113年台聯大轉學考-微積分A3A4A6詳解

 台灣聯合大學113學年度學士班轉學生考試

科目: 微積分
類組別: A3/A4/A6

甲、填充題:共8題，每題8分，共64分

解答:$$\lim_{x \to 2}{\displaystyle\int_4^{x^2} {\sin t\over t}\,dt\over x-2} =\lim_{x \to 2}{\displaystyle \left(\int_4^{x^2} {\sin t\over t}\,dt\right)'\over (x-2)'} =\lim_{x \to 2} {2\sin x^2\over x} =\bbox[red, 2pt]{\sin 4}$$

解答:$$\text{Let }\cases{u=e^{-2x}\\ dv=\cos x\,dx},\text{ then}\cases{du=-2e^{-2x}\,dx\\ v=\sin x} \Rightarrow I= \int  e^{-2x}\cos x\,dx =e^{-2x}\sin x+2\int e^{-2x}\sin x\,dx\\ \text{Again, let }\cases{u=e^{-2x}\\ dv = \sin x\,dx}, \text{ then }\cases{du=-2e^{-2x}\,dx \\ v=-\cos x} \Rightarrow I=e^{-2x}(\sin x-2\cos x)-4I\\ \Rightarrow I={1\over 5}e^{-2x}(\sin x-2\cos x) \Rightarrow \int_0^{\infty}  e^{-2x}\cos x\,dx = \left. \left[ {1\over 5}e^{-2x}(\sin x-2\cos x) \right] \right|_0^\infty =\bbox[red, 2pt]{2\over 5}$$

解答:$$u=e^x-1 \Rightarrow du=e^x\,dx \Rightarrow I=\int_0^{\ln 17} {e^x\sqrt{e^x-1} \over e^x+15}\,dx = \int_0^{16} {\sqrt u\over u+16} \,du\\ v=\sqrt u \Rightarrow dv ={du\over 2\sqrt u} = {1\over 2v}du\Rightarrow I=\int_0^4 {2v^2\over v^2+16}dv =2 \int_0^4\left( 1-{16\over v^2+16}\right)\,dv \\ w={v\over 4} \Rightarrow dw={1\over 4}dv \Rightarrow I= 8-32 \int_0^4 {1\over v^2+16} \,dv =8-32\int_0^1 {1\over 4(w^2+1)}dw \\=8-8\int_0^1{1\over w^2+1}\,dw =8-8 \left. \left[ \tan^{-1} w \right] \right|_0^1 =\bbox[red, 2pt]{8-2\pi}$$

解答:$$\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow J=\left| {\partial(x,y)\over \partial(r,\theta}\right| =r \Rightarrow \iint_D (x^2+y^2)^{3/2}\,dA= \int_0^{\pi/3} \int_0^3 r^4\,drd\theta \\= \int_0^{\pi/3} {243\over 5}\,d\theta= \bbox[red, 2pt]{{81\over 5}\pi}$$

解答:$$\cases{u=x+y\\ v=x-y} \Rightarrow \cases{x=(u+v)/2\\ y=(u-v)/2} \Rightarrow \left|{\partial (x,y)\over \partial (u,v)} \right| =\begin{Vmatrix}1/2 & 1/2 \\1/2 & -1/2 \end{Vmatrix} = {1\over 2}\\ \iint_R (x+y)e^{x^2-y^2}\,dA ={1\over 2} \int_0^3 \int_0^2 ue^{uv} \,dv du={1\over 2} \int_0^3 \left. \left[ e^{uv} \right] \right|_0^2 \, du ={1\over 2} \int_0^3 \left( e^{2u}-1\right) \, du \\={1\over 2} \left. \left[ {1\over 2}e^{2u}-u \right] \right|_0^3 =\bbox[red, 2pt]{{1\over 4}(e^6-7)}$$

解答:$$\cases{x=\rho \sin \phi \cos \theta\\ y=\rho \sin \phi\sin \theta\\ z=\rho \cos \phi} \Rightarrow J=\left|{\partial(x,y,z)\over \partial (\rho, \phi,\theta} \right| =\rho^2\sin \phi \\ \Rightarrow \iiint_E x^2\,dV= \int_0^\pi \int_0^\pi \int_0^2 \rho^4 \sin^3 \phi \cos^2 \theta \,d\rho\,d\phi\,d\theta= {32\over 5}\int_0^\pi \int_0^\pi \sin^3 \phi \cos^2 \theta  \,d\phi\,d\theta \\={128\over 15} \int_0^\pi\cos^2 \theta  \, d\theta = \bbox[red, 2pt]{{64\over 15}\pi}$$

解答:$$\cases{x(t)=\cos t\\ y(t)=\sin t\\ z(t)= \ln(\cos t)} \Rightarrow \cases{x'(t)=-\sin t\\ y'(t)=\cos t\\ z'(t)= -\tan t}\\ \Rightarrow \text{ length of the curve }=\int_0^{\pi/4} \sqrt{\sin^2 t+ \cos^2 t+ \tan^2 t}\,dt =\int_0^{\pi/4} \sqrt{1+ \tan^2 t}\,dt \\ =\int_0^{\pi/4} \sec t\,dt =\left. \left[ \ln|\sec t+\tan t| \right] \right|_0^{\pi/4} = \bbox[red, 2pt]{\ln (1+\sqrt 2)}$$

解答:$$\mathbf r(t)=(t-\sin t)\mathbf i+(1-\cos t)\mathbf j \Rightarrow \mathbf r'(t)=(1-\cos t)\mathbf i+ \sin t\mathbf j \\ \Rightarrow \int_C \mathbf F\cdot d\mathbf r =\int_0^{2\pi} ((t-\sin t)\mathbf i +(3-\cos t)\mathbf j)\cdot ((1-\cos t)\mathbf i+ \sin t\mathbf j)\,dt \\= \int_0^{2\pi} ((t-\sin t)(1-\cos t)+(3-\cos t)(\sin t))\,dt=\int_0^{2\pi}  (t-t\cos t+2\sin t)\,dt \\= \left. \left[ {1\over 2}t^2-t\sin t -3\cos t \right] \right|_0^{2\pi} = \bbox[red, 2pt]{2\pi^2}$$

乙、計算、證明題：共3題，每題12分，共36分。

解答:$$\textbf{(a)}\; a_n={(-1)^n x^n \over 3^{2n}(2n)!} \Rightarrow \lim_{n\to \infty} \left|{a_{n+1}\over a_n} \right| =\lim_{n\to \infty} \left|{a_{n+1}\over a_n} \right| =\lim_{n\to \infty} \left|{(-1)^{n+1} x^{n+1} \over 3^{2n+2}(2n+2)!} \cdot { 3^{2n}(2n)! \over (-1)^n x^n  } \right| \\\quad =\lim_{n\to \infty}  \left| {x\over 3^2(2n+2)(2n+1) }\right| =0 \Rightarrow \text{the radius of convergence: }\bbox[red, 2pt]{\infty} \\\textbf{(b)}\;a_n={1\over n(1+\ln^2 n)} \Rightarrow a_{n+1}={1\over (n+1)(1+\ln^2(n+1))} \Rightarrow a_n \gt a_{n+1} \\ \text{and }\lim_{n\to \infty} a_n=0, \text{by alternating series test, it is convergent} \\ \Rightarrow 2^na_{2^n} ={2^n\over 2^n(1+\ln^2 2^n)} ={1\over 1+n^2(\ln 2)^2} \lt {1\over (\ln 2)^2(1+n^2)} \lt {1\over (\ln 2)^2 n^2} \text{ convergent} \\ \text{By Cauchy Condensation Test,} \sum_{n=1}^\infty\left| {(-1)^n\over n(1+\ln^2 n)}\right| \text{ is convergent. }\\ \text{That is , it is }\bbox[red, 2pt]{\text{absolutely convergent}}. $$

解答:$$\textbf{(a)}\;L=(\tan x)^{\cos x} \Rightarrow \ln L=\cos x \ln (\tan x) ={\ln(\tan x)\over 1/\cos x} \\ \lim_{x\to (\pi/2)^-}{\ln(\tan x)\over 1/\cos x} =\lim_{x\to (\pi/2)^-}{(\ln(\tan x))' \over (1/\cos x)'} = \lim_{x\to (\pi/2)^-}{\cos x\over \sin^2 x} =0 \\ \Rightarrow \lim_{x\to (\pi/2)^-} L= e^0=\bbox[red, 2pt]1 \\\textbf{(b)}\;y=\alpha x \Rightarrow \lim_{(x,y) \to (0,0)}{y^2 \sin^2 x\over x^4+y^4} =\lim_{x\to 0}{\alpha^2 x^2 \sin^2 x\over (\alpha^4+1) x^4} ={\alpha^2 \over \alpha^4+1}\lim_{x\to 0} {\sin^2 x\over x^2} \\={\alpha^2 \over \alpha^4+1}\lim_{x\to 0} {(\sin^2 x)'\over (x^2)'}= {\alpha^2 \over \alpha^4+1}\lim_{x\to 0} {\sin 2 x\over 2x}  ={\alpha^2 \over \alpha^4+1} \text{ is dependent on the value of }\alpha \\ \Rightarrow \bbox[red, 2pt]{\text{ the limit does not exist}}$$

解答:$$f(x,y)=(x^2+y^2)e^{-x} \Rightarrow \cases{f_x= e^{-x}(2x-x^2-y^2)\\ f_y=2ye^{-x}} \Rightarrow \cases{f_{xx}= e^{-x}(x^2+y^2-4x+2)\\ f_{xy}=-2ye^{-x} \\ f_{yy}= 2e^{-x}} \\ \Rightarrow D(x,y)=f_{xx}f_{yy}-(f_{xy})^2 =2e^{-x}(x^2-y^2-4x+2)\\ \cases{f_x=0\\ f_y=0} \Rightarrow \cases{(x,y)=(0,0)\\ (x,y)=(2,0)} \Rightarrow \cases{D(0,0)=4\gt 0 \Rightarrow f_{xx}(0,0)= 2\gt 0\Rightarrow f(0,0)=0\\ D(2,0)= -4e^{-2} \lt 0} \\ \Rightarrow \bbox[red, 2pt]{ \cases{\text{local minimum: }0\\ \text{saddle point: }(2,0)}}$$

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解題僅供參考, 其他歷年試題及詳解