115年清華科管碩士班-微積分詳解

 國立清華大學115學年度碩士班考試入學

系所班組別:科技管理研究所乙組
考試科目:微積分

解答:$$\textbf{(a) }\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n \sqrt{1 - \left(\frac{k}{n}\right)^2} = \int_0^1 \sqrt{1 - x^2} \, dx = 半徑為1的四分之一圓面積=\bbox[red, 2pt]{\pi \over 4} \\ \textbf{(b) } \lim_{x\to0^+} \sqrt{x}\ln x = \lim_{x\to0^+} \frac{\ln x}{x^{-1/2}}  = \lim_{x\to0^+} \frac{{d\over dx}\ln x}{ {d\over dx}x^{-1/2}} = \lim_{x\to0^+} \frac{\frac{1}{x}}{-\frac{1}{2}x^{-3/2}} = \lim_{x\to0^+} \left( -2x^{1/2} \right) = \bbox[red, 2pt]0  \\ \textbf{(c) }u=\sqrt x \Rightarrow dx = 2udu \Rightarrow I=\int_1^3 e^{\sqrt{x}} \, dx = \int_1^{\sqrt{3}} e^u \cdot 2u \, du = 2 \int_1^{\sqrt{3}} u e^u \, du \\ \qquad = 2 \left. \left[ ue^u-e^u \right] \right|_1^{\sqrt 3} = \bbox[red, 2pt]{2(\sqrt 3-1)}e^{\sqrt 3} \\\textbf{(d) }改變積分順序: I= \int_0^1 \int_{\sqrt y}^1 \sqrt{x^3+1}\, dxdy =\int_0^1 \int_0^{x^2} \sqrt{x^3 + 1} \, dy \, dx = \int_0^1 x^2 \sqrt{x^3 + 1} \, dx \\ \qquad u=x^3+1 \Rightarrow du =3x^2 \,dx \Rightarrow I= \frac{1}{3} \int_1^2 u^{1/2} \, du = \frac{1}{3} \left[ \frac{2}{3} u^{3/2} \right]_1^2   = \bbox[red, 2pt]{\frac{2(2\sqrt{2} - 1)}{9}} \\ \textbf{(e) }f(x)=x^x \Rightarrow \ln f(x)= x\ln x \Rightarrow {f'(x)\over f(x)} =\ln x+1 \Rightarrow f'(x) =f(x)(\ln x+1)=x^x(\ln x+1) \\ \quad \Rightarrow f'(2)=2^2(\ln 2+1) = \bbox[red, 2pt]{4(\ln 2+1)} \\\textbf{(f) } xy+y^3=1 \Rightarrow 0\cdot y(0)+y^3(0)=1 \Rightarrow y^3(0)=1 \Rightarrow y(0)=1\\ \quad 又 \Rightarrow y+xy'+3y^2y'=0 \Rightarrow y'=-{y\over x+3y^2} \Rightarrow y'(0)=-{y(0) \over 0+3y^2(0)}= \bbox[red, 2pt]{-{1\over 3}} \\ \textbf{(g) }y = \int_1^x \sqrt{\sqrt{t} - 1} \, dt \Rightarrow y'=\sqrt{\sqrt x-1} \Rightarrow  曲線長度 L = \int_a^b \sqrt{1 + [y']^2} \, dx = \int_1^{16} \sqrt{\sqrt{x}} \, dx \\\quad = \int_1^{16} x^{1/4} \, dx = \left. \left[ {4\over 5}x^{5/4} \right] \right|_1^{16}= \bbox[red, 2pt]{124\over 5} \\\textbf{(h) } L= \sum_{n=0}^\infty (\ln 2)^n \left(1 + \frac{1}{n!}\right) = \sum_{n=0}^\infty (\ln 2)^n + \sum_{n=0}^\infty \frac{(\ln 2)^n}{n!} \\\quad \sum_{n=0}^\infty (\ln 2)^n ={1\over 1-\ln 2}, \quad\sum_{n=0}^\infty \frac{(\ln 2)^n}{n!}  =e^{\ln 2 } =2 \Rightarrow L= \bbox[red, 2pt]{{1\over 1-\ln 2}+2}$$

解答:$$\textbf{(a) }\text{若函數 $f(x)$ 在閉區間 $[a, b]$ 上連續，且在開區間 $(a, b)$ 上可微分，}\\ \text{則在 $(a, b)$ 內至少存在一個數 $c$，使得：$f'(c) = \frac{f(b) - f(a)}{b - a}$} \\ \textbf{(b) }f(x)=\sin x, \text{根據均值定理，在介於 $a$ 與 $b$ 的開區間內存在某個數 $c$，使得：} \\\qquad f'(c)={\sin a-\sin b\over a-b} \Rightarrow \sin a-\sin b=(a-b)\cos c \Rightarrow |\sin a-\sin b| =|a-b||\cos c |\le |a-b|\\ \Rightarrow |\sin a-\sin b|\le |a-b|\quad \bbox[red, 2pt]{故得證}$$

解答:$${x^2\over a^2}+ {y^2\over b^2}=1 \Rightarrow 面積為ab\pi，又通過(3,4) \Rightarrow {9\over a^2}+{16\over b^2}=1\\ 算幾不等式:{9\over a^2}+{16\over b^2} \ge 2\sqrt{{9\over a^2}\cdot {16\over b^2}} =2\cdot {12\over ab} \Rightarrow 1\ge {24\over ab} \Rightarrow ab\ge 24\\ 此時{9\over a^2}={16\over b^2}={1 \over 2} \Rightarrow \cases{a=3\sqrt 2\\ b=4\sqrt 2} \Rightarrow 面積最小的橢圖方程式: \bbox[red, 2pt]{ \frac{x^2}{18} + \frac{y^2}{32} = 1 }$$

解答:$$\textbf{(a) } V = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \pi y^2 \, dx = \pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 x \, dx   = \pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos(2x)}{2} \, dx = \frac{\pi}{2} \left[ x + \frac{1}{2}\sin(2x) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}= \bbox[red, 2pt]{\pi^2\over 2} \\ \textbf{(b) }y=\cos x \Rightarrow y'=-\sin x \Rightarrow S= \int_a^b 2\pi y \sqrt{1+(y')^2}\,dx  = 2 \int_0^{\frac{\pi}{2}} 2\pi \cos x \sqrt{1 + \sin^2 x} \, dx\\ = 4\pi \int_0^{\frac{\pi}{2}} \cos x \sqrt{1 + \sin^2 x} \, dx  = 4\pi \int_0^1 \sqrt{1 + u^2} \, du \quad (u=\sin x) \\=  4\pi \left[ \frac{1}{2} u \sqrt{1 + u^2} + \frac{1}{2} \ln \left( u + \sqrt{1 + u^2} \right) \right]_0^1  = \bbox[red, 2pt]{2\pi \left( \sqrt{2} + \ln(1 + \sqrt{2}) \right) }$$

解答:$$\cases{x=0\\ y=0} \Rightarrow \cases{(i) f(0+0)=f(0)g(0) \\(ii) f(0)=1+0\cdot g(0)} \Rightarrow f(0)=1  \Rightarrow f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \\=\lim_{h \to 0} \frac{(1 + hg(h)) - 1}{h} = \lim_{h \to 0} \frac{hg(h)}{h} = \lim_{h \to 0} g(h) = 1  \Rightarrow f'(0)=1 \\  f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h}  = f(x) \cdot \lim_{h \to 0} \frac{f(h) - 1}{h} =f(x)\cdot f'(0)=f(x) \\ \Rightarrow f'(x)=f(x)\quad \bbox[red, 2pt]{故得證}$$

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解題僅供參考，碩士班歷年試題及詳解