111年身障四等-微積分詳解

111年身心障礙人員考試

考 試 別： 身心障礙人員考試
等 別： 四等考試
類 科： 氣象
科 目： 微積分

解答：

(一) $$L= \lim_{x\to 1^+}\;(\ln x)^{x-1}  \Rightarrow \ln L=\lim_{x\to 1^+}\;(x-1)\ln (\ln x) = \lim_{x\to 1^+} \cfrac{\ln(\ln x)}{1\over x-1} \\ = \lim_{x\to 1^+} \cfrac{(\ln(\ln x))'}{({1\over x-1})'} =\lim_{x\to 1^+} \cfrac{1\over x\ln x}{-{1\over (x-1)^2}}= \lim_{x\to 1^+}\cfrac{-(x-1)^2}{x\ln x}= \lim_{x\to 1^+}\cfrac{(-(x-1)^2)'}{(x\ln x)'} \\ =\lim_{x\to 1^+}\cfrac{-2(x-1) }{1+\ln x} ={0\over 1}= 0 \Rightarrow L=e^0=1 \Rightarrow \lim_{x\to 1^+}\;(\ln x)^{x-1} =\bbox[red, 2pt]{1}$$ (二) $$f(x)=(\ln x)^{x-1}  =e^{(x-1)\ln(\ln x)}\Rightarrow f'(x)= \left(\ln(\ln x)+  {x-1\over x\ln x}\right) e^{(x-1)\ln(\ln x)} \\ =\left(\ln(\ln x)+  {x-1\over x\ln x}\right)(\ln x)^{x-1} =\bbox[red, 2pt]{(\ln x)^{x-1}\ln(\ln x)- (\ln x)^{x-2}- {1\over x}(\ln x)^{x-2}}$$

解答：

(一) $$取\cases{u=\ln(x^2-1) \Rightarrow du =2x/(x^2-1)\\ dv=dx \Rightarrow v=x} \Rightarrow \int \ln \sqrt{x^2-1}\;dx ={1\over 2}\int \ln (x^2-1)\;dx \\= {1\over 2}\left(x\ln(x^2-1)-\int {2x^2\over x^2-1} dx\right)= {1\over 2}\left(x\ln(x^2-1)-\int 2+{2 \over x^2-1} dx\right) \\= {1\over 2} x\ln(x^2-1)-x -{1\over 2}\int  {1 \over x-1} -{1\over x+1} dx = \bbox[red, 2pt]{{1\over 2} x\ln(x^2-1)-x-{1\over 2} (\ln |x-1|-\ln|x+1|) +C }$$ (二) $$\int {x+7\over x^2+2x-3}\,dx =\int {2\over x-1}-{1\over x+3}\,dx = \bbox[red, 2pt]{2\ln |x-1|-\ln|x+3| +C}$$ (三) $$u=\sqrt x \Rightarrow du={1\over 2\sqrt x}dx ={1\over 2u}dx \Rightarrow 2udu = dx \Rightarrow \int {1\over \sqrt x+4}\,dx =2\int {u\over u+4}\,du \\ =2\int 1-{4\over u+4}\,du = 2(u-4\ln(u+4))+C = \bbox[red, 2pt]{2\sqrt x-8\ln(\sqrt x+4)+C}$$

解答： $$\text{利用 lagrange 運算子求極值}，取h(x,y)=g(x,y)-{7\over 4} ，假設\cases{{\partial f\over \partial x} =\lambda {\partial h\over \partial x} \\ {\partial f\over \partial y} =\lambda {\partial h\over \partial y} \\h(x,y)=0} \\ \Rightarrow \cases{y-1= \lambda(2x+y) \cdots(1)\\ x-1= \lambda(x+2y) \cdots(2)\\ x^2+xy+y^2 -{7\over 4}=0 \cdots(3)}，{(1)\over (2)} \Rightarrow {y-1\over x-1} ={2x+y\over x+2y} \Rightarrow (x-y)(2(x+y)-1)=0\\ 若x=y代入(3) \Rightarrow x^2+x^2+x^2-{7\over 4}=0 \Rightarrow x^2={7\over 12} \Rightarrow x=\pm \sqrt{7\over 12} \Rightarrow (x,y)=(\pm {\sqrt 7\over 2\sqrt 3},\pm {\sqrt 7\over 2\sqrt 3});\\ 若2(x+y)-1=0 \Rightarrow x+y={1\over 2} \Rightarrow x^2+xy+y^2 = (x+y)^2-xy = {1\over 4}-xy = {7\over 4} \Rightarrow xy=-{3\over 2} \\ \Rightarrow y=-{3\over 2x} \Rightarrow x+y=x-{3\over 2x}={1\over 2} \Rightarrow 2x^2-x-3=0 \Rightarrow (2x-3)(x+1)=0 \\ \Rightarrow (x,y)=(3/2,-1),(-1,3/2) \\ \Rightarrow \cases{f({\sqrt 7\over 2\sqrt 3},  {\sqrt 7\over 2\sqrt 3}) = ({\sqrt 7\over 2\sqrt 3}-1)^2 ={19\over 12}-{\sqrt 7\over \sqrt 3}\\  f(-{\sqrt 7\over 2\sqrt 3},  -{\sqrt 7\over 2\sqrt 3}) = (-{\sqrt 7\over 2\sqrt 3}-1)^2 ={19\over 12}+ {\sqrt 7\over \sqrt 3} \\ f({3\over 2},-1) =  f(-1,{3\over 2})={1\over 2} \cdot (-2)=-1 } \Rightarrow \bbox[red, 2pt]{\cases{最大值={19\over 12}+{\sqrt {21}\over 3}\\ 最小值=-1}}$$

解答： $$圖形為半徑為2的球體與角錐體的交集\\利用球坐標\cases{x= \rho \sin \phi \cos \theta\\ y= \rho \sin \phi \sin \theta\\z=\rho \cos \phi} \Rightarrow D=\int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^2\sin \phi \,d\rho\, d\phi \,d\theta =\int_0^{2\pi} \int_0^{\pi/4}   {8\over 3}\sin \phi  \, d\phi \,d\theta\\ =\int_0^{2\pi}     {8\over 3}(1-{\sqrt 2\over 2}) \,d\theta ={8\over 3}(1-{\sqrt 2\over 2}) \cdot 2\pi = \bbox[red, 2pt]{{8\over 3}(2-\sqrt 2)\pi}$$
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