115年中山大學企管碩士-微積分詳解

國立中山大學 115學年度碩士班考試入學招生考試試題

科目名稱:微積分【企管系企管甲班碩士班甲組選考、乙組選考、丙組選考】

 

解答:$$\cases{\Delta x=15\\ n=10} \text{ and }\cases{y_0=0, y_1=25, y_2=40, y_3=70,y_4=80, y_5=90,\\y_6=65 ,y_7=50, y_8=60,y_9=35,y_{10}=0} \\\textbf{(i) Trapezoidal Rule }\\\qquad Area \approx{\Delta x\over 2}[y_0+ 2(y_1+\cdots+y_{n-1})+y_n] ={15\over 2}(2(25+40+\cdots+60+35)) =\bbox[red, 2pt]{7725} \\\textbf{(ii) Simpson's Rule} \\ \qquad Area \approx {\Delta x\over 3}[y_0+ 4(y_1+y_3+\cdots+y_{n-1}) +2(y_2+y_4+\cdots+y_{n-2}) +y_n] \\\qquad ={15\over 3}(4(25+70+90+50+35)+ 2(40+80+65+60)) =\bbox[red, 2pt]{7850}$$

解答:$$u=\sqrt x \Rightarrow du={1\over 2\sqrt x}dx \Rightarrow dx=2\sqrt x\,du =2u\,du \Rightarrow I=\int_0^1 e^{-\sqrt x}\,dx =\int_0^1 2ue^{-u}\,du \\\cases{v=2u\\ dw=e^{-u}\,du} \Rightarrow \cases{dv=2du\\ w=-e^{-u}} \Rightarrow I= \left. \left[ -2ue^{-u} \right] \right|_0^1+\int_0^1 2e^{-u}\,du=-2e^{-1}+ \left. \left[ -2e^{-u} \right] \right|_0^1 \\=-2e^{-1}-2e^{-1}+2 = \bbox[red, 2pt]{2-{4\over e}}$$

解答:$$f(x) \gt g(x) \text{ for }x\in[-1,1] \Rightarrow Area= \int_{-1}^1 (f(x)-g(x))\,dx =\int_{-1}^1 (x^2-2x+1+e^x)\,dx \\= \left. \left[ {1\over 3}x^3-x^2 +x+e^x\right] \right|_{-1}^1={1\over 3}+e-(-{7\over 3}+e^{-1}) =\bbox[red, 2pt]{{8\over 3}+e-e^{-1}}$$

解答:$$u= \sqrt x \Rightarrow du={1\over 2\sqrt x}dx={1\over 2u}dx \Rightarrow dx=2u\,du \Rightarrow \int_0^1 {1\over x+\sqrt x}\,dx = \int_0^1 {2u\over u^2+u}\,du \\=2\int_0^1 {1\over u+1}\,du =2 \left. \left[ \ln(u+1) \right] \right|_0^1 = \bbox[red, 2pt]{2 \ln 2}$$

解答:$$\cases{u=x \\ dv=\sqrt{x+1} \,dx} \Rightarrow \cases{du=dx \\ v={2\over 3}(x+1)^{3/2}} \Rightarrow \int x(x+1)^{0.5} \,dx= {2\over 3}x(x+1)^{3/2} -{2\over 3}\int (x+1)^{3/2}\,dx \\= \bbox[red, 2pt]{{2\over 3}x(x+1)^{3/2}-{4\over 15}(x+1)^{5/2}+C}$$

解答:

$$f(t)=3t^4+4t^3 \Rightarrow f'(t)=12t^3+12t^2 \Rightarrow f''(t)=36t^2+24t\\ f(t)=0 \Rightarrow t^3(3t+4)=0 \Rightarrow t=0,-{4\over 3} \\ f'(t)=0 \Rightarrow 12t^2(t+1)=0 \Rightarrow t=0,-1 \Rightarrow \cases{f''(0)=0\\ f''(-1)=12 \gt 0 \Rightarrow \text{local minimum: }f(-1)=-1} \\ f''(x)=0 \Rightarrow 12t(3t+2)=0 \Rightarrow \text{ inflection points at }t=0,-{2\over 3}$$

解答:$$y'={xe^x\over 2y} \Rightarrow \int 2y\,dy =\int xe^x\,dx \Rightarrow y^2=xe^x-e^x+C \Rightarrow \bbox[red, 2pt]{y=\pm \sqrt{xe^x-e^x+C}}$$

解答:$$u=1-4\sin(7x) \Rightarrow du =-28\cos(7x)\,dx \Rightarrow \int \cos(7x) \sqrt{1-4\sin(7x)} \,dx =\int-{1\over 28} \sqrt u\,du \\=-{1\over 42}u^{3/2}+C =\bbox[red, 2pt]{-{1\over 42}(1-4\sin(7x))^{3/2}+C}$$

解答:$$\textbf{(i) }\ln(1+u) = \sum_{n=1}^\infty (-1)^{n-1}{u^n \over n} \Rightarrow \ln x=\ln(14+(x-14)) =\ln \left( 14 \left[ 1+{x-14\over 14}\right] \right) \\=\ln 14+ \ln \left( 1+{x-14\over 14} \right) =\ln 14+ \sum_{n=1}^\infty (-1)^{n-1}{((x-14)/14)^n \over n}  =\ln 14+ \sum_{n=1}^\infty {(-1)^{n-1} \over n\cdot 14^n} (x-14)^n \\ \Rightarrow (x-14)\ln x = (x-14)\ln(14)+ \sum_{n=1}^\infty {(-1)^{n-1} \over n\cdot 14^n} (x-14)^{n+1} \\ \Rightarrow \bbox[red, 2pt]{f(x)=(x-14)\ln(14)+ \sum_{k=2}^\infty {(-1)^{k-2} \over (k-1)\cdot 14^{k-1}} (x-14)^{k}} \\\textbf{(ii) } \left|{x-14\over 14} \right|\lt 1 \Rightarrow 0\lt x\lt 28 \\ \ln x \Rightarrow x\ne 0; x=28 \Rightarrow  \sum_{k=2}^\infty {(-1)^{k-2}\over k-1}\cdot 14 \text{ converges} \Rightarrow \text{interval of convergence: }\bbox[red, 2pt]{(0,28] }$$

解答:$$V=\int_0^4 \int_0^{5\sqrt x}{2y\over 1+x^2} \,dydx =\int_0^4 {25x\over 1+x^2} \,dx = {25\over 2} \left. \left[ \ln|u| \right] \right|_1^{17}\quad (u=1+x^2) \\={25\over 2}(\ln 17-\ln 1) = \bbox[red, 2pt]{{25\over 2}\ln 17}$$

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