115年政大科技管理碩士班-微積分詳解

 國立政治大學115學年度碩士班暨碩士在職專班招生考試試題

考試科目：微積分
系所別：科技管理與智慧財產研究所 科技管理組

以下各題，每大題10分，共100分。請寫明計算程序，無過程0分計算。

解答:$$\textbf{(a) }\lim_{x\to 2}{\int_2^x {1\over \sqrt{t^2+2t+8}}dt\over x-2} =\lim_{x\to 2}{{d\over dx}\int_2^x {1\over \sqrt{t^2+2t+8}}dt\over {d\over dx}(x-2)} =\lim_{x\to 2}{  {1\over \sqrt{x^2+2x+8}} \over 1} = {1\over \sqrt{16}} = \bbox[red,2pt]{1\over 4} \\\textbf{(b) } \lim_{n\to \infty} \sum_{k=0}^{n-1} {n\over (n+2k)^2} = \lim_{n\to \infty} \sum_{k=0}^{n-1} {1\over n(1+2(k/n))^2} = \int_0^1 {1\over (1+2x)^2} \,dx = \left. \left[ -{1\over 2(1+2x)} \right] \right|_0^1 = \bbox[red, 2pt]{1\over 3}$$

解答:$$\cases{u= \sin x \\ dv=e^x \,dx} \Rightarrow \cases{du=\cos x\,dx\\ v=e^x} \Rightarrow I=e^x \sin x-\underbrace{\int e^x \cos x\,dx}_{I_1} \\ \cases{u=\cos x\\ dv=e^x \,dx } \Rightarrow \cases{du=-\sin x\,dx\\ v=e^x} \Rightarrow I_1=e^x \cos x+\int e^x\sin x \Rightarrow I=e^x \sin x-e^x\cos x-I \\ \Rightarrow 2I=e^x(\sin x-\cos x) \Rightarrow I= \bbox[red, 2pt]{{1\over 2}e^x(\sin x-\cos x) +C}$$

解答:$$a_n={(x-3)^n\over n\cdot 4^n} \Rightarrow \lim_{n\to \infty} \left|{a_{n+1} \over a_n} \right| =\lim_{n\to \infty} \left|{(x-3)^{n+1}\over (n+1)4^{n+1} } \cdot {n4^n \over (x-3)^n }\right| = \lim_{n\to \infty} \left|{ (x-3)n \over 4(n+1)} \right| \\=\lim_{n\to \infty} \left|{ x-3 \over 4} \cdot {n\over n+1} \right| =\left|{ x-3 \over 4}  \right| \lt 1 \Rightarrow |x-3|\lt 4 \Rightarrow -1\lt x\lt 7 \\ x=-1 \Rightarrow a_n={1\over n}(-1)^n \Rightarrow \sum a_n 收斂 (交錯級數審斂法) \\ x=7 \Rightarrow a_n={1\over n} \Rightarrow \sum a_n 為調和級數\Rightarrow 發散 \\ \Rightarrow 收斂區間為\bbox[red, 2pt]{[-1,7)}$$

解答:$$\int {x+3\over \sqrt{9-x^2}}\,dx = \underbrace{\int {x\over \sqrt{9-x^2}}\,dx}_{I_1} +\underbrace{\int {3\over \sqrt{9-x^2}}\,dx}_{I_2} \\ u=9-x^2 \Rightarrow du=-2x\,dx \Rightarrow xdx=-{1\over 2}du \Rightarrow I_1= -{1\over 2} \int {1\over \sqrt{u}}\,du= -{1\over 2}\cdot 2\sqrt u+c_1=-\sqrt u+c_1 \\ \qquad \Rightarrow I_1=-\sqrt{9-x^2}+c_1 \\ x=3\sin \theta \Rightarrow dx=3\cos \theta \,d\theta \Rightarrow I_2= \int {9\cos \theta \over \sqrt{9-9\sin^2 \theta}}\,d\theta =\int{9\cos \theta\over 3\cos \theta}\,d\theta =3\theta+c_2 \\ \qquad \Rightarrow I_2=3\sin^{-1}{x\over 3}+c_2 \\ \Rightarrow I_1+I_2= \bbox[red, 2pt]{-\sqrt{9-x^2}+3\sin^{-1}{x\over 3}+c_3}$$

解答:$$\cases{x(t)=\cos^3 t\\ y(t) =\sin^3 t} \Rightarrow \cases{x'(t)=-3\cos^2 t \sin t \\ y'(t)=3\sin^2 t\cos t} \Rightarrow L= \int_0^\pi \sqrt{(-3\cos^2 t \sin t )^2+ (3\sin^2 t\cos t)^2}\,dt \\ = \int_0^\pi \sqrt{9\cos^2 t\sin^2 t(\cos^2t+\sin^2 t)}\,dt = \int_0^\pi \sqrt{9\cos^2 t\sin^2t}\,dt= \int_0^\pi \sqrt{{9\over 4}\sin^2(2t)}\,dt \\=\int_0^\pi {3\over 2}|\sin(2t)|\,dt = 2\int_0^{\pi/2} {3\over 2}\sin(2t)\,dt = \left. \left[ -{3\over 2}\cos(2t) \right] \right|_0^{\pi/2}=\bbox[red, 2pt]3$$

解答:$$\textbf{(a) } \int f(x)\,dx=1 \Rightarrow \int_0^\infty kx^2e^{-x}\,dx =1 \Rightarrow k \left. \left[ -e^{-x}(x^2+2x+2) \right] \right|_0^\infty =2k=1 \Rightarrow k= \bbox[red, 2pt]{1\over 2} \\\textbf{(b) }F(x)=\int_0^x f(t)\,dt =\int_0^x{1\over 2}t^2e^{-t}\,dt= {1\over 2} \left. \left[ -e^{-t}(t^2+2t+2) \right] \right|_0^x =1-{1\over 2}e^{-x}(x^2+2x+2) \\\qquad \Rightarrow \bbox[red, 2pt]{F(x) = \begin{cases}1-{1\over 2}e^{-x}(x^2+2x+2),& x\ge 0\\ 0,& x\lt 0 \end{cases}} \\\textbf{(c) }E(X) =\int xf(x)\,dx = \int_0^x x \left( {1\over 2}x^2e^{-x} \right)\,dx ={1\over 2} \int_0^\infty x^3e^{-x}\,dx\\ \quad 利用\Gamma(n) = \int_0^\infty x^{n-1}e^{-x}\,dx=(n-1)!  \Rightarrow E(X)={1\over 2}\cdot 3!= \bbox[red, 2pt]3$$

解答:$$\cases{x=\rho \sin \phi \cos \theta\\ y =\rho \sin \phi \sin \theta\\ z=\rho \cos \phi} \Rightarrow \iiint_E {z^2\over \sqrt{x^2+y^2+z^2}}\,dV= \int_0^{2\pi} \int_0^{\pi/2} \int_2^3 {\rho^2 \cos^2\theta \over \sqrt{ \rho^2}} \rho^2 \sin \phi\,d\rho d\phi d\theta \\ = \int_0^{2\pi} \int_0^{\pi/2} \int_2^3  \rho^3 \cos^2 \phi \sin \phi\,d\rho d\phi d\theta = \left( \int_0^{2\pi}1\,d\theta \right) \cdot \left( \int_0^{\pi/2} \cos^2 \phi \sin \phi\,d \phi \right) \cdot \left( \int_2^3 \rho^3\,d\rho \right) \\=2\pi\cdot {1\over 3}\cdot {65\over 4}= \bbox[red, 2pt]{65\pi\over 6}$$

解答:$$6{dy \over dx}+y^4e^{2x}=0 \Rightarrow 6{dy \over dx}=-y^4e^{2x} \Rightarrow \int {6\over y^4}\,dy = \int -e^{2x}\,dx \Rightarrow -{2\over y^3} =-{1\over 2}e^{2x}+ c_1\\ \Rightarrow {2\over y^3}={1\over 2}e^{2x}+c_2 \Rightarrow y^3={2\over {1\over 2}e^{2x}+c_2} ={4\over e^{2x}+c_3} \Rightarrow \bbox[red, 2pt]{y= \sqrt[3]{4\over e^{2x}+c_3}}$$

解答:$$u=x^2 \Rightarrow du=2xdx \Rightarrow xdx={1\over 2}du \Rightarrow I= \int_0^\infty x^3e^{-x^2} \,dx = \int_0^\infty ue^{-u } {1\over 2}\,du \\ = {1\over 2}\left. \left[ -ue^{-u}-e^{-u} \right] \right|_0^\infty = \bbox[red, 2pt]{1\over 2}$$

解答:$$200x+400y=48000 \Rightarrow x+2y=240 \Rightarrow \cases{f(x,y)=240x^{3/4} y^{1/4} \\g(x,y)=x+2y-240} \Rightarrow \cases{f_x=\lambda g_x\\ f_y= \lambda g_y\\ g=0} \\ \Rightarrow \cases{180x^{-1/4}y^{1/4} =\lambda \\ 60x^{3/4} y^{-3/4}=2\lambda\\ x+2y=240} \Rightarrow {3y\over x} ={1\over 2} \Rightarrow x=6y \Rightarrow 6y+2y=240 \Rightarrow y=30 \Rightarrow x=180 \\ \Rightarrow f(180,30)= 240\cdot 180^{3/4}\cdot 30^{1/4} =240\cdot 6^{3/4} \cdot 30^{3/4}\cdot 30^{1/4} =240\cdot 6^{3/4}\cdot 30= \bbox[red, 2pt]{7200\cdot 6^{3/4}}\\ 這是商學院的考題,可以改用\text{Cobb-Douglas production function } \\對於f(x,y)=Ax^\alpha y^\beta, 在預算P_x x+P_y y=M下，最大產能分配比例為\\ \cases{x={\alpha\over \alpha+\beta}\cdot {M\over P_x } ={3\over 4}\cdot {48000\over 200}=180\\ y={\beta\over \alpha+\beta}\cdot {M\over P_y} ={1\over 4}\cdot {48000\over 400}=30} \Rightarrow 結果相同$$

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解題僅供參考，碩士班歷年試題及詳解