2021-2022秋高数上期末

(仅供参考, 有误不许打我)

一

$AACBB$

1-5

$x\in (0,1)$ 时, $\frac{1}{2}\le \frac{1}{1+x^8}\le 1$

$\therefore f(1)=3-{\int }_0^1\frac{dt}{1+t^8}>0,f(0)=-2<0$,

且 $f^{\prime }(x)=5-\frac{1}{1+x^8}>0,\therefore f(x)$ 在 $(0,1)\uparrow$

二

2-1

$7200$

$f^{(6)}(0)x^n$ 中 $n\le 5$ 时, 为 0

$x^n$ 中 $\otimes n>6$ 时, $x=0\to$ 为 0

$x^6$ 求 6 次导, 常数

$f(x)$ 中 $x^6$ 的系数: $4+3+2+1=10$

$60x^5$, $300x^4$, $1200x^3$, $3600x^2$, $7200x$, $7200$

2-2

$\frac{3}{8}$

$$\begin{array}{rl}& {\int }_0^{\pi }\cos 4xdx\\ & ={\int }_0^{\pi }(\frac{1+\cos 2x}{2}{)}^{2}dx\\ & =\frac{1}{8}{\int }_0^{\pi }(3+4\cos 2x+\cos 4x)dx\\ & =\frac{3}{8}\pi +\frac{1}{4}{\int }_0^{\pi }\cos 2xd2x+\frac{1}{32}{\int }_0^{\pi }\cos 4xd4x\\ & {{=\frac{3}{8}\pi +\frac{1}{4}\sin 2x]}_0^{\pi }+\frac{1}{32}\sin 4x]}_0^{\pi }=\frac{3}{8}\pi \end{array}$$

平均值 = $\frac{\frac{3}{8}\pi }{\pi }=\frac{3}{8}$

2-3

套公式

2-4

$\frac{3}{2}\ln 2$

$$\begin{array}{rl}& \text{ 原式 }=\underset{x\to \mathrm{\infty }}{lim}{e}^{x\ln (\frac{x+a}{x-a})}=e^{\underset{x\to \mathrm{\infty }}{lim}x\ln (\frac{x+a}{x-a})}\\ & \underset{x\to \mathrm{\infty }}{lim}x\ln (\frac{x+a}{x-a})=\underset{x\to \mathrm{\infty }}{lim}\frac{\ln (\frac{x+a}{x-a})}{\frac{1}{x}}\stackrel{l^{\prime }H}{=}\frac{\frac{x-a}{x+a}\cdot \frac{-2a}{(x-a{)}^2}}{-\frac{1}{x^2}}\\ & \underset{x\to \mathrm{\infty }}{lim}\frac{2a{x}^2}{x^2-a^2}=\underset{x\to \mathrm{\infty }}{lim}\frac{2a}{1-\frac{a^2}{x^2}}=2a\\ & \therefore e^{2a}=8,2a=\ln 8=3\ln 2\Rightarrow a=\frac{3}{2}\ln 2\end{array}$$

2-5

$\frac{2\sqrt{2}}{\pi }$

令 $\frac{1}{n}=dx(n\to \mathrm{\infty })$

原式 $=|{\int }_0^1\sqrt{1+\cos \pi x}dx|=|\frac{2}{\pi }{\int }_0^1\sqrt{2{\cos }^2\frac{\pi x}{2}}d\frac{\pi x}{2}|=-\frac{2}{\pi }\sqrt{2}\cos \frac{\pi x}{2}{]}_0^1=\frac{2\sqrt{2}}{\pi }>0$

三

$$\begin{array}{rl}& ds=\sqrt{1+(\frac{dx}{dy}{)}^2}dy=\sqrt{1+(\frac{1}{\sqrt{y}}{)}^2}dy=\sqrt{1+\frac{1}{y}}=\sqrt{\frac{1+y}{y}}\\ & {\int }_1^{3}2\pi \cdot 2\sqrt{y}\cdot \frac{\sqrt{1+y}}{\sqrt{y}}dy{=4\pi {\int }_1^3\sqrt{1+y}dy=4\pi \cdot \frac{2}{3}(1+y{)}^{\frac{3}{2}}]}_1^3\\ & =\frac{8}{3}\pi (8-2\sqrt{2})=\frac{16}{3}\pi (4-\sqrt{2})\end{array}$$

$\therefore$ 旋转面的面积为 $\frac{16}{3}\pi (4-\sqrt{2})$

四

$y=x{\ln }^{2}x+Cx$

解: $x{y}^{\prime }-y=2x\ln x$

$$\begin{array}{rl}& y^{\prime }-\frac{y}{x}=2\ln x\\ & p(x)=-\frac{1}{x},v(x)=e^{\int p(x)dx}=e^{\int (-\frac{x}{x})dx}=e^{-\ln x}=\frac{1}{x}\\ & \therefore \frac{1}{x}{y}^{\prime }-\frac{1}{x^2}y=\frac{2\ln x}{x}\\ & (\frac{y}{x}{)}^{\prime }=\frac{2\ln x}{x},\frac{y}{x}=\int \frac{2\ln x}{x}dx=\int \frac{2\ln x}{1}d(\ln x)={\ln }^{2}x+C\\ & \Rightarrow y=x{\ln }^{2}x+Cx\end{array}$$

五

$a=e^{\frac{1}{e}}$

解:

$$\{\begin{array}{l}\text{ y equals}\to x=log(a)x\\ \text{Same slope}\to 1=\frac{1}{x\ln a}\end{array}$$

$\Rightarrow x=\frac{\ln x}{\ln a}=\frac{1}{\ln a}\therefore \ln x=1,x=e$

$$x=\frac{1}{\ln a}=e,a=e^{\frac{1}{e}}$$

六

解: $R\sin \theta +\sqrt{2}R\sqrt{1+\cos \theta }=R(\sin \theta +\sqrt{2}\cdot \sqrt{2}\cos \frac{\theta }{2})$

$$\begin{array}{r}=2R\cos \frac{\theta }{2}(1+\sin \frac{\theta }{2})\end{array}$$

令 $t=\frac{\theta }{2}\in [0,\frac{\pi }{2}]$，则：

$$\begin{array}{rl}(\cos t(1+\sin t){)}^{\prime }& =-\sin t+\cos 2t\\ & =-2{\sin }^{2}t-\sin t+1\\ & =(-2\sin t+1)(\sin t+1)\end{array}$$$$\text{ max时, }\sin t=\frac{1}{2}\Rightarrow t=\frac{\pi }{6},\theta =\frac{\pi }{3}$$

即 $\mathrm{△}$ 为等边 $\mathrm{△}ABC$ 时, $\mathrm{△}ABC$ max $=3\sqrt{3}R$

七

$P<1$

解:

$P\le 0$ 时, 其…为反常积分且收敛

$p>0$ 时

a: $\underset{x\to 0^{+}}{lim}\frac{\frac{e^{-x}}{x^p}}{\frac{1}{x^p}}=1$, 且 ${\int }_0^1\frac{1}{x^p}dx$ 在 $p<1$ 时收敛

当 $x\to 0^{+}$时, $\frac{e^{-x}}{x^P}\sim \frac{1}{x^P}$, 积分 ${\int }_0^1\frac{e^{-x}}{x^p}$ 和积分 ${\int }_0^1\frac{1}{x^p}$ 同时敛散

(极限比较判别法)

$\therefore {\int }_0^1\frac{e^{-x}}{x^p}dx$ 在 $p<1$ 时收敛

b: $\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{e^{-x}}{x^p}}{\frac{1}{x^2}}=\underset{x\to \mathrm{\infty }}{lim}\frac{x^{2-p}}{e^x}=0$

$2-p<0$ 时, 显然

$2-p>0$ 时, 洛必达求多次即可得

$\therefore 0<\frac{e^{-x}}{x^p}<\frac{1}{x^2}$, 且 $\frac{1}{x^2}$ 收敛

综上: $P\in (0,1)\cup (-\mathrm{\infty },0]$

八

(1) $\frac{-e}{2}$

(2) $\frac{3}{2}$

8-1

$\frac{-e}{2}$

解: $x\to 0$ 时, 为零比零, 使用洛必达

$$\begin{array}{rl}& ((1+x{)}^{\frac{1}{x}}{)}^{\prime }\\ & =(e^{\frac{1}{x}\ln (1+x)}{)}^{\prime }\\ & =(\frac{\ln (1+x)}{x}{)}^{\prime }\cdot e^{\frac{1}{x}\ln (1+x)}\\ & =\frac{\frac{x}{1+x}-\ln (x+1)}{x^2}\cdot e^{\frac{1}{x}\ln (1+x)}\end{array}$$

因此洛必达法则：原式

$$\begin{array}{rl}& =\frac{\frac{\frac{x}{1+x}-\ln (1+x)}{x^2}{e}^{\frac{1}{x}\ln (x+1)}-0}{1}\\ & =e\underset{x\to 0}{lim}\frac{\frac{x}{1+x}-\ln (1+x)}{x^2}(0/0)\\ & =e\underset{x\to 0}{lim}\frac{\frac{1}{(1+x{)}^2}-\frac{1}{1+x}}{2x}\\ & =e\underset{x\to 0}{lim}\frac{\frac{-x}{(1+x{)}^2}}{2x}\\ & =e\underset{x\to 0}{lim}\frac{-1}{2(1+x{)}^2}\\ & =\frac{-e}{2}\end{array}$$

(2) 解: $x\to 0$ 时

$$\begin{array}{rl}& (\cos x+1)\to 2\\ & \text{ 原式 }=\frac{1}{2}\underset{x\to 0}{lim}\frac{x}{\ln (1+x)}\cdot \frac{3\sin x+x^2\cos \frac{1}{x}}{x}\\ & =\frac{1}{2}\times 1\times 3=\frac{3}{2}\end{array}$$

九

9-1

$\frac{2}{3}$

$$\begin{array}{rl}& \text{ (1) }{\int }_{\frac{1}{e}}^e\frac{{\ln }^{2}x}{x}dx\\ & ={\int }_{\frac{1}{e}}^e{\ln }^{2}xd(\ln x)\\ & ={\frac{1}{3}{\ln }^{3}x|}_{\text{e }}^{\frac{1}{e}}\\ & =\frac{1}{3}-(-\frac{1}{3})=\frac{2}{3}\end{array}$$

9-2

$\frac{1}{4}+\frac{\pi }{8}$

$$\begin{array}{rl}& \text{ (2) }{\int }_1^{\sqrt{2}}\frac{1}{x^3\sqrt{x^2-1}}dx\\ & \text{ 令 }x=\sec \theta \text{. }\\ & dx=\sec \theta \tan \theta d\theta \\ & x=\sqrt{2}\theta =\frac{\pi }{4}\\ & x=1\theta =0\\ & \text{ 原 }={\int }_0^{\frac{\pi }{4}}\frac{\sec \theta \tan \theta }{{\sec }^3\theta \tan \theta }d\theta \\ & ={\int }_0^{\frac{\pi }{4}}{\cos }^2\theta d\theta \\ & ={\int }_0^{\frac{\pi }{4}}\frac{\cos 2\theta +1}{2\cdot 2}d2\theta \\ & ={\frac{1}{4}\sin 2\theta |}_0^{\frac{\pi }{4}}+\frac{1}{2}\cdot \frac{\pi }{4}\\ & =\frac{1}{4}+\frac{\pi }{8}\end{array}$$

9-3

$\frac{1}{20}+\frac{-\ln 5}{80}$

$$\begin{array}{rl}& \text{ (3) }{\int }_1^{\mathrm{\infty }}\frac{1}{x^6(x^5+4)}dx={\int }_1^{\mathrm{\infty }}\frac{-\frac{1}{5}}{x^5+4}d{x}^{-5}\\ & \text{ 令 }{x}^{-5}=t\\ & \text{ 则 }\int \frac{1}{x^6(x^5+4)}dx=-\frac{1}{5}\int \frac{1}{\frac{1}{t}+4}dt=-\frac{1}{5}\int \frac{t}{4t+1}dt\\ & =-\frac{1}{5}\int (\frac{1}{4}-\frac{\frac{1}{4}}{4t+1})dt\\ & =-\frac{t}{20}+\frac{1}{80}\int \frac{1}{t+\frac{1}{4}}dt\\ & =-\frac{t}{20}+\frac{1}{80}\ln (t+\frac{1}{4})+C\\ & =-\frac{x^{-5}}{20}+\frac{1}{80}\ln (x^{-5}+\frac{1}{4})+C\\ & {{\therefore \text{ 原式 }=-\frac{x^{-5}}{20}]}_1^{\mathrm{\infty }}+\frac{\ln (x^{-5}+\frac{1}{4})}{80}]}_1^{\mathrm{\infty }}\\ & =(0-\frac{1}{20})+\frac{-\ln 4-\ln \frac{5}{4}}{80}=\frac{1}{20}+\frac{-\ln 5}{80}\end{array}$$

9-4

$$\frac{4\sqrt{3}}{9}\arctan (\frac{2}{\sqrt{3}}x+\frac{1}{\sqrt{3}})+\frac{8}{9}\cdot \frac{x+\frac{1}{2}}{\frac{4}{3}(x+\frac{1}{2}{)}^2+1}+C$$$$\begin{array}{rl}& \text{ (4) }\int \frac{1}{(1+x+x^2{)}^2}dx\\ & \text{ 令 }\frac{3}{4}{\tan }^2\theta =(x+\frac{1}{2}{)}^2\\ & =\int \frac{1}{((x+\frac{1}{2}{)}^2+\frac{3}{4}{)}^2}d(x+\frac{1}{2})\text{ 则 }\frac{\sqrt{3}}{2}\tan \theta =(x+\frac{1}{2})\\ & =\int \frac{\frac{16}{9}{d}^{\frac{\sqrt{3}}{2}}\tan \theta }{({\tan }^2\theta +1{)}^2}=\frac{\sqrt{3}}{2}{\int }^{\frac{16}{9}}\frac{{\sec }^2\theta }{{\sec }^4\theta }d\theta =\frac{16}{9}\cdot \frac{\sqrt{3}}{2}\int {\cos }^2\theta d\theta \\ & =\frac{8\sqrt{3}}{9}\int \frac{\cos 2\theta +1}{2}d\theta =\frac{4\sqrt{3}}{9}\theta +\frac{2\sqrt{3}}{9}\sin 2\theta +C\\ & \because \theta =\arctan (\frac{2}{\sqrt{3}}(x+\frac{1}{2}))=\arctan (\frac{2}{\sqrt{3}}x+\frac{1}{\sqrt{3}})\\ & \therefore \text{ 原式 }=\frac{4\sqrt{3}}{9}\arctan (\frac{2}{\sqrt{3}}x+\frac{1}{\sqrt{3}})+\frac{8}{9}\cdot \frac{x+\frac{1}{2}}{\frac{4}{3}(x+\frac{1}{2}{)}^2+1}+C\end{array}$$

社恐, 害怕……仅供参考

ID: 一张看上去像宠物小精灵的画