108年聯合大學轉學考-微積分詳解

國立聯合大學108學年度暑假轉學生招生考試

科目：微積分

一、選擇題(單選題，每題5分)

解答：$$\int_0^1 {1 \over 1+x^2}\,dx =\left.\left[ \tan^{-1}x \right]\right|_0^1 =\tan^{-1} 1，故選\bbox[red, 2pt]{(B)}$$

解答：$$\int_2^4 {1 \over 1-x^2}\,dx = \int_2^4 {1 \over(1-x)(1+x)}\,dx= {1\over 2}\int_2^4 {1 \over 1-x}+ {1\over 1+x}\,dx = {1\over 2} \left.\left[ -\ln |1-x| +\ln|1+x|  \right]\right|_2^4 \\={1\over 2}(-\ln 3+\ln 5-\ln 3)={1\over 2}\ln 5-\ln 3，故\bbox[blue, 2pt]{無解}$$

解答：$$利用複變的柯西積分:z=\cos\theta +i\sin\theta =e^{i\theta} \Rightarrow \cases{dz = ie^{i\theta}d\theta =izd\theta \Rightarrow d\theta ={1\over iz}dz \\\cos \theta=(e^{i\theta}+e^{-i\theta}) /2 =(z+1/z)/2}\\ 因此\int_0^{2\pi }{2\over \sqrt 2-\cos \theta}d\theta = \oint_C {2\over \sqrt 2-(z+1/z)/2}\cdot {1\over iz}\,dz =\oint_C{4i\over (z-(\sqrt 2-1))(z-(\sqrt 2+1))}dz,其中C為單位圓 \\ ={4i\over -2} \times 2\pi i =\bbox[red, 2pt]{4\pi}$$

解答：$$\lim_{\theta\to 1} {\sin(\theta-1) \over \theta-1} =\lim_{\theta\to 1} {(\sin(\theta-1))' \over (\theta-1)'} =\lim_{\theta\to 1} {\cos(\theta-1) \over  1} = 1，故選\bbox[red, 2pt]{(C)}$$

解答：$$\lim_{x\to \pi/2} {\sec x \over 1+\tan x} =\lim_{x\to \pi/2} {1 \over \cos x+\sin x} =1，故選\bbox[red, 2pt]{(B)}$$

解答：$${x^2\over 1}-{y^2\over 4}=1 為一雙曲線\Rightarrow {x^2\over 1}-{y^2\over 4}={z\over 3}為雙曲拋物面，故選\bbox[red, 2pt]{(D)}$$

解答：$$y={1\over x^2-4x+3} \Rightarrow y'=-{2x-4\over (x^2-4x+3)^2} \Rightarrow y'(0) = -{-4\over 9} ={4\over 9}，故選\bbox[red, 2pt]{(C)}$$

解答：$$f(x,y,z) = 2x^2 +3y^2+z^2 \Rightarrow \cases{f_x=4x\\ f_y=6y\\ f_z=2z} \Rightarrow (f_x,f_y,f_z)|_{(2,1,3)} =(8,6,6) \\ \Rightarrow 方向導數=(8,6,6)\cdot {(1,0,-2)\over |(1,0,-2)|} ={1\over \sqrt 5}\left((8,6,6)\cdot (1,0,-2) \right) =-{4\over \sqrt 5}，故選\bbox[red, 2pt]{(D)}$$

解答：$$f(x,y,z)=x^2+y^2+z^2 \Rightarrow \vec u=(f_x,f_y,f_z) =(2x,2y,2z) \\\Rightarrow {\vec u\over |\vec u|} ={(2x,2y,2z)\over \sqrt{4(x^2+y^2+ z^2)}} ={(2x,2y,2z)\over \sqrt{4 \cdot 9}} =({x\over 3}, {y\over 3},{z\over 3})，故選\bbox[red, 2pt]{(C)}$$

解答：$$(x+1){dy\over dx} =x(y^2+1) \Rightarrow \int {1\over y^2+1}dy = \int {x\over x+1}dx \Rightarrow \tan^{-1}y = x-\ln(x+1)+C\\ 將初始值y(0)=0 代入 \Rightarrow C=0 \Rightarrow y= \tan(x-\ln(x+1))，故選\bbox[red, 2pt]{(A)}$$

二、計算題(非選擇題，每題10分)

解答：$$令\cases{u=t^n \\ dv=e^t\,dt} \Rightarrow \cases{du =nt^{n-1}\,dt\\ v=e^t} \Rightarrow \int t^ne^t\,dt = t^ne^t-n\int t^{n-1}e^t\,dt\\ 因此\int t^3e^t\,dt = t^3e^t-3\int t^2e^t\,dt =t^3e^t-3(t^2e^t -2\int te^t\,dt) = t^3e^t -3t^2e^t+ 6\int te^t\,dt \\= t^3e^t -3t^2e^t+ 6(te^t-\int e^t\,dt) =\bbox[red, 2pt]{t^3e^t -3t^2e^t+ 6te^t-6e^t+C}$$

解答：$$I=\int e^{ax}\cos(bx)\,dx = {1\over b}e^{ax}\sin(bx) -{a\over b}\int e^{ax}\sin(bx)\,dx \left( \because\cases{u=e^{ax} \Rightarrow du=ae^{ax}\\ dv=\cos(bx)dx \Rightarrow v={1\over b}\sin(bx)}\right)\\ ={1\over b}e^{ax}\sin(bx) -{a\over b}\left(-{1\over b}e^{ax}\cos (bx)+ {a\over b}I\right)\left( \because\cases{u=e^{ax} \Rightarrow du=ae^{ax}\\ dv=\sin (bx)dx \Rightarrow v=-{1\over b}\cos(bx)}\right) \\ ={1\over b}e^{ax}\sin(bx) +{a\over b^2} e^{ax}\cos(bx)-{a^2\over b^2}I \Rightarrow (1+{a^2\over b^2}) I= {1\over b}e^{ax}\sin(bx) +{a\over b^2} e^{ax}\cos(bx) \\ \Rightarrow I= {b^2\over a^2+b^2}\left({1\over b}e^{ax}\sin(bx) +{a\over b^2} e^{ax}\cos(bx)\right) =\bbox[red, 2pt]{{b\over a^2+b^2} e^{ax}\sin (bx)+ {a\over a^2+b^2} e^{ax}\cos(bx)+C}$$

解答：$${dy\over dx}+3x^2y= 6x^2 \Rightarrow {dy\over dx}= 3x^2(2-y) \Rightarrow \int {1\over 2-y}dy = \int 3x^2 dx \Rightarrow -\ln |2-y|=x^3+C_1\\ \Rightarrow 2-y=C_2e^{-x^3} \Rightarrow y=2-C_2e^{-x^3}，將y(0)=1 代入左式\Rightarrow 1=2-C_2 \Rightarrow C_2=1\\ \Rightarrow \bbox[red, 2pt]{y=2-e^{-x^3}}$$

解答：$$\lim_{x\to 0}{\tan x-x\over x^3} =\lim_{x\to 0}{(\tan x-x)'\over (x^3)'} =\lim_{x\to 0}{\sec^2 x-1\over 3x^2} = \lim_{x\to 0}{\tan^2 x \over 3x^2} = \lim_{x\to 0}{(\tan^2 x)' \over (3x^2)'}\\ =\lim_{x\to 0}{2\tan  x\sec^2 x \over 6x} = \lim_{x\to 0}{(\tan  x\sec^2 x)' \over (3x)'} = \lim_{x\to 0}{ \sec^4x +2\tan^2 x\sec^2 \over 3} \\={1\over 3} \lim_{x\to 0} \left({1\over \cos^4 x} +{2\sin^2 x\over \cos^4 x}\right) =\bbox[red, 2pt]{1\over 3}$$

解答：$$\iint_D x^3dydz +x^2ydzdx + x^2zdxdy \Rightarrow \mathbf{F}=(x^3,x^2y,x^2z)，再利用散度定理\\ \Rightarrow \iint_D x^3dydz +x^2ydzdx + x^2zdxdy = \iiint\nabla \cdot \mathbf F\,dV= \iiint (3x^2+x^2+x^2)\,dV =\iiint 5x^2\,dV\\ 利用圓柱坐標: \cases{x=r\cos \theta\\ y= r\sin \theta \\ z=z},其中\cases{0\le r\le 2\\ 0\le \theta \le 2\pi\\ 0\le z\le 6}; \\因此\iiint 5x^2\,dV = \int_0^6 \int_0^{2\pi} \int_0^2 5r^3\cos^2\theta \,drd\theta dz = 20\int_0^6 \int_0^{2\pi}\cos^2\theta \,d\theta dz\\ =20\int_0^6 \int_0^{2\pi}{\cos 2\theta +1\over 2} \,d\theta dz = 20\pi \int_0^6 dz = \bbox[red,2pt]{120\pi}$$

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