112年政大轉學考-微積分(二)詳解

國立政治大學112學年度轉學生招生考試

科目:微積分(二)

系所別:應用數學系二年級

解答: $$\textbf{(a) }  \lim_{x\to 0}{\int_2^{3x+2} \ln(t-1)\,dt\over x^2} =  \lim_{ x\to 0}{{d\over dx}\int_2^{3x+2} \ln(t-1)\,dt\over {d\over dx}x^2} = \lim_{ x\to 0}{ 3\ln(3x+1)\over 2x}\\\quad  =\lim_{x\to 0}{ {d\over dx}(3\ln(3x+1)) \over {d\over dx}(2x)} =\lim_{x\to 0} { {9\over 3x+1}\over 2} =\bbox[red, 2pt]{9\over 2}\\ \textbf{(b) } u=\cos x \Rightarrow du=-\sin x \,dx \Rightarrow \int{\sin x\over 1+\cos^2 x}\,dx =\int {-1\over 1+u^2}\,du =- \tan^{-1} u+C\\\quad =\bbox[red, 2pt]{-\tan^{-1} \cos x+C} \\\textbf{(c) }   \cases{u= \arctan(x)\\ dv=4dx} \Rightarrow \cases{du={dx\over 1+x^2} \\ v=4x} \Rightarrow \int 4 \arctan(x)\,dx =4x \arctan(x)- \int{4x\over 1+x^2}\,dx \\\quad =4x \arctan(x)-2\ln(1+x^2) +C \Rightarrow \int_0^1 4 \arctan(x)\,dx = \left. \left[4x \arctan(x)-2\ln(1+x^2) \right] \right|_0^1 \\\quad =\bbox[red, 2pt]{\pi-2\ln 2} \\\textbf{(d) }   \cases{ u= \csc(x)\\ dv=\csc^2 (x)dx} \Rightarrow \cases{du=-\csc(x)\cot(x)\,dx \\ v=-\cot x}\\\quad \Rightarrow I= \int{4\over \sin^3(x)}\,dx= 4\int \csc^3(x)\,dx = -4\csc(x) \cot(x)-4\int \csc(x)\cot^2(x)\,dx \\\quad =-4\csc(x) \cot(x)-4 \int \csc(x)(\csc^2(x)-1)\,dx =-4\csc(x) \cot(x)-I+4\int \csc(x)\,dx \\\quad \Rightarrow 2I=-4\csc(x) \cot(x)+4\int {1\over \sin x}\,dx I=-4\csc(x) \cot(x)+2 \int{1\over \sin(x/2)\cos(x/2)} \,dx \\ \Rightarrow I=-2\csc(x) \cot(x)+\int {\tan^2(x/2) \over \tan^2(x/2) \sin(x/2)\cos(x/2)}\,dx \\\quad =-2\csc(x) \cot(x)+\int {\sec^2(x/2) \over  \tan(x/2) }\,dx =\bbox[red, 2pt]{-2\csc(x) \cot(x)+ 2\ln\left(\left| \tan{x\over 2} \right|\right)+C}$$

解答: $$\textbf{(a) }{n\over \sqrt{3+n^p}} \lt {n\over \sqrt{n^p}} ={1\over n^{(p/2-1)}}\\ \text{By p-series test, if }{p\over 2}-1\gt 1 {(\bbox[red,2pt]{p\gt 4})}, \text{then } \sum_{n=1}^\infty {1\over n^{p/2-1}} \text{ will converge.} \\ \text{And by comparison-test, the series }\sum_{n=1}^\infty {n\over \sqrt{3+n^p}} \text{ will converge}\\ \textbf{(b) }   \lim_{n\to \infty}   \sum_{i=1}^n {3\pi\over 4n} \sec^2 \left({i \pi\over 4n} \right) =\int_0^1 {3\pi\over 4}\sec^2{\pi x\over 4}\,dx =3\left. \left[ \tan{\pi x\over 4} \right] \right|_0^1 =\bbox[red, 2pt]3$$

解答: $$\frac{d y}{dx} ={\sqrt{1-y^2} \over 1+x^2} \Rightarrow \int {1\over \sqrt{1-y^2}}dy = \int {1\over 1+x^2}\,dx  \Rightarrow \arcsin y= \arctan x +C \\ \Rightarrow y= \sin \left( \arctan x +C \right) \Rightarrow y(0)= \sin C=0 \Rightarrow C=0 \Rightarrow y= \sin(\arctan x) \\ \Rightarrow \bbox[red, 2pt]{y={x\over \sqrt{1+x^2}}}$$

解答: $$\cases{x=\rho \sin \phi \cos \theta\\ y=\rho \sin \phi \sin \theta\\ z= \rho \cos \phi} \Rightarrow   z=\rho \cos \phi=\sqrt{x^2+y^2}=\rho \sin \phi \Rightarrow \phi={\pi\over 4} \\ \Rightarrow V= \int_0^{\pi/2} \int_{\pi /4}^{\pi/2} \int_0^\sqrt{12} \rho^2 \sin \phi \,d\rho \,d\phi \,d\theta =8\sqrt 3\int_0^{\pi/2} \int_{\pi/4}^{\pi/2} \sin \phi \,d\phi\,d\theta =4\sqrt 6 \int_0^{\pi /2} 1\,d\theta = \bbox[red, 2pt]{2\sqrt 6\pi}$$

解答: $$r(t)=e^t \mathbf i+(1+t) \mathbf j \Rightarrow r'(t)=e^t \mathbf i+  \mathbf j \\ \Rightarrow \int_C \mathbf F\cdot d\mathbf r = \int_0^1 ((e^t(1+t)^2 +2(1+t)) \mathbf i +(e^{2t}(1+t)+2e^t+2)\mathbf j) \cdot (e^t \mathbf i+\mathbf j)\,dt \\ =\int_0^1 (e^{2t}(1+t)(2+t)+2e^t(t+2)+2)\,dt =\left. \left[ {1\over 2} e^{2t} (t+1)^2+ 2e^t(t+1)+2t\right] \right|_0^1 = \bbox[red, 2pt]{2e^2+4e-{1\over 2}}$$

解答: $$r(t)=(2\cos t)\mathbf i+(2\sin t)\mathbf j \Rightarrow C:x^2+y^2=4 \\ \Rightarrow \int_C -y^3dx+x^3dy = \iint_D \left(\frac{\partial }{\partial x}x^3-\frac{\partial }{\partial y}(-y^3) \right) \,dA =\iint_D 3(x^2+y^2) dA\\ =3 \int_0^{2\pi} \int_0^2 r^2\cdot r\,drd\theta =3 \int_0^{2\pi}4\,d\theta = \bbox[red, 2pt]{24\pi}$$

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解題僅供參考，轉學考歷年試題及詳解