110年警大碩士班－微積分詳解

中央警察大學 110 學年度碩士班入學考試試題

所 別：消防科學研究所
科 目：微積分(同等學力加考)

解答：$$\mathbf{(一)}\;u=x-1 \Rightarrow \int x\sqrt{x-1}\,dx =\int (u+1)\sqrt{u}\,du = \int u^{3/2}+ u^{1/2}\,du = {2\over 5}u^{5/2}+{2\over 3}u^{3/2}+C \\\quad = {2\over 5}(x-1)^{5/2}+{2\over 3}(x-1)^{3/2}+C= {2\over 3}(x-1)^{3/2}({3\over 5}(x-1)+1)+C \\\quad = \bbox[red, 2pt]{{2\over 3}(x-1)^{3/2}({3\over 5}x+{2\over 5})+C} \\\mathbf{(二)}\; \cases{u= x\\ dv =\sqrt{x-1}dx} \Rightarrow \cases{du=dx \\ v={2\over 3}(x-1)^{3/2}} \Rightarrow \int x\sqrt{x-1}\,dx = {2\over 3}x(x-1)^{3/2}-{2\over 3}\int (x-1)^{3/2}\,dx \\\quad = {2\over 3}x(x-1)^{3/2}-{2\over 3}\cdot {2\over 5}(x-1)^{5/2} +C ={2\over 3}(x-1)^{3/2}(x-{2\over 5}(x-1))+C\\\quad =\bbox[red,2pt]{{2\over 3}(x-1)^{3/2}({3\over 5}x+{2\over 5})+C}\\ \mathbf{(三)}\;兩\bbox[red,2pt]{結果相同}$$

解答：$$\mathbf{(一)}\;f(x)={1\over x} \Rightarrow f(-x)=-{1\over x}=-f(x) \Rightarrow f(x)為奇函數\Rightarrow \int_{-1}^1 {1\over x}\,dx =\bbox[red, 2pt]0\\ \mathbf{(二)}\; \int 2^x\,dx = \int e^{x\ln 2}\,dx = {1\over \ln 2} e^{x\ln 2}+C = \bbox[red, 2pt]{{1\over \ln 2}2^x+C} \\\mathbf{(三)}\; \int 0\,dx =\bbox[red,2pt]{C為一常數}$$

解答：$$\mathbf{(一)}\;兩圖形\cases{x^2-y=0\\ x^2+y=8} 的交點\cases{A(-2,4)\\ B(2,4)} \Rightarrow 所圍面積＝\int_{-2}^2 (8-x^2)-x^2\,dx =\int_{-2}^2 8-2x^2 \,dx \\\qquad  =\left. \left[ 8x-{2\over 3}x^3 \right] \right|_{-2}^2 =32-{32\over 3} =\bbox[red,2pt]{ 64\over 3} \\ \mathbf{(二)}\; \int_0^{2\pi} \int_0^{2(1+\cos\theta)} r\,drd\theta = \int_0^{2\pi}  {1\over 2}({2(1+\cos\theta)})^2 \,d\theta= \int_0^\pi 4(1+\cos\theta)^2 \,d\theta \\ \qquad = \int_0^\pi 4+ 8\cos\theta +4\cos^2\theta\,d\theta =\int_0^\pi 6+ 8\cos\theta +2\cos 2\theta\,d\theta = \left. \left[6\theta +8\sin \theta +\sin 2\theta \right] \right|_0^\pi \\\qquad = \bbox[red,2pt]{6\pi}$$

解答：$$\mathbf{(一)}\;x=\sqrt{2t^2+1} \Rightarrow y=x^2-4x= 2t^2+1-4\sqrt{2t^2+1} \Rightarrow {dy\over dt}= 4t-{8t \over \sqrt{2t^2+1}}\\ \qquad \Rightarrow \left. {dy\over dt}\right|_{t=2} =8-{16\over 3} =\bbox[red, 2pt]{8\over 3}\\\mathbf{(二)}\; y=6 x^{2/3}-4x^{-1/2 } \Rightarrow y'=4x^{-1/3}+2x^{-3/2} \Rightarrow y'(1)=6 \\ \qquad \Rightarrow 切線：y=6(x-1)+2 \Rightarrow \bbox[red,2pt]{y=6x-4}$$

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解題僅供參考，其他歷年試題及詳解