114年警察大學碩士班-微積分詳解

中央警察大學114學年度碩士班入學考試試題

所別:消防科學研究所、交通管理研究所
科目:微積分(同等學力加考)

解答:$$\textbf{(一) }F(\mu) =\int_{-\infty}^\infty f(t)e^{-j2\pi \mu t}\,dt = \int_{-1/2}^{1/2} e^{-j2\pi \mu t}\,dt =\int_{-1/2}^{1/2} (\cos(2\pi \mu t)-j\sin(2\pi \mu t))\,dt \\ \qquad = \left. \left[ {1 \over 2\pi \mu} \sin(2\pi \mu t)+{j\over 2\pi \mu } \cos(2\pi \mu t)\right]  \right|_{-1/2}^{1/2} = {2\over 2\pi \mu} \sin(\pi \mu) =\bbox[red, 2pt]{\sin (\pi \mu) \over \pi\mu} \\\textbf{(二) } \cases{u=x^2\\ dv= \sin x\,dx} \Rightarrow \cases{du=2x \,dx\\ v=-\cos x} \Rightarrow I=\int x^2 \sin x\,dx = -x^2\cos x +2\int x\cos x\,dx \\\qquad \cases{u=x\\ dv= \cos x\,dx } \Rightarrow \cases{du=dx\\ v=\sin x} \Rightarrow I=-x^2\cos x+2\left(x\sin x-\int \sin x\,dx \right) \\\qquad =\bbox[red,2pt]{-x^2\cos x+2\left(x\sin x+\cos x  \right)+C}$$

解答:

$$\textbf{(一) }\int_0^\pi x\sin x\,dx = \left. \left[ \sin x-x\cos x\right] \right|_0^\pi =\bbox[red, 2pt]{\pi} \\\textbf{(二) }\int_0^\pi y^2\pi \,dx= \pi\int_0^\pi x^2 \sin^2 x \,dx= {\pi \over 2}\int_0^\pi x^2 (1- \cos 2x) \,dx\\ \qquad={\pi\over 2} \left. \left[ {x^3\over 3}-{x^2 \sin(2x)\over 2} -{x\cos(2x)\over 2} +{\sin(2x) \over 4}\right] \right|_0^{\pi} = \bbox[red, 2pt]{{\pi^4\over 6}-{\pi^2 \over 4}} \\ \textbf{(三) }A質心x坐標={\int_0^\pi x^2\sin x\,dx \over \int_0^\pi x\sin x\,dx } ={\pi^2-4\over \pi} \Rightarrow A質心繞y軸旋轉周長=2\pi\cdot {\pi^2-4\over \pi} =2(\pi^2-4) \\\qquad \Rightarrow A繞y軸旋轉體積=2(\pi^2-4)\cdot \pi = \bbox[red, 2pt]{2\pi(\pi^2-4)}$$

解答:$$1kg的火箭至少需要1\times 9.8=9.8牛頓，只有1牛頓的推力,無法發射成功!! \bbox[cyan,2pt]{題目有疑義}$$

解答:$$\textbf{(一) } e^2為常數\Rightarrow {d\over dx}e^2 = \bbox[red, 2pt]0\\ \textbf{(二) } {d\over dx}2^x = {d\over dx}e^{x\ln 2} = \ln 2 e^{x \ln x} = \bbox[red, 2pt]{2^x\ln 2} \\\textbf{(三) }{d\over dx} (\sin x)^x ={d\over dx} e^{x\ln (\sin x)}=(\ln (\sin x)+ x\cot x) e^{x\ln(\sin x)} = \bbox[red, 2pt]{(\ln (\sin x)+ x\cot x)(\sin x)^x}$$

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解題僅供參考，其他碩士班考試詳解