107年台北大學轉學考-微積分詳解

 國立臺北大學107學年度日間學士班暨進修學士班轉學生招生考試

科目：微積分

解答：$$\mathbf{1.}\;\cases{u=\tan^{-1}x\\ dv =dx} \Rightarrow \cases{du ={1\over 1+x^2}dx\\ v=x} \Rightarrow \int \tan^{-1}x\,dx = x\tan^{-1} x-\int {x\over 1+x^2}\,dx \\ \quad= \bbox[red, 2pt]{x\tan^{-1}x-{1\over 2}\ln (1+x^2)+C} \\ \mathbf{2.}\; I= \int_0^b \int_0^{\sqrt{a^2-a^2y^2/b^2}} 1\,dxdy =\int_0^b \sqrt{a^2-{a^2y^2\over b^2}}\,dy =\int_0^b a\sqrt{1-({y\over b})^2}\,dy\\\quad 取y=b\sin\theta \Rightarrow dy =b\cos\theta d\theta \Rightarrow I=\int_0^{\pi /2} ab\cos^2\theta\,d\theta =\left. \left[ {ab\over 4}\sin(2\theta)+{ab\over 2}\theta \right]\right|_0^{\pi/2} ={ab\over 4}\pi \\\quad \Rightarrow 欲求之面積=4I = \bbox[red,2pt]{ab\pi} \\\mathbf{3.}\;\cases{u=\ln x\\ dv=dx} \Rightarrow \cases{du ={1\over x}\,dx\\ v=x} \Rightarrow \int  \ln x\,dx = x\ln x-\int 1\,dx = x\ln x-x\\ \quad \Rightarrow \int_0^1 \ln x= \left.\left[ x\ln x-x \right]\right|_0^1 =\bbox[red,2pt]{-1} \\\mathbf{4.}\; \int_1^2 {1\over x}\,dx = {2-1\over 6}({1\over 1}+4\cdot {1\over (1+2)/2}+ {1\over 2}) ={1\over 6}(1+{8\over 3}+{1\over 2})= \bbox[red, 2pt]{25\over 36}$$

解答：$$\mathbf{1.}\; 假設 |(3x-1)-5|= 3|x-2| \lt \varepsilon  \Rightarrow |x-2|\lt {\varepsilon \over 3}\\ 因此取\delta= {\varepsilon \over 3},則\forall x滿足|x-2|\lt \delta，可以得到 |(3x-1)-5|\lt \varepsilon，因此\lim_{x\to 2}(3x-1)=5 ，\bbox[red, 2pt]{故得證} \\\mathbf{2.}\; f(x)=\sqrt{4-x^2} \Rightarrow f(c)=\sqrt{4-c^2},其中c,x\in[-2,2]\\ \Rightarrow |f(c)-f(x)| =|\sqrt{4-c^2}-\sqrt{4-x^2}|  \le \sqrt{|c^2-x^2|} \le \sqrt{|x-c|} \cdot \sqrt{|x|+|c|} \\ \le 2\sqrt{|x-c|} ,因此令2\sqrt{|x-c|}=\varepsilon \Rightarrow |x-c|={\varepsilon ^2 \over 4}\\所以取\delta \lt {\varepsilon ^2 \over 4} \Rightarrow |x-c|\lt\delta \Rightarrow |f(x)-f(c)| \lt \varepsilon \Rightarrow f(x)在區間[-2,2]連續，\bbox[red,2pt]{故得證} \\ \mathbf{3.}\; \lim_{h\to 0}{f(x+h)-f(x)\over h} = \lim_{h\to 0}{(x+h)^2-(x+h)+1-x^2+x-1\over h} = \lim_{h\to 0}{2hx +h^2-h\over h} \\\quad = \lim_{h\to 0} (2x-1+h)=2x-1 \Rightarrow f(x)可微,\forall x\in \mathbb R,\bbox[red,2pt]{故得證} \\ \mathbf{4.}\;均值定理f'(c)={f(100)-f(0)\over 100-0} =f(100)\ge 6 \Rightarrow f(100)最小值為\bbox[red, 2pt]6\\ \mathbf{5.}\; x^4+y^4= 8xy^2 \Rightarrow 4x^3+ 4y^3 y'=8y^2 +16xyy' \Rightarrow (4y^3-16xy)y'=8y^2-4x^3\\\quad \Rightarrow y'={8y^2-4x^3\over 4y^3-16xy} = \bbox[red, 2pt]{2y^2-x^3\over y(y^2-4x)}\\\mathbf{6.}\;y=\cos(x\cos x) \Rightarrow y'=-\sin(x\cos x)\cdot (\cos x-x\sin x) =\bbox[red,2pt]{(x\sin x-\cos x)\sin (x\cos x)} \\\mathbf{7.}\, y={x\over x+{a\over x}} \Rightarrow y'={1\over x+{a\over x}}-{x\over (x+{a\over x})^2}(1-{a\over x^2}) =\bbox[red, 2pt]{2a \over x(x+{a\over x})^2}$$

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