107年高師大轉學考-微積分詳解

國立高雄師範大學 107 學年度學士班轉學生招生考試試題

系所別：電子工程學 系 二 年 級
科目：微 積 分

解答：$$\mathbf{(a)}\; \lim_{x\to 0}{\tan x\over x} =\lim_{x\to 0}{(\tan x)'\over (x)'} =\lim_{x\to 0}{\sec^2 x\over 1} = \bbox[red, 2pt]1 \\\mathbf{(b)}\; \lim_{x\to \infty}{-1\over x} \le \lim_{x\to \infty}{\sin x\over x} \le \lim_{x\to \infty}{1\over x} \Rightarrow 0\le \lim_{x\to \infty}{\sin x\over x}\le 0 \Rightarrow \lim_{x\to \infty}{\sin x\over x}= \bbox[red, 2pt]0$$

解答：$$y=(x^2+1)^3 \Rightarrow y'=3(x^2+1)^2 (2x) = \bbox[red, 2pt]{6x(x^2+1)^2}$$

解答：$$y=f(x)=\sqrt{2x-3} \Rightarrow x= {y^2+3\over 2} \Rightarrow f^{-1}(x)= \bbox[red,2pt]{{1\over 2}(x^2＋3）}$$

解答：$$\int_0^2 |2x-1|\,dx =\int_0^{1/2} 1-2x\,dx +\int_{1/2}^2 2x-1\,dx = \left.\left[ x-x^2\right]\right|_0^{1/2} + \left.\left[ x^2-x \right]\right|_{1/2}^2 \\={1\over 4}+2+{1\over 4}= \bbox[red,2pt]{5\over 2}$$

解答：$$\int{2\over x}\,dx =\bbox[red, 2pt]{2\ln|x|+C}$$

解答：$$\mathbf{(a)}\;取\cases{u=\cos(2x) \Rightarrow du -2\sin(2x)\,dx\\ dv=e^{-x}\,dx \Rightarrow v=-e^{-x}} \\ \qquad \Rightarrow \int e^{-2x}\cos (2x)\,dx = -e^{-x}\cos(2x)-2 \int e^{-x}\sin(2x)\,dx\\\qquad 同理，取\cases{u= \sin(2x) \Rightarrow du =2\cos(2x)\,dx \\ dv=e^{-x}\,dx \Rightarrow v=-e^{-x}} \\\qquad \Rightarrow \int e^{-x}\sin(2x)\,dx = -e^{-x}\sin(2x)+ 2\int e^{-x}\cos(2x)\,dx \\ \qquad 因此\int e^{-2x}\cos (2x)\,dx ＝ -e^{-x}\cos(2x)+ 2e^{-x}\sin(2x)-4 \int e^{x}\cos (2x)\,dx\\ \qquad \Rightarrow 5\int e^{-2x}\cos (2x)\,dx ＝ -e^{-x}\cos(2x)+ 2e^{-x}\sin(2x) \Rightarrow \int e^{-2x}\cos (2x)\,dx \\\qquad ＝\bbox[red,2pt]{{1\over 5}e^{-x}(2\sin(2x)-\cos(2x))+C} \\\mathbf{(b)}\;令\sin u={x-2\over 2} \Rightarrow \cos u\,du = {1\over 2}\,dx \Rightarrow \int {1\over \sqrt{4x-x^2}}\,dx =\int{1\over 2\sqrt{1-({x-2\over 2})^2}}\,dx\\\qquad = \int { 2\cos u\over 2 \sqrt{1-\sin^2 u}}\,du =\int {2\cos u\over 2\cos u}\,du = \int 1\,du = u +C= \bbox[red, 2pt]{\sin^{-1}{x-2\over 2}+C} \\ \mathbf{(C)}\; 取\cases{u=\sec x \Rightarrow du = \sec x\tan x\,dx\\ dv= \sec^2 x\,dx \Rightarrow v=\tan x} \Rightarrow I=\int \sec^3 x\,dx =\sec x\tan x-\int \sec x\tan^2 x\,dx \\\qquad = \sec x\tan x-\int \sec x(\sec^2x -1)\,dx = \sec x\tan x-\int \sec^3 x+\int \sec x\,dx\\ \qquad \Rightarrow 2I= \sec x\tan x-\int \sec x\,dx =\sec x\tan x-\ln|\tan x+\sec x | \\ \qquad \Rightarrow I= \bbox[red, 2pt]{{1\over 2}(\sec x\tan x+\ln |\sec x+\tan x|)+C} \\\mathbf{(d)}\; u=\cos(2x) \Rightarrow du = -2\sin(2x)\,dx \\\Rightarrow \int \sin^3(2x)\sqrt{\cos (2x)}\,dx = \int \sin(2x)(1-\cos^2(2x)) \sqrt{\cos(2x)}\,dx \\\qquad = \int (1-u^2)\sqrt u\cdot -{1\over 2}\,du =-{1\over 2} \int u^{1/2}-u^{5/2}\,du =-{1\over 2}({2\over 3}u^{3/2}-{2\over 7}u^{7/2})={1\over 7}u^{7/2}-{1\over 3}u^{3/2} \\\qquad =\bbox[red, 2pt]{{1\over 7} \cos^{7/2}(2x)-{1\over 3}\cos^{3/2}(2x)+C}$$

解答：$$y\to 0 \Rightarrow x\to \infty,題目可能有誤$$

解答：$$f(x)=\cos x \Rightarrow f'(x)=-\sin x\Rightarrow f''(x)=-\cos x \Rightarrow f'''(x)= \sin x \Rightarrow \cases{f(\pi/4) =\sqrt 2/2\\ f'(\pi/4) =-\sqrt 2/2\\ f''(\pi/4) = -\sqrt 2/2\\ f'''(\pi/4) =\sqrt 2/2\\ } \\ \Rightarrow \text{the Taylor series for f(x) centered at }\pi/4 =\bbox[red, 2pt]{{\sqrt 2\over 2}-{\sqrt 2\over 2}(x-{\pi \over 4}) -{\sqrt 2\over 4}(x-{\pi\over 4})^2 +\cdots}$$

解答：$$\int_1^2 \int_1^x (2x^2y^{-2}+ 2y)\,dydx = \int_1^2 \left. \left[ -2x^2y^{-1}+ y^2\right] \right|_1^x \,dx =\int_1^2 -2x+3x^2 -1\,dx \\= \left. \left[ -x^2+x^3-x \right]\right|_1^2 =2+1=\bbox[red, 2pt]3$$

解答：$${1\over 2}\int_{\pi/3}^{\pi/3} r^2\,d\theta ={1\over 2}\int_{\pi/3}^{\pi/3} 9\cos^2(3\theta)\,d\theta = {9\over 4} \int_{\pi/3}^{\pi/3} \cos(6\theta)+1\,d\theta ={9\over 4}\left.\left[ {1\over 6}\sin (6\theta)+\theta\right] \right|_{\pi/3}^{\pi/3} \\={9\over 4}\times {2\over 3}\pi =\bbox[red,2pt]{3\pi \over 2}$$

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