國立臺北大學115學年度日間學士班轉學招生考試
學制系級:統計學系日間學士班2、3年級
科目:微積分
解答:$$\textbf{(a) } f(x) = x^2 - x - \ln x \Rightarrow f'(x)=2x-1-{1\over x}={(2x+1)(x-1) \over x} =0 \\ \quad \Rightarrow \bbox[red, 2pt]{x=1}\quad (\ln x\Rightarrow x\gt 0) \\ \textbf{(b) } \begin{cases} 0\lt x\lt 1 \Rightarrow f'(x)\lt 0 \Rightarrow f(x) \text{ is decreasing} \\ x\gt 1 \Rightarrow f'(x)\gt 0 \Rightarrow f(x) \text{ is increasing}\end{cases} \Rightarrow \bbox[red, 2pt]{\text{ local minimum: }f(1)=0} \\ \textbf{(c) }f''(x)=2+{1\over x^2}\gt 0 \Rightarrow \bbox[red, 2pt]{f(x)\text{ is concave upward, }x\in (0, \infty); \text{ NO inflection points}} \\\textbf{(d) }$$
解答:$$\textbf{(a) } \int_{0}^{1} \cos(\pi x / 2) dx = \left. \left[ {2\over \pi}\sin(\pi x/2) \right] \right|_0^1= \bbox[red, 2pt]{2\over \pi} \\ \textbf{(b) }\cases{u =(\ln x)^2 \\ dv= dx} \Rightarrow \cases{du=(2\ln x)/x \,dx\\ v=x} \Rightarrow I=\int (\ln x)^2 dx = x(\ln x)^2 - 2\int \ln x dx \\ \cases{u=\ln x \\ dv=dx} \Rightarrow \cases{du=dx/x\\ v=x} \Rightarrow I=x(\ln x)^2-2 \left[ x\ln x -\int 1\,dx\right]= x(\ln x)^2-2x\ln x+2x+C \\ \quad \Rightarrow \int_1^2 (\ln x)^2\,dx = \left. \left[ x(\ln x)^2-2x\ln x+2x \right] \right|_1^2 = \bbox[red, 2pt]{2(\ln 2)^2 - 4\ln 2 + 2 } \\ \textbf{(c) } \int_{0}^{1} \frac{2}{2x^2+3x+1} dx = \int_{0}^{1} \left( \frac{4}{2x + 1} - \frac{2}{x + 1} \right) dx = \left[ 2\ln\vert{}2x + 1\vert{} - 2\ln\vert{}x + 1\vert{} \right]_{0}^{1} \\ \quad = 2\ln\left(\frac{3}{2}\right) - 2\ln\left(\frac{1}{1}\right) = \bbox[red, 2pt]{2\ln\left(\frac{3}{2}\right) }$$
解答:$$\textbf{(a) }u= {1+e^x} \Rightarrow du=e^x\,dx \Rightarrow \int e^x \sqrt{1+e^x}\,dx = \int \sqrt u\,du ={2\over 3}u^{3/2}+C= \bbox[red, 2pt]{{2\over 3}(1+e^x)^{3/2}+C} \\ \textbf{(b) } \cases{u=x^2+2x\\ dv=\cos x\,dx} \Rightarrow \cases{du =(2x+2)\,dx\\ v=\sin x} \Rightarrow I= \int(x^2+2x) \cos x\,dx \\\quad = (x^2 + 2x)\sin x - \int (2x + 2)\sin x \, dx \\ \qquad \cases{u=2x+2\\ dv=\sin x\,dx } \Rightarrow \cases{du =2\,dx\\ v=-\cos x} \Rightarrow I=(x^2 + 2x)\sin x - \left[ -(2x+2)\cos x+ \int 2\cos x\,dx \right] \\\quad =(x^2+2x)\sin x+(2x+2)\cos x-2\sin x+C= \bbox[red, 2pt] {(x^2 + 2x - 2)\sin x + (2x + 2)\cos x + C} \\ \textbf{(c) }\cases{u= \sin(3x) \\dv=e^{2x}\, dx} \Rightarrow \cases{du= 3\cos(3x)\,dx \\ v={1\over 2}e^{2x}} \Rightarrow I = \int e^{2x} \sin(3x) \, dx \\\quad = \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2}\int e^{2x}\cos(3x) \, dx \\ \quad \cases{u=\cos(3x) \\dv=e^{2x}\,dx} \Rightarrow \cases{du=-3\sin(3x) \\ v={1\over 2}e^{2x}} \Rightarrow I= \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2} \left[ {1\over 2}e^{2x}\cos(3x) +{3\over 2} \int e^{2x} \sin(3x)\right] \\=\frac{1}{2}e^{2x}\sin(3x)-{3\over 4}e^{2x}\cos(3x)-{9\over 4}I \Rightarrow \frac{13}{4}I = \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{4}e^{2x}\cos(3x) \\\quad \Rightarrow I = \frac{2}{13}e^{2x}\sin(3x) - \frac{3}{13}e^{2x}\cos(3x) + C = \bbox[red, 2pt]{\frac{e^{2x}}{13} (2\sin(3x) - 3\cos(3x)) + C}$$
解答:$$\textbf{(a) }f(x) = \frac{x^2}{9 + x^6} \Rightarrow f(-x)=f(x) \Rightarrow \text{ $f(x)$ is even} \Rightarrow I= \int_{-\infty}^{\infty} \frac{x^2}{9 + x^6} \,dx = 2\int_{0}^{\infty} \frac{x^2}{9 + x^6} \,dx\\ \quad u=x^3 \Rightarrow du=3x^2 \,dx \Rightarrow I= 2 \int_{0}^{\infty} \frac{1}{9 + u^2} \cdot \frac{du}{3} = \frac{2}{3} \int_{0}^{\infty} \frac{1}{3^2 + u^2} \,du \\\quad =\frac{2}{3} \left[ \frac{1}{3} \arctan\left(\frac{u}{3}\right) \right]_{0}^{\infty} ={2\over 9}\cdot {\pi\over 2}= {\pi\over 9} \Rightarrow { \text{ convergent}} \text{ and } \bbox[red, 2pt]{\pi\over 9} \\ \textbf{(b) }\cases{f(x)= \displaystyle {x+1\over \sqrt{x^4-x}} \\g(x)=\displaystyle {1\over x}} \Rightarrow \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{\frac{x + 1}{\sqrt{x^4 - x}}}{\frac{1}{x}} = \lim_{x \to \infty} \frac{1 + \frac{1}{x}}{\sqrt{1 - \frac{1}{x^3}}} =1\\ \quad \int_{1}^{\infty} g(x) \,dx \text{ is divergent} \Rightarrow \int_1^\infty f(x)\,dx \text{ is }\bbox[red, 2pt]{\text{divergent}}$$
解答:$$ V = \int_{a}^{b} 2\pi r h \, dx = \int_{0}^{1} 2\pi (x)(x^{1/3}) \, dx = 2\pi \int_{0}^{1} x^{4/3} \, dx = 2\pi \left[ \frac{3}{7}x^{7/3} \right]_{0}^{1} =\bbox[red, 2pt]{ \frac{6\pi}{7}}$$
解答:$$\textbf{(a) } -\frac{1}{2\sqrt{n}} \le \frac{(-1)^n}{2\sqrt{n}} \le \frac{1}{2\sqrt{n}} \text{ and }\cases{\lim_{n\to \infty} {1\over 2\sqrt n} =0 \\\lim_{n\to \infty} -{1\over 2\sqrt n} =0 } \\\qquad \text{By Squeeze Theorem, } \lim_{n \to \infty} \frac{(-1)^n}{2\sqrt{n}} = 0 . \bbox[red, 2pt]{\text{Convergence and the limit=0}} \\ \textbf{(b) } 0 \le \cos^2 n \le 1 \Rightarrow 0 \le \frac{\cos^2 n}{2^n} \le \frac{1}{2^n} \text{ and } \lim_{n \to \infty} \frac{1}{2^n} = 0 \\\qquad \text{By Squeeze Theorem, } \lim_{n \to \infty} \frac{\cos^2n}{2^{n}} = 0 . \bbox[red, 2pt]{\text{Convergence and the limit=0}} $$
解答:$$\textbf{(a) } a_n = \frac{x^n}{n!} \Rightarrow \lim_{n \to \infty} \left\vert{} \frac{a_{n+1}}{a_n} \right\vert{} = \lim_{n \to \infty} \left\vert{} \frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}} \right\vert{} = \lim_{n \to \infty} \left\vert{} \frac{x}{n+1} \right\vert{} = \vert{}x\vert{} \lim_{n \to \infty} \frac{1}{n+1} = 0 \\ \qquad \Rightarrow \bbox[red, 2pt]{\cases{\text{radius of convergence: }\infty \\ \text{interval of convergence: }(-\infty, \infty)}} \\ \textbf{(b) } a_n = n!(2x - 1)^n \Rightarrow \lim_{n \to \infty} \left\vert{} \frac{a_{n+1}}{a_n} \right\vert{} = \lim_{n \to \infty} \left\vert{} \frac{(n+1)!(2x - 1)^{n+1}}{n!(2x - 1)^n} \right\vert{} = \vert{}2x - 1\vert{} \lim_{n \to \infty} (n+1) \\\qquad =\begin{cases}0,& x=1/2\\ \infty,& \text{otherwise.} \end{cases} \Rightarrow \bbox[red, 2pt]{\cases{\text{radius of convergence: }0 \\ \text{interval of convergence: }[1/2,1/2]}}$$
解答:$$e^z=xyz \Rightarrow {\partial\over \partial x}e^z ={\partial\over \partial x}(xyz) \Rightarrow {\partial z\over \partial x}e^z =yz + xy{\partial z\over \partial x} \Rightarrow \left( e^z-xy \right) {\partial z\over \partial x}=yz \\ \Rightarrow \bbox[red, 2pt]{\frac{\partial z}{\partial x} = \frac{yz}{e^z - xy}} \\{\partial\over \partial y}e^z ={\partial\over \partial y}(xyz) \Rightarrow {\partial z\over \partial y}e^z = xz + xy{\partial z\over \partial y} \Rightarrow (e^z - xy) \frac{\partial z}{\partial y} = xz \Rightarrow \bbox[red, 2pt]{\frac{\partial z}{\partial y} = \frac{xz}{e^z - xy}}$$
解答:$$\textbf{(a) } \int_0^1 \int_0^{e^y} \sqrt{1+e^y} dx dy = \int_0^1 e^y \sqrt{1+e^y}\,dy = \int_2^{1+e} \sqrt{u} du , u=1+e^y\\= \left[ \frac{2}{3} u^{3/2} \right]_2^{1+e} = \frac{2}{3} \left( (1+e)^{3/2} - 2^{3/2} \right) = \bbox[red, 2pt]{\frac{2}{3} \left( (1+e)\sqrt{1+e} - 2\sqrt{2} \right)} \\ \textbf{(b) } I= \iint_D x^3 dA = \int_1^e \int_0^{\ln x} x^3 dy dx = \int_1^e x^3 \ln x dx \\ \qquad \cases{u=\ln x\\ dv= x^3\,dx} \Rightarrow \cases{du = dx/x\\ v=x^4/4} \Rightarrow I= $\left[ \frac{1}{4} x^4 \ln x - \frac{1}{16} x^4 \right]_1^e \\ = \left( \frac{1}{4} e^4 \ln e - \frac{1}{16} e^4 \right) - \left( \frac{1}{4} 1^4 \ln 1 - \frac{1}{16} 1^4 \right) = \bbox[red, 2pt]{\frac{3e^4 + 1}{16}} \\ \textbf{(c) } I= \int_0^4 \int_{\sqrt{x}}^2 \frac{1}{y^3+1} dy dx = \int_0^2 \int_0^{y^2} \frac{1}{y^3+1} dx dy = \int_0^2 {y^2\over y^3+1}\,dy \\ \qquad u= y^3+1 \Rightarrow du =3y^2 \,dy \Rightarrow I= \int_1^9 {1\over 3u}\,du = {1\over 3} \left. \left[ \ln |u| \right] \right|_1^9 = {1\over 3} \ln 9= \bbox[red, 2pt]{{2\over 3}\ln 3} \\ \textbf{(d) } \iint_D x y^2 dA = \int_{-1}^1 \int_0^{\sqrt{1-y^2}} x y^2 dx dy = \int_{-1}^1 \frac{1}{2} (y^2 - y^4) dy = \frac{1}{2} \left[ \frac{1}{3} y^3 - \frac{1}{5} y^5 \right]_{-1}^1 \\ = \frac{1}{2} \left( \frac{2}{15} - (-\frac{2}{15}) \right) = \bbox[red, 2pt] {\frac{2}{15} }$$
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