\int }_a^{b}g(x)\text{(112)}& & \tan xx,x\in [0,\frac{\pi }{2}]\text{(113)}& & \frac{\tan x}{x}1\frac{x}{\tan x},x\in [0,\frac{\pi }{4}]\text{(114)}& & \mathrm{根}\mathrm{据}\mathrm{保}\mathrm{序}\mathrm{性}\text{(115)}& & {\int }_0^{\frac{\pi }{4}}\frac{\tan x}{x}dx{\int }_0^{\frac{\pi }{4}}1dx=\frac{\pi }{4}{\int }_0^{\frac{\pi }{4}}\frac{x}{\tan x},x\in [0,\frac{\pi }{4}]\text{(116)}& & \mathrm{证}：{\int }_0^{\frac{\pi }{4}}\frac{\tan x}{x}\mathrm{与}1\mathrm{的}\mathrm{关}\mathrm{系}\text{(117)}& & \mathrm{积}\mathrm{分}\mathrm{中}\mathrm{值}\mathrm{定}\mathrm{理}\text{(118)}& & {\int }_0^{\frac{\pi }{4}}\frac{\tan x}{x}=f(\xi )(\frac{\pi }{4}-0)=\frac{\tan \xi }{\xi }\ast \frac{\pi }{4},\xi \in [0,\frac{\pi }{4}]\text{(119)}& & \mathrm{根}\mathrm{据}\frac{\tan x}{x}\mathrm{在}x\in [0,\frac{\pi }{4}]\mathrm{上}\mathrm{单}\mathrm{调}\mathrm{递}\mathrm{增}\text{(120)}& & 0<f(\xi )<\frac{4}{\pi },0<{\int }_0^{\frac{\pi }{4}}\frac{\tan x}{x}<1\text{(121)}& & \mathrm{选}(B)\end{array}$$

三、积分上限函数

$$\begin{array}{}\text{(122)}& & \mathrm{如}\mathrm{果}f(x)\mathrm{在}\mathrm{区}\mathrm{间}[a,b]\mathrm{上}\mathrm{连}\mathrm{续},\mathrm{则}\mathrm{\Phi }(x)={\int }_a^{b}f(t)dt\mathrm{在}[a,b]\mathrm{上}\mathrm{可}\mathrm{导},\mathrm{且}{\int }_a^{b}f(t)dt)\text{(123)}& & ({\int }_a^{x}f(t)dt{)}^{\prime }=f(x),({\int }_a^{x^2}f(t)dt{)}^{\prime }=f(x^2)\ast 2x\text{(124)}& & \mathrm{如}\mathrm{果}f(x)\mathrm{在}\mathrm{区}\mathrm{间}[a,b]\mathrm{上}\mathrm{连}\mathrm{续},{\varphi }_1(x),{\varphi }_2(x)\mathrm{为}\mathrm{可}\mathrm{导}\mathrm{函}\mathrm{数},\mathrm{则}\mathrm{\Phi }(x)={\int }_a^{b}f(t)dt\mathrm{在}[a,b]\mathrm{上}\mathrm{可}\mathrm{导},\mathrm{且}({\int }_{{\varphi }_1(x)}^{{\varphi }_2(x)}f(t)dt{)}^{\prime }\text{(125)}& & =f[{\varphi }_2(x)]\ast {\varphi }_2^{\prime }(x)-f[{\varphi }_1(x)]\ast {\varphi }_1^{\prime }(x)=({\int }_{{\varphi }_1(x)}^{0}f(t)dt+{\int }_{{\varphi }_2(x)}^{0}f(t)dt{)}^{\prime }\text{(126)}& & \mathrm{设}\mathrm{函}\mathrm{数}f(x)\mathrm{在}[-l,l]\mathrm{上}\mathrm{连}\mathrm{续},\mathrm{则}\text{(127)}& & \mathrm{如}\mathrm{果}f(x)\mathrm{为}\mathrm{奇}\mathrm{函}\mathrm{数},\mathrm{那}\mathrm{么}{\int }_0^{x}f(t)dt\mathrm{必}\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数}\text{(128)}& & \mathrm{如}\mathrm{果}f(x)\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数},\mathrm{那}\mathrm{么}{\int }_0^{x}f(t)dt\mathrm{必}\mathrm{为}\mathrm{奇}\mathrm{函}\mathrm{数}\end{array}$$

$$\begin{array}{}\text{(129)}& & \mathrm{任}\mathrm{取}x\in [a,b),\mathrm{取}\mathrm{\Delta }x0,\mathrm{使}x+\mathrm{\Delta }x\in [a,b)\text{(130)}& & \frac{\mathrm{\Delta }F}{\mathrm{\Delta }x}=\frac{F(x+\mathrm{\Delta }x)-F(x)}{\mathrm{\Delta }x}=\frac{1}{\mathrm{\Delta }x}[{\int }_a^{x+\mathrm{\Delta }x}f(t)dt-{\int }_a^{x}f(t)dt]=\frac{1}{\mathrm{\Delta }x}{\int }_x^{x+\mathrm{\Delta }x}f(t)dt=f(x+\sigma \mathrm{\Delta }x)\to f(x)(\mathrm{\Delta }x\to 0^{+})\end{array}$$

推论：

$$\begin{array}{}\text{(131)}& & \mathrm{若}f(x)、{\varphi }^{\prime }(x)、\psi (x)\mathrm{于}[a,b]\mathrm{上}\mathrm{连}\mathrm{续},\mathrm{则}\text{(132)}& & (1)({\int }_a^{\varphi (x)}f(t)dt{)}^{\prime }=f(\varphi (x)){\varphi }^{\prime }(x)\text{(133)}& & (2)({\int }_b^{\psi (x)}f(t)dt{)}^{\prime }=-f(\psi (x)){\psi }^{\prime }(x)\text{(134)}& & (3)({\int }_{\psi (x)}^{\varphi (x)}f(t)dt{)}^{\prime }=f(\varphi (x)){\varphi }^{\prime }(x)-f(\psi (x)){\psi }^{\prime }(x)\end{array}$$

例题

$$\begin{array}{}\text{(135)}& & 1.\mathrm{设}\mathrm{函}\mathrm{数}f(x)\mathrm{在}R\mathrm{上}\mathrm{连}\mathrm{续},\mathrm{且}\mathrm{是}\mathrm{奇}\mathrm{函}\mathrm{数},\mathrm{则}\mathrm{其}\mathrm{原}\mathrm{函}\mathrm{数}\mathrm{均}\mathrm{是}\mathrm{偶}\mathrm{函}\mathrm{数}.\mathrm{当}f(x)\mathrm{是}\mathrm{偶}\mathrm{函}\mathrm{数}\mathrm{时}？\mathrm{是}\mathrm{周}\mathrm{期}\mathrm{函}\mathrm{数}？\text{(136)}& & \mathrm{证}：\text{(137)}& & \mathrm{令}{F}_0(x){\int }_0^{x}f(t)dt,x\in R\text{(138)}& & F_0(-x)={\int }_0^{-x}f(t)dt\stackrel{t=-u}{=}{\int }_0^{x}f(-u)d(u)={\int }_0^{x}f(u)du=F_0(x)\Rightarrow F_0(x)\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数}\end{array}$$

$$\begin{array}{}\text{(139)}& & \mathrm{求}\mathrm{变}\mathrm{现}\mathrm{积}\mathrm{分}\mathrm{导}\mathrm{数}\text{(140)}& & (1)F(x)={\int }_x^{e^{-x}}f(t)dt\text{(141)}& & (2)F(x)={\int }_0^{x^2}(x^2-t)f(t)dt\text{(142)}& & (3)F(x)={\int }_0^{x}f(x^2-t)dt\text{(143)}& & (4)\mathrm{设}\mathrm{函}\mathrm{数}y=y(x)\mathrm{由}\mathrm{参}\mathrm{数}\mathrm{方}\mathrm{程}\{\begin{array}{ll}& x=1+2t^2\\ & y={\int }_1^{1+2\ln t}\frac{e^u}{u}du\end{array}(t1),\mathrm{求}\frac{d^{2}y}{d{x}^2}{|}_{x=9}\text{(144)}& & \mathrm{解}:\text{(145)}& & (1)F(x{)}^{\prime }=({\int }_x^{e^{-x}}f(t)dt{)}^{\prime }=f(e^{-x})(-e^{-x})-f(x)\text{(146)}& & (2)F(x{)}^{\prime }=({\int }_0^{x^2}(x^2-t)f(t)dt{)}^{\prime }=({\int }_0^{x^2}{x}^{2}f(t)dt-{\int }_0^{x^2}tf(t)dt{)}^{\prime }\text{(147)}& & =2x{\int }_0^{x^2}f(t)dt+x^{2}f(x^2)2x-x^{2}f(x^2)2x=2x{\int }_0^{x^2}f(t)dt\text{(148)}& & (3)F(x)={\int }_0^{x}f(x^2-t)dt=-\frac{1}{2}{\int }_0^{x}f(x^2-t^2)d(x^2-t^2)\stackrel{u=x^2-t^2}{=}-\frac{1}{2}{\int }_0^{x}f(u)du\text{(149)}& & F(x{)}^{\prime }=\frac{1}{2}f(x^2)2x=xf(x^2)\text{(150)}& & (4)\frac{dy}{dx}=\frac{\frac{e^{1+2\ln t}}{1+2\ln t}\frac{2}{t}}{4t^2}=\frac{e}{2(1+2\ln t)}\text{(151)}& & \frac{d^{2}y}{d{x}^2}=\frac{d(\frac{dy}{dx})}{dx}=\frac{e}{2}(-\frac{\frac{2}{t}}{(1+2\ln t{)}^2})\frac{1}{4t}\end{array}$$

$$\begin{array}{}\text{(152)}& & 2.\mathrm{求}\mathrm{变}\mathrm{现}\mathrm{积}\mathrm{分}\mathrm{的}\mathrm{积}\mathrm{分}:\text{(153)}& & (1)\mathrm{设}f(x)={\int }_0^x\frac{\sin t}{\pi -t}dt,\mathrm{求}{\int }_0^{\pi }f(x)dx\text{(154)}& & \mathrm{解}:\text{(155)}& & {\int }_0^{\pi }f(x)dx={\int }_0^{\pi }{\int }_0^x\frac{\sin t}{\pi -t}dtdx\text{(156)}& & =x{\int }_0^x\frac{\sin t}{\pi t}{|}_0^{\pi }-{\int }_0^{\pi }x\frac{\sin x}{\pi -x}dx\text{(157)}& & =\pi {\int }_0^{\pi }\frac{\sin x}{\pi t}+{\int }_0^{\pi }\frac{[(\pi -x)-\pi ]\sin x}{\pi -x}dx={\int }_0^{\pi }\sin xdx=2\text{(158)}& & (2)\underset{x\to \mathrm{\infty }}{lim}\frac{({\int }_0^x{e}^{t^2}dt{)}^2}{{\int }_0^x{e}^{2t^2}dt}=\underset{x\to \mathrm{\infty }}{lim}\frac{(2{\int }_0^x{e}^{t^2}dt)e^{x^2}}{e^{2x^2}}=\underset{x\to \mathrm{\infty }}{lim}\frac{2{\int }_0^x{e}^{t^2}}{e^{x^2}}=\underset{x\to \mathrm{\infty }}{lim}\frac{1}{2x}=0\end{array}\$\$$$

\begin{align}
&(3)设f(x)连续,\phi(x)=\int_0^1{f(tx)dt},且\lim_{x\rightarrow0}\frac{f(x)}{x}=A(常数),求\phi'(x)并讨论\phi'(x)在x=0处的连续性\
&当x\neq0时\
&令u=tx,t\in[0,1],u=tx\in[0,x],\phi(x)=\int_0^{1f(tx)dt\overset{tx=u}{=}\int_0}x{f(u)d(\frac{u}{x})}=\frac{\int_0^xf(u)du}{x}\
&\phi'(x)=\frac{xf(x)-\int_0^xf(u)du}{x2}\
&当x=0时,f(0)=0,\phi(0)=f(0)=0,\phi'(0)=\lim_{x\rightarrow0}\frac{\phi(x)\phi(0)}{x-0}=\lim_{x\rightarrow0}\frac{\int_0^xf(u)du}{x2}=\lim_{x\rightarrow 0}\frac{f(x)}{2x}=\frac{1}{2}A\
&\lim_{x\rightarrow0}\phi'(x)=\lim_{x\rightarrow 0}{\frac{xf(x)-\int_0^xf(u)du}{x2}}=A-\frac{1}{2}A=\frac{1}{2}A=\phi'(0)\Leftrightarrow\phi'(x)在x=0处连续\
\end{align}

$$\mathrm{注}\mathrm{解}：$$

\begin{align}
&注意变限积分进行正逆运算时上下限的映射\
&例如F(x)=\int_0^{x{f(t)dt}\overset{t=-u}{=}\int_{-a}}f(-u)d(-u)\
\end{align}

$$\text{\#\#\# 四、定积分的计算 \#\#\#\# 1.牛顿莱布尼茨公式}$$

\int_a^{bf(x)dx=F(x)|_a}b=F(b)-F(a)

$$\text{\#\#\#\# 2.换元积分法}$$

\int_a^{bf(x)dx=\int_\alpha}\beta{f(\Phi(t))\Phi'(t)dt}

$$\text{\#\#\#\# 3.分部积分法}$$

\int_a^budv=uv|_ab-\int_a^bvdu

$$\text{\#\#\#\# 4.奇偶性和周期性}$$

\begin{align}
&直接使用奇偶性周期性定义证明\
&(1)设f(x)为[-a,a]上的连续函数(a0),则\
&\int_{-a}{a}f(x)dx=$$\{\begin{array}{lll}0,& f(x)\mathrm{奇}\mathrm{函}\mathrm{数}\text{2}{\int }_0^{a}f(x)dx,& f(x)\mathrm{偶}\mathrm{函}\mathrm{数}\end{array}$$\
&证：\int_{-a}^{0{f(x)dx}\overset{x=-t}{=}\int_0}a{f(-t)d(-t)}=-\int_{0}^{{a}f(t)d(t)=-\int_0}a{f(x)dx}\
\end{align}

$$$$

\begin{align}
&(2)设f(x)是以T为周期的连续函数,则对\forall A，有\int_a^{{a+T}f(x)=\int_0}T{f(x)dx}\
&\int_a^{{a+T}f(x)dx\overset{x=a+t}{=}\int_0}T{f(a+t)d(a+t)}=\int_0^{a+t}f(a+t)dt\
\end{align}

$$$$

$$\begin{array}{}\text{(159)}& & \mathrm{\Phi }:x\in [a,b]\to y\in [c,d],\mathrm{令}\frac{x-a}{b-a}=\frac{y-c}{d-c},y=c+\frac{d-c}{b-a}(x-a)\end{array}$$\

$$\text{ \#\#\#\# 5.奇偶函数积分后的奇偶性(奇偶函数求导后的奇偶性) \#\#\#\#\# 1.奇偶函数求导后的奇偶性}$$

$$\begin{array}{}\text{(160)}& & (1)f(x)\mathrm{为}\mathrm{奇}\mathrm{函}\mathrm{数}:& f(-x)=-f(x)& \iff f^{\prime }(-x)(-1)=-f^{\prime }(x)& \iff f^{\prime }(-x)=f^{\prime }(x)& \iff f^{\prime }(x)\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数}& (2)f(x)\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数}:& f(-x)=f(x)& \iff f^{\prime }(-x)=f^{\prime }(x)& \iff f^{\prime }(-x)(-1)=f^{\prime }(x)& \iff f^{\prime }(-x)=-f^{\prime }(x)& \iff f^{\prime }(x)\mathrm{为}\mathrm{奇}\mathrm{函}\mathrm{数}\end{array}$$

$$\text{\#\#\#\#\# 2.奇偶函数求积分后的奇偶性}$$

$$\begin{array}{}\text{(161)}& & \mathrm{设}F(x)\mathrm{为}f(x)\mathrm{的}\mathrm{原}\mathrm{函}\mathrm{数}& (1)f(x)\mathrm{为}\mathrm{奇}\mathrm{函}\mathrm{数}:& f(-x)=-f(x)& \iff \int f(-x)dx=-\int f(x)dx& \iff -\int f(-x)d(-x)=-\int f(x)dx& \iff F(-x)=F(x)& \iff F(x)\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数}& (2)f(x)\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数}:& f(-x)=f(x)& \iff \int f(-x)dx=\int f(x)dx& \iff -\int f(-x)d(-x)=\int f(x)dx& \iff F(-x)=-F(x)& \iff F(x)\mathrm{为}\mathrm{奇}\mathrm{函}\mathrm{数}\end{array}$$

$$\text{\#\#\#\#\# 3.奇偶函数复合后的奇偶性}$$

$$\begin{array}{}\text{(162)}& & \text{exist}f(x),g(x),F(x)=f(g(x))& \mathrm{设}f(x)\mathrm{为}\mathrm{奇}\mathrm{函}\mathrm{数}& (1)g(x)\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数}& F(-x)=f(g(-x))=f(g(x))=F(x),F(x)\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数}& (2)g(x)\mathrm{为}\mathrm{奇}\mathrm{函}\mathrm{数}& F(-x)=f(g(-x))=f(-g(x))=-f(g(x))=-F(x),F(x)\mathrm{为}\mathrm{奇}\mathrm{函}\mathrm{数}& \mathrm{设}f(x)\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数}& (1)g(x)\mathrm{为}\mathrm{奇}\mathrm{函}\mathrm{数}& F(-x)=f(g(-x))=f(g(x))=F(x),F(x)\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数}& (2)g(x)\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数}& F(-x)=f(g(-x))=f(g(x))=F(x),F(x)\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数}& \mathrm{注}\mathrm{解}:\mathrm{外}\mathrm{偶}\mathrm{全}\mathrm{偶},\mathrm{外}\mathrm{奇}\mathrm{奇}\mathrm{偶}\end{array}$$

$$\mathrm{例}\mathrm{题}：$$

\begin{align}
&1.设M=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\frac{\sin x}{1+x^2}\cos4xdx},N=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(\sin x^3+\cos4x)dx},P=\int_{-\frac{\pi}{2}}^{{\frac{\pi}{2}}(x}2\sin^3x-\cos4x)dx,则\
&(A)N<P<M(B)M<P<N(C)N<M<P(D)P<M<N\
&根据对称性判断\
&M:f_M(x)为奇函数，F_M(x)为偶函数\
&N:N=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(\sin x^3+\cos4x)dx}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin ^{3xdx+\int_{-\frac{\pi}{2}}}{2}}\cos ^4xdx\
&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin ^{3xdx=0,\int_{-\frac{\pi}{2}}}{2}}\cos ^4xdx\geq 0,\Rightarrow N\geq 0\
&P:P=\int_{-\frac{\pi}{2}}^{{\frac{\pi}{2}}(x}2\sin^3x-\cos4x)dx=\int_{-\frac{\pi}{2}}^{{\frac{\pi}{2}}x}2\sin^{3xdx-\int_{-\frac{\pi}{2}}}{2}}\cos^4xdx\
&\int_{-\frac{\pi}{2}}^{{\frac{\pi}{2}}x}2\sin^{3xdx=0,\int_{-\frac{\pi}{2}}}{2}}\cos^4xdx\geq0,\Rightarrow P\leq0\
&\Leftrightarrow P<M<N,\space\space选(D)\
\end{align}

$$$$

\begin{align}
&2.设f(x)=\begin{cases}&kx,0\leq x\leq \frac{1}{2}a\&c,\frac{1}{2}a<x\leq a\end{cases},求F(x)=\int_0^xf(t)dt,x\in[0,a]\
&F(x)=\begin{cases}&\int_0^{xktdt=\frac{1}{2}kt}2|_0^{x=\frac{1}{2}kx}2,0\leq x\leq \frac{1}{2}a\&\int_0^{{\frac{1}{2}a}ktdt+\int_{\frac{1}{2}a}}c cdt=\frac{1}{8}ka^2+c2-\frac{1}{2}ac,\frac{1}{2}a<x\leq a\end{cases}\
\end{align}

$$$$

$$\begin{array}{}\text{(163)}& & 3.\mathrm{证}\mathrm{明}：{\int }_0^{2\pi }f(|\cos x|)dx=4{\int }_0^{\frac{\pi }{2}}f(|\cos x|)dx\end{array}$$

$$\text{ \#\#\#\# 6.已有公式}$$

\begin{align}
&(1)\int_0^{{\frac{\pi}{2}}{\sin}nxdx=\int_0^{{\frac{\pi}{2}}\cos}n xdx=\begin{cases}\frac{n-1}{n}\frac{n-3}{n-2}...\frac{1}{2}\frac{\pi}{2},&n为偶数\frac{n-1}{n}\frac{n-3}{n-2}...*\frac{2}{3},&n为大于1的奇数\end{cases}}\
&(2)\int_0^{\pi}xf(\sin x)dx=\frac{\pi}{2}\int_0^{\pi}f(\sin x)dx(f(x)为连续函数)\
\end{align}

$$\text{\#\#\#\# 7.与定积分有关的证明 \#\#\#\# 8.经典例题： \#\#\#\#\# 例题1:}$$

\begin{align}
&\lim_{n\rightarrow \infty}{(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n})}\
&法1：夹逼定理+基本不等式\
&\frac{1}{1+x}<\ln(x+1)<x\
&令x=\frac{1}{n}\
&得\frac{1}{n+1}=\frac{\frac{1}{n}}{\frac{1}{n}+1}<\ln(\frac{1}{n}+1)=\ln(n+1)-\ln(n)<\frac{1}{n}\
&得\frac{1}{n+2}<ln(n+2)-ln(n+1)<\frac{1}{n+1}\
&得\frac{1}{n+n}<\ln(n+n)-\ln(n+n-1)<\frac{1}{n+n-1}\
&得\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}<ln(2n)-ln(n)=ln2\
&法2：\lim_{n\rightarrow \infty}{(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n})}中\
&\frac{1}{n+1}中n为主体，1为变体\
&\frac{变体}{主体}\rightarrow^{n \rightarrow{\infty}}$$\{\begin{array}{l}0,\mathrm{次}(\mathrm{夹}\mathrm{逼}\mathrm{定}\mathrm{理})\text{A}\ne 0,\mathrm{同}(\mathrm{定}\mathrm{积}\mathrm{分})\end{array}$$\
&\lim_{\lambda \rightarrow 0}{\sum_{i=1}^{n}{f(\xi_i)\Delta x_i}=\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{i=1}^{{n}f(\xi_i)(b-a)}=\int_0}1\frac{1}{1+x}=\ln(1+x)|_{0}^{1}=\ln2\
\end{align}

$$\text{\#\#\#\#\# 例题2}$$

\begin{align}
&设f(x)=\int_0^{\pi}{\frac{\sin x}{\pi-t}dt},计算\int_0^{\pi}f(x)dx.\
&法1：分部积分+换元法\
&原式=xf(x)|_0^{\pi}-\int_0dx}\
&=\pi{\int_0^{{\pi}{\frac{\sin{t}}{\pi-t}dt}-\int_0}{\pi-x}}dx}\
&=\int_0^{\pi}{\frac{(\pi-x)\sin x}{\pi-x}dx}=2\
&法2：\
&原式=\int_0^{\pi{f(x)d(x-{\pi})}=(x-\pi)f(x)|_0}-\int_0^{\pi}{\frac{(x-\pi)\sin x}{\pi-x}dx}=2\
&法3：二重积分转化为累次积分\
&原式=\int_0^{\pi}{\int_0\frac{x\sin t}{\pi-t}dt}dx\
\end{align}

$$\text{\#\#\#\#\# 例题3  }$$

$$\begin{array}{}\text{(164)}& & \mathrm{法}1：\mathrm{构}\mathrm{造}\mathrm{辅}\mathrm{助}\mathrm{函}\mathrm{数}& \mathrm{根}\mathrm{据}\mathrm{题}\mathrm{意}f(1)=f(-1)=1,f(0)=-1\Rightarrow f(x)\mathrm{为}\mathrm{偶}\mathrm{函}\mathrm{数},f\mathrm{最}\mathrm{低}\mathrm{点}\mathrm{函}\mathrm{数}\mathrm{值}\mathrm{为}-1& \mathrm{可}\mathrm{以}\mathrm{构}\mathrm{造}\mathrm{符}\mathrm{合}\mathrm{题}\mathrm{意}\mathrm{的}\mathrm{辅}\mathrm{助}\mathrm{函}\mathrm{数}f(x)=2x^2-1& \mathrm{法}2：\mathrm{根}\mathrm{据}\mathrm{函}\mathrm{数}\mathrm{的}\mathrm{性}\mathrm{质}\mathrm{直}\mathrm{接}\mathrm{判}\mathrm{断}\end{array}$$

$$\text{ \#\#\#\#\# 例题4 AZZK.jpg)}$$

\begin{align}
&因为\lim_{x\rightarrow 0}{\frac{ax-\sin x}{\int_b^{x{\frac{\ln{1+t}3}}{t}dt}}}=c(c\neq 0)\
&所以\lim_{x\rightarrow 0}{ax-\sin x}=0并且\lim_{x \rightarrow 0}{\int_b^{x{\frac{\ln{1+t}3}}{t}dt}}=0\
&化简,使用洛必达法则上下求导\
&\lim_{x\rightarrow 0}{\frac{ax-\sin x}{\int_b^{x{\frac{\ln{1+t}3}}{t}dt}}}=\lim_{x\rightarrow 0}{\frac{a-\cos x}{\frac{\ln{1+x^3}}{x}}}=\lim_{x\rightarrow 0}{\frac{a-\cos x}{x^2}}\
&\Rightarrow a=1,c=\frac{1}{2},b=0\
\end{align}

$$\text{\#\# 反常积分 \#\#\# 一、无穷区间上的反常积分}$$

\begin{align}
&(1)\int_a^{+\infty}{f(x)}dx=\lim_{t\rightarrow +\infty}{\int_{a}^{t}f(x)dx}\
&(2)\int_{-\infty}^{b}{f(x)}dx=\lim_{t\rightarrow -\infty}{\int_{t}^{b}f(x)dx}\
&(3)\int_{-\infty}^{{0}{f(x)}dx和{\int_{0}}f(x)dx}都收敛,则{\int_{-\infty}^{+\infty}f(x)dx}收敛\
&且{\int_{-\infty}^{{+\infty}f(x)dx}=\int_{-\infty}}dx+{\int_{0}^{+\infty}f(x)dx}\
&如果其中一个发散,结果也发散\
&常用结论：\int_a^{{+\infty}{\frac{1}{x}p}dx}\begin{cases}&p1,收敛\&p\leq1 ,发散\end{cases},(a0)\
\end{align}

$$\text{\#\#\# 二、无界函数的反常积分}$$

\begin{align}
&如果函数f(x)在点a的任一领域内都无界,那么点a为函数f(x)的瑕点(也称为无界点).无界函数的反常积分也成为瑕积分\
&(1)设函数f(x)在(a,b]上连续,点a为f(x)的瑕点.如果极限\lim_{t\rightarrow a^+}{\int_{t}{f(x)dx}}\exist,\
&则称此极限为函数f(x)在区间[a,b]上的反常区间,记作\int_{a}^{{b}f(x)dx,即\int_{a}}f(x)dx=\lim_{t\rightarrow a^+}{\int_{t}{f(x)dx}}\
&这时也称反常积分\int_a^{b{f(x)dx}收敛,如果上述极限不存在，则反常积分\int_a}b{f(x)dx}发散\
&(2)设函数f(x)在[a,b)上连续,点b为函数f(x)的瑕点,则可以类似定义函数f(x)在区间[a,b]上的反常积分\int_a^bf(x)dx=\lim_{t\rightarrow b^-}{\int_atf(x)dx}\
&设函数f(x)在[a,b]上除点c(a<c<b)外连续,点c为函数f(x)的瑕点,如果反常积分\int_a^{c{f(x)dx}和\int_c}b{f(x)dx}都收敛\
&则称反常积分\int_a^{b{f(x)dx}收敛,且\int_a}b{f(x)dx}=\int_a^{c{f(x)dx}+\int_c}b{f(x)dx}\
&如果至少一个发散,则称\int_a^b{f(x)dx}发散\
&常用结论：\
&\int_a^{b{\frac{1}{(x-a)}p}}\begin{cases}&p<1,收敛\&p\geq 1,发散\end{cases}\
&\int_a^{b{\frac{1}{(x-a)}p}}\begin{cases}&p<1,收敛\&p\geq 1,发散\end{cases}\
\end{align}

$$\text{\#\#\# 三、例题 \#\#\#\#\# 例题1 }$$

$$\text{begin{align} \&intfrac{1}{ln^{alpha}x}d(ln x)rightarrow^{ln x=u}int{frac{du}{u^{alpha+1}}}begin{cases}\&{alpha-1< 1}\&{alpha+11}end{cases}Rightarrow 0<alpha<2 end{align}}$$

$$\text{\#\# 定积分的应用 \#\#\#\# 微元法}$$

$$\begin{array}{}\text{(165)}& & \end{array}$$

$$\text{\#\#\# 一、几何应用 \#\#\#\# 1.平面图形的面积}$$

\begin{align}
&(1)若平面域D由曲线y=f(x),y=g(x)(f(x)\geq g(x)),x=a,x=b(a<b)所围成,则平面域D的面积为\
&S=\int_a^b{[f(x)-g(x)]dx}\
&(2)若平面域D由曲线由\rho=\rho(\theta),\theta=\alpha,\theta=\beta(\alpha<\beta)所围成,则其面积为S=\frac{1}{2}\int_{\alpha}^{\beta}{\rho2(\theta)d\theta}
\end{align}

$$\text{\#\#\#\# 2.旋转体的体积}$$

\begin{align}
&若区域D由曲线y=f(x)(f(x)\geq 0)和直线x=a,x=b(0\leq a<b)及x轴所围成,则\
&(1)区域D绕x轴旋转一周所得到的旋转体体积为V_x=\pi\int_a^b{f2(x)dx}\
&(2)区域D绕y轴旋转一周所得到的旋转体体积为V_y=2\pi\int_a^b{xf(x)dx}\
&(3)区域D绕y=kx+b轴旋转一周所得到的旋转体体积为V=2\pi\int_D\int{r(x,y)d\sigma}\
&例如：求y=x,y=x^2在第一象限的封闭图形绕转轴的体积\
\end{align}

$$![img](https://cdn.jsdelivr.net/gh/blueflylabor/images@1.0/U1$$

\begin{align}
&V_x=2\pi\int_D\int yd\sigma=2\pi\int_0^1{dx}\int_{x2}^{x}ydy\
&V_y=2\pi\int_D\int xd\sigma=2\pi\int_0^1{dx}\int_{x2}^{x}xdy\
&V_{x=1}=2\pi\int_D\int (1-x)d\sigma\
&V_{y=2}=2\pi\int_D\int (2-y)d\sigma\
\end{align}

$$\text{\#\#\#\# 3.曲线弧长}$$

\begin{align}
&(1)C:y=y(x),a\leq x\leq b,s=\int_a^b{\sqrt{1+y'2}dx}\
&(2)C:\begin{cases}&x=x(t)\&y=y(t)\end{cases},\alpha \leq t\leq \beta,s=\int_{\alpha}^{{\beta}{\sqrt{x'}2+y'^2}dx}\
&(3)C:\rho=\rho(\theta),\alpha \leq \theta\leq \beta,s=\int_{\alpha}^{{\beta}{\sqrt{\rho}2+\rho'^2}dx}\
\end{align}

$$\text{\#\#\#\# 4.旋转体侧面积}$$

\begin{align}
&曲线y=f(x)(f(x)\geq 0)和直线x=a,x=b(0\leq a<b)及x轴所围成的区域绕x轴旋转所得到的旋转体的侧面积为\
&S=2\pi\int_a^{b{f(x)\sqrt{1+f'}2(x)}dx}\
\end{align}

$$\text{\#\#\# 二、物理应用 \#\#\#\# 1.压力 \#\#\#\# 2.变力做功 \#\#\#\# 3.引力（较少考） \#\#\#\# 例题1}$$

\begin{align}
&分析题意可知,该容器由x^2+y2=1的圆和x^2+(y-1)2=1的偏心圆组成\
&根据图像的对称性可以避免不同表达式带来的困难\
&对圆的小带子进行积分，带子长度为x，积分区间为-1到\frac{1}{2}，\int_{-1}^{\frac{1}{2}}{\pi x^2dy}\
&由于图像的对称性，将积分结果乘二\
&(1)V=2\pi\int_{-1}^{{\frac{1}{2}}{x}2}dy=2\pi\int_{-1}^{{\frac{1}{2}}{(1-y}2)dy}=\frac{9\pi}{4}\
\end{align}

$$![\mathrm{屏}\mathrm{幕}\mathrm{截}\mathrm{图}2021-04-19203327](https://cdn.jsdelivr.net/gh/blueflylabor/images@1.0/$$

\begin{align}
&(2)W=FS=GS=mgS=\rho VSg\
&上部为W_1=\int_{\frac{1}{2}}^{2}(2y-y2)(2-y)dy\rho g\
&下部为W_2=\int^{{\frac{1}{2}}_{-1}(1-y}2)(2-y)dy*\rho g\
&W=W_1+W_2\
\end{align}

$$\text{ \#\#\#\# 例题2 }$$

\begin{align}
&F_p=PA=\rho ghA\
&将图像分为上部和下部，上部为矩形区域和下部的抛物线围成的面积区域，对其进行依次求解\
&P_1=2\rho gh\int_1^{h+1}{h+1-y}dy=\rho gh^2\
&P_2=2\rho gh\int_0^1{(h+1-y)\sqrt{y}dy=4\rho g(\frac{1}{3}h+\frac{2}{15})}\
&\frac{P_1}{P_2}=\frac{4}{5}\Rightarrow h=2,h=-\frac{1}{3}(舍去)
\end{align}

$$.jpg)$$