110年升官等-微積分詳解

110年公務人員升官等考試

等 級： 薦任
類科（ 別）： 物理
科 目： 微積分

解答：$$\displaystyle \lim_{x\to \infty} \sqrt{x^2+2x+3}-x =  \lim_{x\to \infty} \cfrac{(x^2+2x+3)-x^2}{\sqrt{x^2+2x+3}+x} =  \lim_{x\to \infty} \cfrac{ 2x+3 }{\sqrt{x^2+2x+3}+x}\\ =\lim_{x\to \infty} \cfrac{ (2x+3)' }{(\sqrt{x^2+2x+3}+x)'} =\lim_{x\to \infty} \cfrac{ 2  }{{x+1\over \sqrt{x^2+2x+3}}+1} ={2\over 1+1}=\bbox[red, 2pt]{1}$$

解答：$$f(x)={3x-5\over \sqrt{3-2x}} \Rightarrow f'(x)= {3\over \sqrt{3-2x}} +{3x-5\over  (3-2x)^{3/2}} ={9-6x+3x-5 \over  (3-2x)^{3/2}} ={4-3x\over (3-2x)^{3/2}} \\ \Rightarrow f''(x)={3-3x\over (3-2x)^{5/2}}；因此f'(x)=0 \Rightarrow x={4\over 3} \Rightarrow f''({4\over 3}) ={-1\over (1/3)^{5/2}} \lt 0\\ 又x\in ({4\over 3},{3\over 2}) \Rightarrow f'(x) \lt 0，即f(x)為遞減，當x\in ({4\over 3},{3\over 2})；\\ 又x\in (-\infty,{4\over 3}) \Rightarrow f'(x) \gt 0，即f(x)為遞增，當x\in (-\infty,{4\over 3} )；\\ 因此f({4\over 3})=-\sqrt 3為\bbox[red,2pt]{極大值亦為最大值}，其坐標(x,y)=\bbox[red,2pt]{({4\over 3},-\sqrt 3)}$$

解答：$$F(x)= \int_{-x}^x \sqrt{1-t^2}\;dt \Rightarrow F'(x)= \sqrt{1-x^2}\cdot {dx\over dx}-\sqrt{1-(-x)^2}\cdot {d(-x)\over dt}= 2\sqrt{1-x^2} \\ \Rightarrow F''(x)=\bbox[red,2pt]{-2x\over \sqrt{1-x^2}}$$

解答：$$\cases{u=\ln x\\ dv =x^2dx} \Rightarrow \cases{du={1\over x}dx\\ v={1\over 3}x^3}\Rightarrow \int  x^2\ln(x)\;dx= {1\over 3}x^2\ln(x)-\int {1\over 3}x^2\;dx= {1\over 3}x^2\ln(x)- {1\over 9}x^3+C\\ \Rightarrow \int_{e^{-1}}^1  x^2\ln(x)\;dx= \left. \left[{1\over 3}x^2\ln(x)- {1\over 9}x^3 \right]\right|_{e^{-1}}^1 =0-{1\over 9}-\left({1\over 3}e^{-3}-{1\over 9}e^{-3} \right) = \bbox[red,2pt]{{4\over 9}e^{-3}-{1\over 9}}$$

解答：$$\int_0^1 \int_y^1 ye^{x^3}\;dxdy = \int_0^1\int_0^x ye^{x^3}\;dydx =\int_0^1 {1\over 2}x^2e^{x^3}\;dx =\left. \left[ {1\over 6}e^{x^3}\right]\right|_0^1 =\bbox[red, 2pt]{{1\over 6}(e-1)}$$

解答：$$令\cases{u=x^2y\\ v=y/x} \Rightarrow \cases{x=(u/v)^{1/3} \\y=(uv^2)^{1/3}} \Rightarrow J(u,v)= \begin{vmatrix} {\partial x \over \partial u} & {\partial x \over \partial v} \\ {\partial y \over \partial u} & {\partial y \over \partial v} \end{vmatrix} =\begin{vmatrix} {1\over 3}u^{-2/3}v^{-1/3} & -{1\over 3}u^{1/3}v^{-4/3} \\ {1\over 3}u^{-2/3}v^{2/3} & {2\over 3}u^{1/3}v^{-1/3} \end{vmatrix}\\={1\over 3}u^{-1/3}v^{-2/3}；同時\cases{1\le x^2y \le 2\\ x\le y\le 3x} \Rightarrow \cases{1\le u\le 2\\ 1\le v\le 3}\\ 因此\iint_\Omega xy^3\;dxdy = \int_1^3\int_1^2 (u/v)^{1/3}\cdot (uv^2) \cdot {1\over 3}u^{-1/3}v^{-2/3}\;dudv  = \int_1^3\int_1^2   {1\over 3}u v \;dudv\\ =\int_1^3 {1\over 2}v \;dv =\bbox[red,2pt]{2}$$

解答：$$C:(-1,0,1)\to(1,2,2) \Rightarrow \cases{x(t)=-1+2t\\ y(t)=2t\\ z(t)=1+t},t\in [0,1] \Rightarrow \cases{dx=2dt\\ dy=2dt\\ dz=dt} \\ \Rightarrow \int_C 3x^2y^2 \cdot dx+2x^3y\cdot dy+2z\cdot dz= \int_0^1 3(2t-1)^2(2t)^2(2dt)+2(2t-1)^3 (2t)(2dt) +2(t+1)dt \\ =\int_0^1 24t^2(2t-1)^2+ 8t(2t-1)^3 +2(t+1)\;dt\\ =\left. \left[24({4\over 5}t^5-t^4+{1\over 3}t^3) +8({8\over 5}t^5-3t^4+2t^3-{1\over 2}t^2)+t^2+2t\right] \right|_0^1\\= 24({4\over 5}-1+{1\over 3})+8({8\over 5}-3+2-{1\over 2})+1+2 ={16\over 5}+ {4\over 5}+3= \bbox[red,2pt]{7}$$

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