高数强化NO16|广义积分|抽象函数的积分|运用递推公式

抽象函数的积分

1. 含有抽象函数导数的积分
   

$$\begin{array}{rl}& \text{解：}{\int }_0^3(x^2+x)f^{\prime \prime \prime }(x)dx={\int }_0^3(x^2+x)(f^{\prime \prime }(x){)}^{\prime }dx\\ & =(x^2+x)f^{\prime \prime }(x){|}_0^3-{\int }_0^3(2x+1)f^{\prime \prime }(x)dx\\ & =12f^{\prime \prime }(3)-\left[(2x+1)f^{\prime }(x){|}_0^3-{\int }_0^{3}2f^{\prime }(x)dx\right]\\ & =12f^{\prime \prime }(3)-\left(7f^{\prime }(3)-f^{\prime }(0)-2f(x){|}_0^3\right)\\ & =12f^{\prime }(3)-7f^{\prime }(3)+f^{\prime }(0)+2f(3)-2f(0)\\ & \\ & \text{点}(3,2)\text{为}f(x)\text{的一个拐点，又}f(x)\text{二阶可导，故}{f}^{\prime \prime }(3)=0.\\ & l_1\text{过点}(2,4)\text{和}(3,2)\text{，故}{l}_1\text{的斜率为}\frac{2-4}{3-2}=-2,\text{则}{f}^{\prime }(2)=-2.\\ & l_2\text{过点}(2,4)\text{和}(0,0)\text{，故}{l}_2\text{的斜率为}\frac{4-0}{2-0}=2,\text{则}{f}^{\prime }(0)=2.\\ & f(x)\text{过}(0,0)\text{和}(2,2),\text{则}f(0)=0,f(2)=2.\\ & \\ & \text{故上式}=12\cdot 0-7\cdot (-2)+2+2\cdot 2-2\cdot 0=20\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{被积函数中出现抽象函数的导数，一定要使用分部积分；}\\ & 2.\text{出现}{f}^{\prime }(x)dx\text{，凑成}{f}^{\prime }(x)dx=df(x)\text{；出现}{f}^{\prime \prime }(x)dx\text{，凑成}{f}^{\prime \prime }(x)dx=d{f}^{\prime }(x)\text{；可}\\ & \text{能会使用多次分部积分。}\end{array}$$
2. 含有变限积分的积分

$$\begin{array}{rl}{\int }_0^{1}f(x)dx& ={\int }_0^{1}x\cdot \frac{f(x)}{x}dx={\int }_0^1{\left(\frac{1}{2}{x}^2\right)}^{\prime }\cdot \frac{f(x)}{x}dx\\ & =\frac{1}{2}{x}^2\cdot \frac{f(x)}{x}{|}_0^1-{\int }_0^1\frac{1}{2}{x}^2\cdot {\left(\frac{f(x)}{x}\right)}^{\prime }dx\\ & =-{\int }_0^1\frac{1}{2}{x}^2\cdot \frac{\sin x^2}{x}dx=-{\int }_0^1\frac{1}{2}x\sin x^{2}dx=\frac{1}{4}(\cos 1-1)\end{array}$$
$$\begin{array}{rl}& \text{改题：}\\ & f(x)=x^3{\int }_{\sqrt{x}}^x\frac{\sin t^2}{t}dt,\text{则}{\int }_0^{\sqrt{\pi }}f(x)dx=\\ & \\ & \text{解：}{\int }_0^{\sqrt{\pi }}f(x)dx={\int }_0^{\sqrt{\pi }}{x}^3\cdot \frac{f(x)}{x^3}dx={\int }_0^{\sqrt{\pi }}{\left(\frac{1}{4}{x}^4\right)}^{\prime }\cdot \frac{f(x)}{x^3}dx\\ & =\frac{1}{4}{x}^4\cdot \frac{f(x)}{x^3}{|}_0^{\sqrt{\pi }}-{\int }_0^{\sqrt{\pi }}\frac{1}{4}{x}^4\cdot \frac{\sin x^2}{x}dx\\ & =-\frac{1}{4}{\int }_0^{\sqrt{\pi }}{x}^3\sin x^{2}dx\\ & =-\frac{1}{8}{\int }_0^{\sqrt{\pi }}{x}^2\sin x^{2}d(x^2)\\ & \stackrel{t=x^2}{=}-\frac{1}{8}{\int }_0^{\pi }t\sin tdt\\ & =-\frac{1}{8}\cdot \frac{\pi }{2}\cdot {\int }_0^{\pi }\sin tdt\\ & =-\frac{1}{8}\cdot \frac{\pi }{2}\cdot 2\\ & =-\frac{\pi }{8}\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{被积函数中含有变限积分，由于变限积分易求导，将变限积分作为分部积分法中的}u\text{即可。}\\ & 2.\text{值得注意的是，这里是令}{\int }_1^x\frac{\sin t^2}{t}dt\text{这一变限积分单独作}u\text{，而不是令}f(x)\text{作}u\text{。}\end{array}$$

运用递推公式

证：(1) an+1−an=∫01(xn+1−xn)1−x2dx<0,故{an}单调减少.an=∫01xn1−x2dx=∫01xn−1⋅x1−x2dx=∫01xn−1[(1−x2)32⋅23(−12)]′dx=−13∫01xn−1[(1−x2)32]′dx=−13xn−1(1−x2)32∣01+n−13∫01(1−x2)32xn−2dx=n−13∫01(1−x2)1−x2xn−2dx=n−13(∫011−x2xn−2dx−∫01xn1−x2dx)=n−13(an−2−an), an=n−1n+2an−2n=2,3,⋯ \begin{aligned} &\text{证：}(1)\ a_{n+1}-a_n = \int_{0}^{1} \left(x^{n+1}-x^n\right)\sqrt{1-x^2} dx < 0,\text{故}\{a_n\}\text{单调减少.} \\ \\ &a_n = \int_{0}^{1} x^n \sqrt{1-x^2} dx = \int_{0}^{1} x^{n-1} \cdot x\sqrt{1-x^2} dx = \int_{0}^{1} x^{n-1} \left[(1-x^2)^{\frac{3}{2}} \cdot \frac{2}{3}\left(-\frac{1}{2}\right)\right]' dx \\ &= -\frac{1}{3}\int_{0}^{1} x^{n-1} \left[(1-x^2)^{\frac{3}{2}}\right]' dx = -\frac{1}{3}x^{n-1}(1-x^2)^{\frac{3}{2}}\bigg|_{0}^{1} + \frac{n-1}{3}\int_{0}^{1} (1-x^2)^{\frac{3}{2}} x^{n-2} dx \\ &= \frac{n-1}{3}\int_{0}^{1} (1-x^2)\sqrt{1-x^2} x^{n-2} dx = \frac{n-1}{3}\left( \int_{0}^{1} \sqrt{1-x^2} x^{n-2} dx - \int_{0}^{1} x^n \sqrt{1-x^2} dx \right) \\ &= \frac{n-1}{3}\left(a_{n-2}-a_n\right),\ a_n = \frac{n-1}{n+2}a_{n-2}\quad n=2,3,\cdots \end{aligned} ​证：(1) an+1​−an​=∫01​(xn+1−xn)1−x2​dx<0,故{an​}单调减少.an​=∫01​xn1−x2​dx=∫01​xn−1⋅x1−x2​dx=∫01​xn−1[(1−x2)23​⋅32​(−21​)]′dx=−31​∫01​xn−1[(1−x2)23​]′dx=−31​xn−1(1−x2)23​​01​+3n−1​∫01​(1−x2)23​xn−2dx=3n−1​∫01​(1−x2)1−x2​xn−2dx=3n−1​(∫01​1−x2​xn−2dx−∫01​xn1−x2​dx)=3n−1​(an−2​−an​), an​=n+2n−1​an−2​n=2,3,⋯​
$$\begin{array}{rl}& (2)1=\frac{a_{n-1}}{a_{n-1}}>\frac{a_n}{a_{n-1}}>\frac{a_{n+1}}{a_{n-1}}=\frac{n}{n+3}\to 1,\text{由夹逼准则可知}\underset{n\to \infty }{\lim }\frac{a_n}{a_{n-1}}=1\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{需要考生掌握两个公式的证明（用分部积分证明），并且记住结论：}\\ & \Gamma \text{函数：}{\int }_0^{+\infty }{x}^n{e}^{-x}dx=n!；\\ & \\ & \text{华里士公式：}{\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx={\int }_0^{\frac{\pi }{2}}{\cos }^{n}xdx=\{\begin{array}{ll}\frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdot \frac{n-5}{n-4}\cdots \frac{1}{2}\cdot \frac{\pi }{2},& n=2k\\ \frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdot \frac{n-5}{n-4}\cdots \frac{2}{3}\cdot 1,& n=2k+1\end{array}\end{array}$$

广义积分

$$\begin{array}{rl}& \text{解：}{\int }_1^{+\infty }f(x)dx={\int }_1^e\frac{1}{(x-1{)}^{\alpha -1}}dx+{\int }_e^{+\infty }\frac{1}{x(\ln x{)}^{\alpha +1}}dx\\ & =\frac{1}{2-\alpha }(x-1{)}^{2-\alpha }{|}_1^e+\frac{1}{-\alpha }(\ln x{)}^{-\alpha }{|}_e^{+\infty }\\ & \\ & \text{要使得}{(x-1{)}^{2-\alpha }|}_{x=1}\text{存在，}2-\alpha >0\\ & \text{要使得}{(\ln x{)}^{-\alpha }|}_{x=+\infty }\text{存在，}-\alpha <0\\ & \text{或}0<\alpha <2\end{array}$$
$$\begin{array}{rl}& \text{若}\alpha =2,{\int }_1^e\frac{1}{x-1}dx=\ln (x-1){|}_1^e\text{不存在}\\ & \\ & \text{若}\alpha =0,{\int }_e^{+\infty }\frac{1}{x\ln x}dx=\ln \ln x{|}_e^{+\infty }\text{不存在}\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}(1)\text{反常积分}{\int }_0^a\frac{1}{x^p(\ln x{)}^q}dx(a<1)\text{，当}p<1\text{时收敛；当}p>1\text{时发散；}p=1\\ & \text{时，}q>1\text{收敛，}q\le 1\text{发散。}\\ & (2)\text{反常积分}{\int }_a^{+\infty }\frac{1}{x^p(\ln x{)}^q}dx(a>1)\text{，当}p>1\text{时收敛；当}p<1\text{时发散；当}p=1\\ & \text{时，}q>1\text{收敛，}q\le 1\text{发散。}\end{array}$$
$$\begin{array}{rl}& x\to 0\text{时，}p\text{越小，}{x}^p\text{越大，}\frac{1}{x^p}\text{越小，收敛.}\end{array}$$
$$\begin{array}{rl}& x\to +\infty \text{时，}p\text{越大，}{x}^p\text{越大，}\frac{1}{x^p}\text{就越小，反常积分收敛.}\end{array}$$

$$\begin{array}{rl}& \text{解：}{\int }_0^{+\infty }\frac{1}{x^a(1+x{)}^b}dx={\int }_0^c\frac{1}{x^a(1+x{)}^b}dx+{\int }_c^{+\infty }\frac{1}{x^a(1+x{)}^b}dx(c>0)\\ & \\ & \text{因为}\underset{x\to 0}{\lim }\frac{x^a(1+x{)}^b}{x^a}=1,\text{故}{\int }_0^c\frac{1}{x^a(1+x{)}^b}dx\text{与}{\int }_0^c\frac{1}{x^a}dx\text{同敛散.}\\ & \\ & {\int }_0^c\frac{1}{x^a}dx=\frac{1}{1-a}{x}^{1-a}{|}_0^c\\ & 1-a>0,a<1,\text{当}a<1\text{时}{\int }_0^c\frac{1}{x^a}dx\text{收敛.}\\ & \\ & \text{故}{\int }_0^c\frac{1}{x^a(1+x{)}^b}dx\text{收敛.}\end{array}$$
$$\begin{array}{rl}& \underset{x\to +\infty }{\lim }\frac{\frac{1}{x^a(1+x{)}^b}}{\frac{1}{x^a\cdot x^b}}=1,\text{故}{\int }_c^{+\infty }\frac{1}{x^a(1+x{)}^b}dx\text{与}{\int }_c^{+\infty }\frac{1}{x^{a+b}}dx\text{同敛散.}\\ & \\ & {\int }_c^{+\infty }\frac{1}{x^{a+b}}dx=\frac{1}{1-a-b}{x}^{1-a-b}{|}_c^{+\infty }\\ & 1-a-b<0,a+b>1\\ & 选C\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{广义积分的比较判别法的使用要点可以概况成三句话：}\\ & \text{一是大收则小收；二是同阶同敛散；}\\ & \text{三是两个特殊积分的敛散性：}{\int }_a^b\frac{1}{(x-a{)}^p}dx：p<1\text{收敛，}p\ge 1\text{发散；}\\ & {\int }_1^{+\infty }\frac{1}{x^p}dx：p>1\text{收敛，}p\le 1\text{发散。}\\ & 2.\text{具体判断时，需要先将积分按照瑕点进行划分，保证一个积分中只含一个瑕点（无穷}\\ & \text{也算一个瑕点），然后在每个积分的瑕点处用上述原则进行判断。}\end{array}$$

$$\begin{array}{rl}& \text{解：}{\int }_0^1\frac{(\ln (1-x){)}^{\frac{2}{m}}}{\sqrt[n]{x}}dx={\int }_0^c\frac{(\ln (1-x){)}^{\frac{2}{m}}}{x^{\frac{1}{n}}}dx+{\int }_c^1\frac{(\ln (1-x){)}^{\frac{2}{m}}}{x^{\frac{1}{n}}}dx(0<c<1)\\ & \\ & x\to 0\text{时，}\ln (1-x)\sim -x,(\ln (1-x){)}^{\frac{2}{m}}\sim (-x{)}^{\frac{2}{m}}=x^{\frac{2}{m}}\\ & \\ & {\int }_0^c\frac{(\ln (1-x){)}^{\frac{2}{m}}}{x^{\frac{1}{n}}}dx\text{与}{\int }_0^c{x}^{\frac{2}{m}-\frac{1}{n}}dx\text{同敛散.}\\ & \\ & {\int }_0^c{x}^{\frac{2}{m}-\frac{1}{n}}dx=\frac{1}{1+\frac{2}{m}-\frac{1}{n}}{x}^{1+\frac{2}{m}-\frac{1}{n}}{|}_0^c\\ & 1+\frac{2}{m}-\frac{1}{n}>0,\text{收敛.}\end{array}$$
$$\begin{array}{rl}& (m\text{和}n\text{均为正整数，}n\ge 1,\frac{1}{n}\le 1,-\frac{1}{n}\ge -1,1-\frac{1}{n}\ge 0,1+\frac{2}{m}-\frac{1}{n}>0\\ & \text{无论}m\text{和}n\text{为什么样的正整数，积分均收敛)}\end{array}$$
$$\begin{array}{rl}& \text{再来考查}{\int }_c^1\frac{(\ln (1-x){)}^{\frac{2}{m}}}{x^{\frac{1}{n}}}dx,\text{因}\underset{x\to 1}{\lim }\frac{\frac{(\ln (1-x){)}^{\frac{2}{m}}}{x^{\frac{1}{n}}}}{(\ln (1-x){)}^{\frac{2}{m}}}=1,\text{故只需考查}{\int }_c^1(\ln (1-x){)}^{\frac{2}{m}}dx\\ & \\ & \underset{x\to 1}{\lim }\frac{(\ln (1-x){)}^{\frac{2}{m}}}{(1-x{)}^{\frac{1}{2}}}=\underset{x\to 1}{\lim }(1-x{)}^{\frac{1}{2}}(\ln (1-x){)}^{\frac{2}{m}}=0,\text{当}x\to 1\text{时，分母大于分子.}\\ & \\ & \text{又}{\int }_c^1\frac{1}{(1-x{)}^{\frac{1}{2}}}dx\text{收敛，根据大收则小收可知，无论}m,n\text{怎样的取值，}\\ & {\int }_c^1\frac{(\ln (1-x){)}^{\frac{2}{m}}}{x^{\frac{1}{n}}}dx\text{收敛. 故选(D)}\end{array}$$