113年高師大轉學考-微積分詳解

 國立高雄師範大學113學年度暑假轉學生招生考試

系所別：電子工程學系 二年級
科 目：微積分（全一頁）

解答:$$\textbf{(a) }\lim_{x\to 0}{1-\cos x\over x} =\lim_{x\to 0}{(1-\cos x)'\over (x)'} = \lim_{x\to 0}{\sin x\over 1} = \bbox[red, 2pt]0 \\\textbf{(b) }\lim_{x\to \infty}{2x^2-4x\over x+1} = \lim_{x\to \infty}\left( 2x-6+{6 \over x+1} \right) = \bbox[red, 2pt]\infty$$

解答:$$f(x) ={9(x^2-3) \over x^3} \Rightarrow f'(x) ={18x \over x^3}-{27(x^2-3) \over x^4} = \bbox[red, 2pt]{-9x^2+81\over x^4}$$

解答:$$3(x^2+y^2)^2=100xy \Rightarrow 6(x^2+y^2)(2x+2yy') =100(y+xy')\\ \Rightarrow 6(3^2+1^2)(2\cdot 3+ 2\cdot 1\cdot y') =100(1+3y') \Rightarrow 60(6+2y')=100(1+3y') \Rightarrow y'=\bbox[red, 2pt]{13\over 9}$$

解答:$$x^2(x^2+y^2) =y^2 \Rightarrow 2x(x^2+y^2 )+x^2(2x+2yy')=2yy' \Rightarrow \sqrt 2\cdot 1+{1\over 2}(\sqrt 2 +\sqrt 2 y')=\sqrt 2y' \\ \Rightarrow {3\sqrt 2\over 2}={\sqrt 2\over 2}y' \Rightarrow y'=3 \Rightarrow \text{ tangent line: }y=3(x-{\sqrt 2\over 2})+{\sqrt 2\over 2} \Rightarrow \bbox[red, 2pt]{3x-y=\sqrt 2}$$

解答:$$\textbf{(a) }u=x^2+4 \Rightarrow du=2xdx \Rightarrow \int {x^3\over 4\sqrt{4+x^2}}\,dx = \int  {(u-4)\over 4\sqrt{u}}\cdot {1\over 2}\,du = \int \left( {1\over 8}\sqrt u-{1\over 2\sqrt u}\right)\,du \\ \quad ={1\over 12}u^{3/2}-\sqrt u+c_1 = \bbox[red, 2pt]{{1\over 12}(x^2+4)^{3/2}-\sqrt{x^2+4}+c_1} \\\textbf{(b) }\int \cos 5x \cos 3x\,dx ={1\over 2} \int \left( \cos 8x+ \cos 2x\right)\,dx ={1\over 2} \left( {1\over 8}\sin 8x +{1\over 2} \sin 2x\right)+c_1 \\\quad = \bbox[red, 2pt]{{1\over 16} \sin 8x+{1\over 4}\sin 2x +c_1} \\\textbf{(c) } \cases{u=e^x\\ dv=\sin x\,dx} \Rightarrow \cases{du=e^x\,dx\\ v=-\cos x} \Rightarrow \int e^x\sin x\,dx = -e^x\cos x+\int e^x \cos x\,dx \\\quad =-e^x \cos x+ e^x\sin x-\int e^x \sin x\,dx \Rightarrow 2\int e^x \sin x\,dx =e^x(\sin x-\cos x) +c_1\\\quad \Rightarrow \int e^x \sin x\,dx= \bbox[red, 2pt]{{1\over 2}e^x (\sin x-\cos x)+c_2} \\\textbf{(d)} \cases{u=\ln x\\ dv=x\,dx} \Rightarrow \cases{du=dx/x\\ v=x^2/2} \Rightarrow \int  x\ln x\,dx = {1\over 2}x^2\ln x-\int{x\over 2}\,dx ={1\over 2}x^2 \ln x-{1\over 4}x^2+c_1 \\ \quad \Rightarrow \int_0^1 x\ln x\,dx = \left. \left[ {1\over 2}x^2 \ln x-{1\over 4}x^2 \right] \right|_0^1 = \bbox[red, 2pt]{-{1\over 4}} \\\textbf{(e) }  \int_0^{\pi/2} \int_0^{\cos x} (1+\sin x)\,dy dx = \int_0^{\pi/2} \cos x (1+\sin x)\, dx = \int_0^{\pi/2}  (\cos x+{1\over 2}\sin 2x)\, dx \\\quad =\left. \left[ \sin x-{1\over 4}\cos 2x \right] \right|_0^{\pi/2} ={5\over 4}-(-{1\over 4}) =\bbox[red, 2pt]{3\over 2}$$

解答:$$\textbf{(a) }f(x)=x^2 \Rightarrow f'(x)=2x \Rightarrow \text{ arc length: }\int_0^\sqrt 2\sqrt{1+(f'(x))^2}\,dx =\int_0^\sqrt 2\sqrt{1+4x^2}\,dx\\\quad \text{Let }2x=\tan \theta, \text{then }2dx=\sec^2 \theta\,d\theta \Rightarrow I=\int \sqrt{1+4x^2}\,dx =\int  \sqrt{1+\tan^2 \theta}\cdot {1\over 2}\sec^2 \theta\,d\theta \\={1\over 2}\int  \sec^3 \theta\,d\theta, \text{now let}  \cases{u=\sec \theta\\ dv=\sec^2\theta d\theta} \text{ then } \cases{du= \sec \theta \tan \theta\,d\theta\\ v=\tan \theta} \\\Rightarrow I={1\over 2}\left(\sec \theta \tan\theta-\int\sec \theta \tan^2 \theta\,d\theta \right) ={1\over 2}\sec \theta\tan \theta-{1\over 2}\int \sec \theta( \sec^2 \theta-1)\,d\theta\\={1\over 2}\sec \theta\tan \theta- I+{1\over 2}\int \sec \theta\,d\theta \\\quad \Rightarrow 2I={1\over 2}\sec \theta\tan \theta+{1\over 2}\ln|\tan \theta+\sec \theta| \Rightarrow I={1\over 4}\left( \sec \theta\tan \theta+ \ln|\tan \theta+\sec \theta|\right) \\\quad \text{ arc length= }{1\over 4}\left. \left[ \sec \theta\tan \theta+ \ln|\tan \theta+\sec \theta| \right] \right|_0^{\tan^{-1}2\sqrt 2} = \bbox[red, 2pt]{{3\over 2}\sqrt 2+{1\over 4}\ln(2\sqrt 2+3)} \\\textbf{(b) }表面積 I=\int_0^{\sqrt 2} 2\pi x\cdot \sqrt{1+4x^2} \,dx \\\quad u=1+4x^2 \Rightarrow du=8xdx \Rightarrow I=\int_1^{9} {1\over 4}\pi \sqrt{u}\,du =\left. \left[{\pi\over 6}u^{3/2} \right] \right|_0^9 = \bbox[red, 2pt] {{13\over 3}\pi}$$

解答:$$f(x)={1\over x} \Rightarrow \cases{f(1)=1 \\ f'(x)=-1/x^2} \Rightarrow \cases{f'(1)=-1\\ f''(x)=2/x^3} \Rightarrow \cases{f''(1)=2\\ f'''(x)=-6/x^4} \Rightarrow f'''(1)=-6 \\ f(x)=f(1)+(x-1)f'(1)+ {1\over 2}(x-1)^2f''(1) +{1\over 6}(x-1)^3 f'''(1)+ \cdots \\\qquad =\bbox[red, 2pt]{1-(x-1)+(x-1)^2}-(x-1)^3+ \cdots$$

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