國立臺中教育大學 115 學年度學士班轉學生招生考試
微積分試題
適用學系： 數學教育學系
一、 選擇題（每題 6%，共 30%）
解答:$$ x = g(y) = \int_y^1 \sqrt{t^3 - 1} \,dt \Rightarrow \frac{dx}{dy} = \frac{d}{dy} \left( \int_y^1 \sqrt{t^3 - 1} \,dt \right) = -\sqrt{y^3 - 1} \\ \Rightarrow L = \int_a^b \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \,dy = \int_1^4 \sqrt{1 + \left(-\sqrt{y^3 - 1}\right)^2} \,dy = \int_1^4 \sqrt{y^3} \,dy = \left[ \frac{2}{5} y^{5/2} \right]_1^4 = \frac{62}{5}\\ ,故選\bbox[red, 2pt]{(B)}$$
解答:$$ x = \frac{y^3}{14} \Rightarrow {dx\over dy} ={3y^2\over 14} \Rightarrow S = \int_a^b 2\pi x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \,dy = \int_0^4 2\pi \left(\frac{y^3}{14}\right) \sqrt{1 + \frac{9y^4}{196}} \,dy \\ = \frac{\pi}{7} \int_0^4 y^3 \sqrt{1 + \frac{9y^4}{196}} \,dy = \frac{\pi}{7} \int_1^{\frac{625}{49}} \sqrt{u} \left(\frac{49}{9} \,du\right) = \frac{\pi}{7} \cdot \frac{49}{9} \int_1^{\frac{625}{49}} u^{1/2} \,du = \frac{7\pi}{9} \left[ \frac{2}{3} u^{3/2} \right]_1^{\frac{625}{49}} \\ = \frac{14\pi}{27} \left( \frac{15625}{343} - 1 \right) = \frac{30564\pi}{1323} = \frac{1132}{49}\pi \quad ( u = 1 + \frac{9y^4}{196} ),故選\bbox[red, 2pt]{(C)}$$
解答:$$\int \frac{5x^3 + 10x^2 + 32x + 10}{x^2(x^2 + 2x + 10)} \,dx = \int \left( \frac{3}{x} + \frac{1}{x^2} + \frac{2x + 3}{x^2 + 2x + 10} \right)\,dx \\= 3\ln |x|-{1\over x} + \int \left( {2x+2 \over x^2+2x+10} +{1\over (x+1)^2+3^2}\right)\,dx \\= 3\ln |x|-{1\over x} +\ln |x^2+2x+10|+ {1\over 3}\tan^{-1}{x+1\over 3},故選\bbox[red, 2pt]{(B)}$$
解答:$$\cosh^{-1}(x) = \ln\left(x + \sqrt{x^2 - 1}\right) \Rightarrow \cosh^{-1}\left(\frac{13}{5}\right) = \ln\left(\frac{13}{5} + \sqrt{\left(\frac{13}{5}\right)^2 - 1}\right) =\ln\left(\frac{13}{5} + \frac{12}{5}\right) \\= \ln\left(\frac{25}{5}\right) = \ln(5) ,故選\bbox[red, 2pt]{(C)}$$
解答:$$y=1=\sqrt{\sin 3x} \Rightarrow \sin 3x=1 \Rightarrow 3x={\pi\over 2} \Rightarrow x={\pi\over 6} \\ \Rightarrow V = \pi \int_{a}^{b} [R(x)^2 - r(x)^2] \, dx = \pi \int_{0}^{\frac{\pi}{6}} \left[ 1^2 - (\sqrt{\sin 3x})^2 \right] \, dx = \pi \int_{0}^{\frac{\pi}{6}} (1 - \sin 3x) \, dx \\ = \pi \left[ x + \frac{1}{3}\cos 3x \right]_{0}^{\frac{\pi}{6}} = \pi \left( \frac{\pi}{6} + 0 - \frac{1}{3} \right) = \frac{\pi^2}{6} - \frac{\pi}{3} ,故選\bbox[red, 2pt]{(A)}$$
二、 填充題（每題 6%，共 30%）
解答:$$ \frac{d}{dx}[xe^x] = \frac{d}{dx}x \cdot e^x +x\cdot \frac{d}{dx}e^x =e^x+xe^x = \bbox[red, 2pt]{e^x(1+x)}$$
解答:$$\frac{d}{dx}[\arctan(3x)] = {{d\over dx}(3x) \over 1+(3x)^2} = \bbox[red, 2pt]{3\over 1+9x^2}$$
解答:$$ \int_{0}^{\pi} 2 \sin x \, dx = [-2 \cos x]_{0}^{\pi} =2-(-2)= \bbox[red, 2pt]4$$
解答:$$u=x^2+1 \Rightarrow du=2x\,dx \Rightarrow \int_{0}^{1} 8x(x^2 + 1)^3 \, dx = \int_{1}^{2} u^3 (4 \, du) = 4 \int_{1}^{2} u^3 \, du \\ =4 \left[ \frac{u^4}{4} \right]_{1}^{2} = \left[ u^4 \right]_{1}^{2}=16-1=\bbox[red, 2pt]{15}$$
解答:$$\int_{0}^{2} \int_{y^2}^{4} 3 \, dx \, dy = \int_{0}^{2} 3(4-y^2) \, dy = \int_{0}^{2} (12 - 3y^2) \, dy = \left[ 12y - y^3 \right]_{0}^{2} =24-8-0=\bbox[red, 2pt]{16}$$
三、 計算證明題（需書寫詳細計算證明過程，每題 10%，共 40%）
解答:$$ I(t) = \int_{0}^{\infty} \frac{\sin x}{x} e^{-tx} \, dx \Rightarrow I'(t) = \int_{0}^{\infty} \frac{\partial}{\partial t} \left( \frac{\sin x}{x} e^{-tx} \right) \, dx = -\int_{0}^{\infty} e^{-tx} \sin x \, dx \\ = - \left[ \frac{e^{-tx}}{t^2 + 1} (-t \sin x - \cos x) \right]_{0}^{\infty} = -\left( 0 - \left( -\frac{1}{t^2 + 1} \right) \right) = -\frac{1}{t^2 + 1} \\ \Rightarrow I(t) = \int -\frac{1}{t^2 + 1} \, dt = -\arctan(t) + C \Rightarrow \lim_{t\to \infty} I(t)= \lim_{t \to \infty} (-\arctan(t) + C) = 0 \\ \Rightarrow -\frac{\pi}{2} + C = 0 \Rightarrow C = \frac{\pi}{2} \Rightarrow I(t) = -\arctan(t) + \frac{\pi}{2} \Rightarrow \int_{0}^{\infty} \frac{\sin x}{x} \, dx =I(0)= \bbox[red, 2pt]{\frac{\pi}{2}}$$
解答:$$\int_{1}^{\infty} \frac{dx}{x^p} = \lim_{t \to \infty} \int_{1}^{t} x^{-p} \, dx \\\textbf{Case I: }p=1 \Rightarrow \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x} \, dx = \lim_{t \to \infty} [\ln\vert{}x\vert{}]_{1}^{t} = \lim_{t \to \infty} (\ln t - \ln 1) = \lim_{t \to \infty} \ln t = \infty \Rightarrow \text{ divergence} \\ \textbf{Case II: }p\gt 1 \Rightarrow 1-p\lt 0 \Rightarrow \lim_{t\to \infty} t^{1-p} =0 \Rightarrow \text{ convergence} \\ \textbf{Case III: }p\lt 1 \Rightarrow 1-p\gt 0\lim_{t\to \infty} t^{1-p} = \infty \Rightarrow \text{ divergence} \\ \text{The integral $\int_{1}^{\infty} \frac{dx}{x^p}$ diverges for }\bbox[red, 2pt]{p\le 1}.$$
解答:$$y=2-x^2=x \Rightarrow x^2+x-2=0 \Rightarrow (x+2)(x-1) =0 \Rightarrow x=1,-2 \\ \Rightarrow A = \int_{-2}^{1} [(2 - x^2) - x] \, dx = \int_{-2}^{1} (2 - x - x^2) \, dx = \left[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{1} = \frac{7}{6} - \left(-\frac{10}{3}\right) = \bbox[red, 2pt]{\frac{9}{2} }$$
解答:$$F(x, y, z) = x^2 + y^2 - z^2 \Rightarrow \nabla F=(F_x,F_y,F_z)=(2x,2y,-2z) \Rightarrow \nabla F(5,12,13)=(10,24,-26) \\ \Rightarrow \text{tangent plane: }10(x-5)+24(y-12)-26(z-13)=0 \Rightarrow \bbox[red, 2pt]{5x + 12y - 13z = 0}$$
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