111年警大碩士班－微積分詳解

中央警察大學 111 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： $$\mathbf{(一)}\;\lim_{n\to \infty }{1\over n}\left[ ({1\over n})^9+ ({2\over n})^9+ ({3\over n})^9+ \cdots +({n \over n})^9 \right] =\lim_{n\to \infty }\sum_{k=1}^n {1\over n}({k\over n})^9 =\int_0^1 x^9\,dx \\ \qquad =\left.{1\over 10}x^{10} \right|_0^1 = \bbox[red, 2pt]{1\over 10} \\\mathbf{(二)}\; u=\cos x \Rightarrow du = -\sin x \,dx \Rightarrow \int \sin x\sin(\cos x)\,dx = \int -\sin u\,du = \cos u+C \\\qquad = \bbox[red, 2pt]{\cos(\cos x)+C} \\\mathbf{(三)}\;\int_0^1 {e^z+1\over e^z+z}\,dz = \left.\left[  \ln(e^z+z)\right] \right|_0^1 = \bbox[red,2pt]{\ln(e+1) }$$

解答：

(一）

$$\cos x=\sin(2x) =2\sin x\cos x \Rightarrow \cos x(1-2\sin x)=0 \Rightarrow \cases{\cos x=0 \\ \sin x=1/2} \Rightarrow \cases{x=\pi/2\\ x=\pi/6} \\ 所圍面積＝\int_0^{\pi/6} \cos x-\sin(2x)\,dx +\int_{\pi/6}^{\pi/2} \sin(2x)-\cos x\,dx \\ =\left.\left[ \sin x+{1\over 2} \cos（2x) \right] \right|_0^{\pi/6} +\left.\left[ -{1\over 2}\cos (2x)-\sin x\right] \right|_{\pi/6}^{\pi/2} =({3\over 4}-{1\over 2})+ (-{1\over 2}+{3\over 4})= \bbox[red,2pt]{1\over 2} \\\mathbf{(二)}\; \cases{x=5 \Rightarrow y=\sqrt{x-1}=2 \\y=0 \Rightarrow 0=\sqrt{x-1} \Rightarrow x=1}\Rightarrow \int_1^5 y^2\pi \,dx =\pi \int_1^5 x-1\,dx =\pi \left. \left[ {1\over 2}x^2-x \right]\right|_1^5 =\bbox[red,2pt]{8\pi}$$

解答 ： $$\mathbf{(一)}\;\int_2^4 {4\over x^2-6x+10}\,dx =\int_2^4 {4\over (x-3)^2+1}\,dx = 4\left.\left[ \tan^{-1}(x-3)\right] \right|_2^4 =4(\tan^{-1} 1-\tan^{-1}(-1)) \\ \qquad =4({\pi\over 4}-(-{\pi\over 4}))= \bbox[red, 2pt]{2\pi} \\\mathbf{(二)}\;\cases{u=\sin^{-1}x\\ dv={1\over \sqrt{1-x^2}}dx} \Rightarrow \cases{du ={dx\over \sqrt{1-x^2}} \\ v=\sin^{-1}x} \Rightarrow I=\int_0^{1/\sqrt 2} {\sin^{-1}x\over \sqrt{1-x^2}}\,dx =\left. (\sin^{-1} x)^2\right|_0^{1/\sqrt 2} -I \\\qquad \Rightarrow 2I=\left. (\sin^{-1} x)^2\right|_0^{1/\sqrt 2} ={\pi^2 \over 16} \Rightarrow I=\bbox[red, 2pt]{\pi^2\over 32} \\ \mathbf{(三)}\;\int_{-2}^2 {4\over \sqrt{16-x^2}}\,dx = \int_{-2}^2 {1\over \sqrt{1-(x/4)^2}}\,dx = \left.\left[4 \sin^{-1}(x/4)\right]\right|_{-2}^2 =4\cdot {\pi\over 3} =\bbox[red,2pt]{4\pi \over 3}$$

 解答： $$r=a(1-\cos \theta) \Rightarrow {dr\over d\theta} = a\sin \theta \Rightarrow 心臟線長度＝\int_0^{2\pi} \sqrt{r^2 +({dr\over d\theta})^2}\,d\theta \\=\int_0^{2\pi} \sqrt{a^2(1-\cos\theta)^2+a^2\sin^2\theta} \,d\theta =\int_0^{2\pi} \sqrt{2a^2-2a^2\cos \theta} \,d\theta =\sqrt 2a \int_0^{2\pi }\sqrt{1-\cos \theta}\,d\theta \\=\sqrt 2a \int_0^{2\pi }\sqrt{2\sin^2 (\theta/2)}\,d\theta =   2a \int_0^{2\pi } \sin  (\theta/2)\,d\theta = 2a \left. \left[ -2\cos(\theta/2)\right]\right|_0^{2\pi} =\bbox[red, 2pt]{8a}\\ 心臟線面積={1\over 2}\int_0^{2\pi} r^2\,d\theta = \int_0^\pi r^2\,d\theta =a^2\int_0^\pi (1-\cos \theta)^2\,d\theta =a^2\cdot {3\over 2}\pi = {3\over 2}a^2\pi \\ \Rightarrow \iint x\,dxdy= \int_0^{2\pi} \int_0^{a(1-\cos\theta)} r\cos\theta\cdot r\,drd\theta =\int_0^{2\pi} {1\over 3}a^3(1-\cos\theta)^3 \cos\theta \,d\theta =-{5\over 4}a^3\pi \\ \Rightarrow 形心x坐標=-{5\over 4}a^3\pi \div ({3\over 2}a^2\pi)=-{5\over 6}a,由於心臟線圖形對稱x軸，因此形心y坐標＝0\\ \Rightarrow 形心＝\bbox[red,2pt]{(-{5\over 6}a,0)}$$

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解題僅供參考，其他歷年試題及詳解