112年成大環工碩士班-微積分詳解

 國立成功大學１１２學年度碩士班招生考試

系所:環境工程學系
科目:微積分

解答: $$\mathbf{1.}\; y=\sin(\sqrt[3] x)+ \sqrt[3]{\sin x} \Rightarrow \bbox[red ,2pt]{y'=\cos(\sqrt[3] x) \cdot {1\over 3}x^{-2/3} +{1\over 3}(\sin x)^{-2/3} \cos x} \\ \mathbf{2.}\; F(x)=\int_0^{e^{2x}}\ln(t+1)\,dt  \Rightarrow F'(x)=\ln(e^{2x}+1)\cdot {d\over dx}e^{2x} \Rightarrow \bbox[red, 2pt]{F'(x)=2e^{2x} \ln(e^{2x}+1)} \\\mathbf{3.}\; f(x)=\arcsin x+\arccos x \Rightarrow f'(x)={1\over \sqrt{1-x^2}} -{1\over \sqrt{1-x^2}} \Rightarrow \bbox[red, 2pt]{f'(x)=0}$$

解答: $$y=x \sqrt{1-x^2} \Rightarrow {dy\over dx}=\sqrt{1-x^2}+{1\over 2}x (1-x^2)^{-1/2}(-2x) =\sqrt{1-x^2}-{x^2 \over \sqrt{1-x^2}} \\ \Rightarrow dy= \bbox[red, 2pt]{\left(\sqrt{1-x^2}-{x^2 \over \sqrt{1-x^2}} \right)dx}$$

解答: $$(x^2+y^2)^2 =4x^2y \Rightarrow (x^2+y^2)(2x+2yy')= 8xy+4x^2y' \\ \Rightarrow (1+1)(2+2y')=8+4y' \Rightarrow 4y'=4 \Rightarrow y'=\bbox[red, 2pt]1$$

解答: $$令\cases{A(a,\sqrt{a-8})\\ B(12,0)} \Rightarrow d= \overline{AB} = \sqrt{(a-12)^2+ a-8} =\sqrt{a^2-23a+136} \\ \Rightarrow 當2a－23＝0,即 a={23\over 2}時,\overline{AB}最小;　此時A=\bbox[red, 2pt]{({23\over 2},{\sqrt{14}\over 2})}最接近(12,0)$$

解答: $$y=x^2\ln{x\over 4} \Rightarrow y'=2x\ln{x\over 4}+x \Rightarrow y''=2\ln{x\over 4}+3\\ y'=0 \Rightarrow x(2\ln{x\over 4}+1)=0 \Rightarrow x={4\over \sqrt e}，(x＝0不在定義域內), \Rightarrow y‘’({4\over \sqrt e})=2\gt 0 \\ \Rightarrow y({4\over \sqrt e})= \bbox[red, 2pt]{-{8\over e}為相對極大值}\\ y''=0 \Rightarrow x=4e^{-3/2} \Rightarrow (4e^{-3/2}, y(4e^{-3/2}))= \bbox[red, 2pt]{(4e^{-3/2},-{24\over e^3})為反曲點}$$

解答: $$\int y^2\pi dx =\int_0^6 e^{x/2} \pi \,dx = \left. \left[ 2\pi e^{x/2} \right] \right|_0^6 = \bbox[red, 2pt]{2\pi (e^3-1)}$$

解答: $$y=31-10(e^{x/20}+e^{-x/20}) \Rightarrow y'=-10({1\over 20}e^{x/20}-{1\over 20} e^{-x/20}) =-{1\over 2}e^{x/20}+{1\over 2}e^{-x/20} \\ \Rightarrow 曲線長= \int_{-20}^{20} \sqrt{1+(y')^2}\,dx= \int_{-20}^{20} \sqrt{{1\over 2}+{1\over 4}e^{x/10} +{1\over 4}e^{-x/10}} \,dx \\=\int_{-20}^{20} ({1\over 2}e^{x/20}+ {1\over 2}e^{-x/20})\,dx =\left. \left[ 10e^{x/20}-10e^{-x/20} \right] \right|_{-20}^{20} =10(e-e^{-1})-10(e^{-1}-e) \\= \bbox[red, 2pt]{20(e-{1\over e})}$$

解答: $$\mathbf{(1)}\; \cases{u=\ln x\\ dv=x^5\,dx} \Rightarrow \cases{du= dx/x\\ v=x^6/6} \Rightarrow \int x^5\ln x \,dx ={1\over 6}x^6\ln x-{1\over 6} \int x^5\,dx  \\ \quad = \bbox[red, 2pt]{{1\over 6}x^6\ln x-{1\over 36}x^6+C} \\ \mathbf{(2)} \; \sin \theta={5\over 3}x \Rightarrow \cos \theta d\theta ={5\over 3}dx \Rightarrow \int_0^{3/5} \sqrt{9-25x^2}\,dx = \int_0^{3/5} 3\sqrt{1-(5x/3)^2}\,dx \\\quad =\int_0^{\pi/2} 3\sqrt{1-\sin^2 \theta}\cdot {3\over 5}\cos \theta \,d\theta =\int_0^{\pi/2} {9\over 5}\cos^2 \theta \,d\theta  =\int_0^{\pi/2} {9\over 10}(\cos(2\theta)+1)\,d\theta \\\quad = \left. \left[ {9\over 20}\sin(2\theta)+{9\over 10}\theta \right] \right|_0^{\pi/2} =\bbox[red, 2pt]{9\pi \over 20} \\ \mathbf{(3)}\;u=2x-1 \Rightarrow du=2dx \Rightarrow \int_1^5 {x\over \sqrt{2x-1}}\,dx = \int_1^9 {(u+1)/2\over \sqrt u}{1\over 2}du \\\quad ={1\over 4}\int_1^9 u^{1/2}+ u^{-1/2} \,du ={1\over 4} \left. \left[ {2\over 3}u^{3/2} +2u^{1/2} \right] \right|_1^9 =\bbox[red, 2pt]{16\over 3}\\ \mathbf{(4)}\; \int \sin(-7x)\cos(6x)\, dx = {1\over 2}\int \sin(-x) +\sin(-13x)\, dx =-{1\over 2} \int \sin(x)+ \sin(13x)\,dx \\ \quad = -{1\over 2}(-\cos x-{1\over 13}\cos(13x))+C = \bbox[red, 2pt]{{1\over 2} \cos x+{1\over 26}\cos(13x)+C} \\ \mathbf{(5)}\;u=8-x \Rightarrow du=-dx \Rightarrow \int_0^8 {3\over \sqrt{8-x}}\,dx = \int_8^0 {3\over \sqrt u}(-du) =\left. \left[-6\sqrt u \right] \right|_8^0 \\\quad = \bbox[red, 2pt]{12\sqrt 2}$$

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