107年台綜大轉學考-微積分C詳解

臺灣綜合大學系統107學年度學士班轉學生聯合招生考試

科目名稱：微積分C

解答：$$f(x)=x^9+ e^{x\ln 5}+ {\ln x\over \ln 2} +{\ln 7\over \ln x} \Rightarrow f'(x)=9x^8 +\ln 5e^{x\ln 5}+ {1\over x\ln 2}-{\ln 7\over x(\ln x)^2} \\ =\bbox[red, 2pt]{9x^8 +5^x\ln 5 + {1\over x\ln 2}-{1\over x\ln x}\cdot \log_x 7 }$$

解答：$$f(x)={\ln x\over \sqrt[3] x} \Rightarrow f'(x)=x^{-4/3}(1-{1\over 3}\ln x)=0 \Rightarrow x=e^3(x\ne 0) \\ \Rightarrow 極大值f(e^3)= \bbox[red,2pt]{3\over e}$$

解答：$$\lim_{x\to 0}{\sin( 2\tan^{-1}x)-\tan(\sin^{-1}2x) \over x^3} =\lim_{x\to 0}{{2x\over 1+x^2}-{2x\over \sqrt{1-4x^2}} \over x^3} =\lim_{x\to 0} \left({2\over x^2(1+x^2)}-{2\over x^2\sqrt{1-4x^2}}\right)\\ 由於\cases{{1\over 1+x^2} = 1-x^2 +x^4-x^6+\cdots \Rightarrow {2\over x^2(1+x^2)}={2\over x^2}-2+2x^2-2x^4+\cdots\\ {1\over \sqrt{1-4x^2}} = 1+2x^2 +6x^4 +\cdots \Rightarrow {2\over x^2\sqrt{1-4x^2}}= {2\over  =x^2}+4+12x^2+\cdots} \\ \Rightarrow \lim_{x\to 0} \left({2\over x^2(1+x^2)}-{2\over x^2\sqrt{1-4x^2}}\right) =\lim_{x\to 0}  \left( -6-10x^2 +\cdots\right) =\bbox[red, 2pt]{-6}$$

解答：$$令y= \lim_{n\to \infty}{1\over n}\sqrt[n]{(n+1)(n+3)\cdots (3n-1)}y= \lim_{n\to \infty} \sqrt[n]{(1+{1\over n}) (1 +{3\over n})\cdots (1+{2n-1\over n})} \\ \Rightarrow \ln y=\lim_{n\to \infty} {1 \over n} \left(\ln {(1+{1\over n}) +\ln (1 +{3\over n}) +\cdots +\ln(1+{2n-1\over n})} \right) \\\qquad = \lim_{n\to \infty} \sum_{k=1}^n {1\over n}\ln(1+{2k-1\over n}) =\lim_{n\to \infty}\left(\sum_{k=1}^{2n} {1\over 2}\ln(1+{k\over n}) -\sum_{k=1}^n{1\over 2}\ln(1+{2k\over n}) \right) \\ \qquad=\int_0^2 {1\over 2}\ln (1+x)\,dx -\int_0^1 {1\over 2}\ln(1+{2x})\,dx \\\qquad ={1\over 2}\left(\left.\left[ (x+1)\ln (x+1)-x\right] \right|_0^2 -\left.\left[(x+{1\over 2})\ln(1+2x)-x \right] \right|_0^1 \right) \\\qquad ={1\over 2}(3\ln 3-2-({3\over 2}\ln 3-1)) =\ln {3^{3/4}\over \sqrt e} \\ \Rightarrow y=\bbox[red,2pt]{{3^{3/4}\over \sqrt e} }$$

解答：$$令\sqrt 2\tan u= x \Rightarrow \sqrt 2\sec^2 u\,du = dx \Rightarrow \int{1\over (x^2+2)^{5/2}}\,dx =\int {\sqrt 2 \sec^2u \over (2\tan^2 u+2)^{5/2}}du \\ =\int {\sqrt 2 \sec^2u \over (2\sec^2 u)^{5/2}}du =\int {1\over 4\sec ^3 u}du ={1\over 4}\int \cos^3 u\,du ={1\over 4}\left({3\over 4}\sin u+{1\over 12}\sin(3u) \right) +C\\={3\over 16}\sin u+ {1\over 48}\sin(3u) +C= {3\over 16}\sin u+ {1\over 48}(3\sin(u)-4\sin^3 u) +C= {1\over 4}\sin u-{1\over 12}\sin^3 u +C\\={1\over 4}\cdot {x\over \sqrt{x^2+2}} -{1\over 12}\cdot {x^3\over (x^2+2)^{3/2}} +C=\bbox[red,2pt]{{x(x^2+3) \over 6(x^2+2)^{3/2}}+C}$$

解答：$${1\over 1+x} = 1-x+x^2-x^3+x^4 -\cdots \\ \Rightarrow \int {1\over 1+x}\,dx = \ln(1+x)= x-{1\over 2}x^2+{1\over 3}x^3 -{1\over 4}x^4+{1\over 5}x^5 -\cdots\\ \Rightarrow {\ln(1+x)\over x} =1-{1\over 2}x +{1\over 3}x^2 -{1\over 4}x^3+ {1\over 5}x^4-\cdots \\\Rightarrow \int_0^1 {\ln(1+x)\over x}\,dx = \left.\left[ x-{1\over 2^2}x^2 +{1\over 3^2}x^3- {1\over 4^2}x^4+ \cdots\right]\right|_0^1 \\ =1-{1\over 2^2}+ {1\over 3^2}- {1\over 4^2} +{ 1\over 5^2}- \cdots\\ 令\cases{S=\sum_{n=1}^\infty{1\over n^2}= 1+{1\over 2^2}+{1\over 3^2}+\cdots ={\pi^2\over 6} \\ S_{odd} =1+{1\over 3^2} +{1\over 5^2}+{1\over 7^2}+\cdots \\ S_{even} ={1\over 2^2} +{1\over 4^2} +{1\over 6^2} + \cdots}\\ \Rightarrow S_{odd} = S-S_{even}=S-{1\over 2^2}\left(1+{1\over 2^2}+{1\over 3^2}+\cdots \right) =S-{1\over 4}S={3\over 4}S ={\pi^2\over 8}  \\ \Rightarrow S_{even}= S-S_{odd}=S-{\pi^2\over 8}={\pi^2 \over 24 } \Rightarrow \int_0^1 {\ln(1+x)\over x}\,dx = S_{odd}-S_{even} = {\pi^2\over 8}-{\pi^2\over 24} =\bbox[red, 2pt]{\pi^2\over 12}$$

解答：

$$面積={1\over 2}\int_0^{2\pi} r^2\,d\theta ={1\over 2}\int_0^{2\pi} \left({3\over 2+\sin \theta} \right)^2\,d\theta ={1\over 2}\cdot 4\sqrt 3\pi = \bbox[red,2pt]{2\sqrt 3\pi}$$

解答：$$將(2,1)代入F(x,y,z)= 2+z^3 +2z-5=0 \Rightarrow z=1\\ 又F(x,y,z) = xy+z^3 +xyz-5 \Rightarrow dF=(y+yz)dx +(x+xz)dy +(3z^2+xy)dz \\ 再將\cases{dx= 1.97-2=-0.03\\ dy = 1.04-1=0.04}代入 dF(2,1,1)=  2\cdot (-0.03)+ 4\cdot 0.04 +5dz =0\\ \Rightarrow dz= -0.02 \Rightarrow f(1.97,2.04)=1-0.02 = \bbox[red,2pt]{0.98}$$

解答：$$令\cases{u=x^2+y^2 \\ v=y^2-x^2} \Rightarrow \begin{vmatrix} {\partial (u,v)\over \partial (x,y)}\end{vmatrix} =\begin{vmatrix} u_x & u_y\\ v_x & v_y\end{vmatrix} =\begin{vmatrix} 2x & 2y\\ -2x & 2y\end{vmatrix} =8xy \Rightarrow \begin{vmatrix} {\partial (x,y)\over \partial (u,v)}\end{vmatrix} ={1\over 8xy} \\ \Rightarrow \iint_R {xy\over y^4-x^4}\,dA =\int_1^4 \int_4^9 {xy \over vu}\cdot {1\over 8xy}\,dudv = \int_1^4 \int_4^9 {1 \over 8vu} \,dudv = \int_1^4 {1\over 8v}\ln{9\over 4}\,dv ={1\over 8}\ln{9\over 4}\cdot \ln 4 \\={1\over 2}\ln {3\over 2}\cdot \ln 2 \Rightarrow 僅考慮第一象限，故欲求之積分值為{1\over 4}\times {1\over 2}\ln {3\over 2}\cdot \ln 2 =\bbox[red,2pt]{{1\over 8}\ln {3\over 2}\cdot \ln 2}$$

解答：$$令\cases{M(x,y)=(x-y)/(x^2+y^2)\\ N(x,y)= (x+y)/(x^2+y^2} \Rightarrow M_y=N_x= (-x^2-2xy+ y^2)/(x^2+y^2)^2\\ 令\Phi(x,y) = \int M\,dx= \int N\,dy = {1\over 2}\ln(x^2+y^2)+ \tan^{-1}(y/x)+C\\ 又\cases{A=r(0)= (2,1)\\ B= r(1)= (1,3)}，則 \int_C Mdx +Ndy = \Phi(B)-\Phi(A) \\= {1\over 2}\ln 10+\tan^{-1}3 -({1\over 2} \ln 5+\tan^{-1}(1/2)) \\= \bbox[red, 2pt]{\ln \sqrt 2+ \tan^{-1}3-\tan^{-1}(1/2)}$$
 

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