1
设切于 \((x_0,\ln x_0)\),则 \(l:y=\frac{1}{x_0}(x-x_0)+\ln x_0(2 \le x_0 \le 6)\)
\[\begin{aligned} S=S(x_0)=&\int_2^6 \left(\frac{x-x_0}{x_0}+\ln x_0-\ln x \right)dx \\ =&\left( \frac{(x-x_0)^2}{2x_0}+x\ln x_0-x\ln x+x \right)\bigg|_2^6 \\ =&\frac{(6-x_0)^2-(2-x_0)^2}{2x_0}+4\ln x_0+4-6\ln 6+2\ln 2 \\ =&\frac{16}{x_0}+4\ln x_0+(2\ln 2-6\ln 6) \end{aligned} \]所以:\(S'(x)=-\frac{16}{x^2}+\frac{4}{x}=\frac{4(x-4)}{x^2}\)
所以取 \(x_0=4\),则 \(l:y=\frac{x}{4}-1+2\ln 2\)
2
(1)
\[\begin{aligned} L =&\int_0^{2\pi} \sqrt{[x'(t)]^2+[y'(t)]^2}dt \\ =&\int_0^{2\pi} \sqrt{(1-\cos t)^2+(\sin t)^2} dt \\ =&\int_0^{2\pi} \sqrt{2-2\cos t} dt \\ =&\int_0^{2\pi} \sqrt{2(2\sin^2 \frac{t}{2})} dt \\ =&4\int_0^{2\pi} \sin \frac{t}{2} d\frac{t}{2} \\ =&-4\cos \frac{t}{2} \bigg|_0^{2\pi} \\ =&-4(-1-1)=8 \end{aligned} \](2)
\[\begin{aligned} S \xlongequal{x'(t) \ge 0}&\int_0^{2\pi}y(t)x'(t)dt \\ =&\int_0^{2\pi}(1-\cos t)^2dt \\ =&8\int_0^{2\pi}\sin^4\frac{t}{2} d\frac{t}{2} \\ =&16\int_0^{\frac{\pi}{2}} \sin^4 tdt \\ =&16 \frac{3!!}{4!!}\frac{\pi}{2} \\ =&3\pi \end{aligned} \]3
\[\begin{aligned} &y'(t)=\left(e^t \int_{\frac{\pi}{2}}^{t}\frac{\cos2u}{e^u}du \right)'=e^t \int_\frac{\pi}{2}^t\frac{\cos 2u}{e^u}du+e^t\frac{\cos 2t}{e^t} \Rightarrow y'(\frac{\pi}{2})=\cos \pi=-1 \\ &x'(t)=e^t\int_\frac{\pi}{2}^t\frac{\sin \frac{u}{3}}{e^u}du+e^t \frac{\sin \frac{u}{3}}{e^t} \Rightarrow x'(\frac{\pi}{2})=\sin \frac{\pi}{6}=\frac{1}{2} \\ &k=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{y'(\frac{\pi}{2})}{x'(\frac{\pi}{2})}=-2 \\ &x(\frac{\pi}{2})=y(\frac{\pi}{2})=0 \\ &l_{切}: y=-2x \\ &l_{法}: y=\frac{x}{2} \end{aligned} \]4
即:\(D=\{(x,y)|(x-1)^2+y^2 \le 1,y \ge x\}\)
\[\begin{aligned} V =& \pi \int_0^1 ((2-(1-\sqrt{1-y^2}))^2-(2-y)^2)dy \\ =& \pi \int_0^1 ( -2-2y^2+4y+2\sqrt{1-y^2} ) dy \\ =& \pi (-2 - \frac{2}{3} +2 +\frac{\pi}{2} ) \\ =& \frac{\pi^2}{2}-\frac{2\pi}{3} \end{aligned} \]5
(1)
设切于点 \((x_0,\sqrt{x_0-2})\),切线为 \(l:x=2\sqrt{x_0-2}y+1\)
所以:\(x_0=2(x_0-2)+1 \Rightarrow x_0=3 \Rightarrow l:x=2y+1\)
\[\begin{aligned} S =& \int_1^{3} \frac{x-1}{2} dx-\int_2^3 \sqrt{x-2} dx \\ =& 1-\left( \frac{(x-2)^\frac{3}{2}}{\frac{3}{2}} \bigg|_2^3 \right) \\ =& \frac{1}{3} \end{aligned} \](2)
\[\begin{aligned} V =& \pi \int_1^{3} (\frac{x-1}{2})^2 dx-\pi \int_2^3 (\sqrt{x-2})^2 dx \\ =& \frac{2\pi}{3}-\frac{\pi}{2} \\ =& \frac{\pi}{6} \end{aligned} \](3)
\[\begin{aligned} V =&\int_1^3 2\pi x \frac{x-1}{2}dx-\int_2^3 2\pi x \sqrt{x-2}dx \\ =& \frac{14\pi}{3}+\frac{52\pi}{15} \\ =& \frac{122\pi}{15} \end{aligned} \]6
7
8
9
(1)
\[\begin{aligned} &xf'(x)=f(x)+\frac{3a}{2}x^2 \\ &\left(\frac{f(x)}{x}\right)'=\frac{3a}{2} \\ &f(x)=(\frac{3a}{2}x+C)x=\frac{3a}{2}x^2+Cx \\ &2=\int_0^1f(x)dx=\left(\frac{a}{2}x^3+\frac{Cx^2}{2}\right) \bigg|_0^1=\frac{a}{2}+\frac{C}{2} \Rightarrow C=4-a \\ &f(x)=\frac{3a}{2}x^2+(4-a)x \end{aligned} \](2)
求得 \(a\) 得范围为:
\[\begin{aligned} &\begin{cases} f(1) \ge 0 \Rightarrow \frac{3a}{2}+4-a \ge 0 \Rightarrow a \ge -8 \\ f'(0) \ge 0 \Rightarrow 4-a \ge 0 \Rightarrow a \le 4 \end{cases} \\ &\Rightarrow -8 \le a \le 4 \end{aligned} \]求得体积表达式:
\[\begin{aligned} V=&\pi \int_0^{1} f(x)^2 dx \\ =&\pi \int_0^1 \left( \frac{9a^2}{4}x^4+3a(4-a)x^3+(4-a)^2x^2 \right)dx \\ =&\pi \left( \left( \frac{9a^2}{20}x^5+ \frac{3a(4-a)}{4}x^4+\frac{(4-a)^2}{3}x^3 \right)\bigg|_0^1\right) \\ =&\pi \left(\frac{9a^2}{20}+\frac{-3a^2+12a}{4}+\frac{a^2-8a+16}{3} \right) \\ \end{aligned} \]求导得:
\[V'(a)=\frac{9a}{10}+\frac{-6a+12}{4}+\frac{2a-8}{3}=\frac{54a-90a+180+40a-160}{60}=\frac{a+5}{15} \]所以 \(a=-5\)
10
(1)
考虑点 \((r,\theta)\)
\[\begin{aligned} &\begin{aligned} dS=& \frac{1}{2} (r+dr)^2d\theta -\frac{1}{2}r^2d\theta \\ =&(rdr+\frac{(dr)^2}{2})d\theta \\ =&(rdr+o(dr))d\theta \\ \end{aligned} \\ &\begin{aligned} dV=& \int_0^{r}2\pi r\sin \theta dS \\ =&\int_0^{r(\theta)} \left(2\pi r^2 \sin \theta dr d\theta +\frac{(dr)^2}{2}d\theta \right) \\ =& 2\pi \sin \theta d\theta \int_0^{r(\theta)}r^2 dr \\ =& \frac{2\pi}{3} r^3(\theta) \sin \theta d\theta \\ \end{aligned} \\ &\begin{aligned} V=&\int_\alpha^{\beta} dV \\ =&\int_\alpha^\beta2\pi r^3(\theta) \sin \theta d\theta \end{aligned} \end{aligned} \]