107年台聯大轉學考-微積分A2詳解

台灣聯合大學系統107學年度學士班轉學生考試

科目：微積分
類組別：A2
甲、填充題：共8題，每題8分，共64分

解答：

$$f(x)=x^2-16 \Rightarrow f(x)\le 0, x\in [-4,4] \Rightarrow 當\cases{a=-4\\ b=4}時, \int_a^b f(x)\,dx 有最小值\\ \Rightarrow 最小值 = \left.\left[ {1\over 3}x^3-16x\right]\right|_{-4}^4 =\bbox[red, 2pt]{-{256\over 3}}$$

解答： $$\int_0^{\pi/2} \sqrt{1-\sin x}\,dx = \int_0^{\pi/2} \sqrt{1-\cos(\pi/2- x)}\,dx = \int_0^{\pi/2} \sqrt{1-(1-2\sin^2 (\pi/4- x/2))}\,dx \\ = \int_0^{\pi/2} \sqrt{2\sin^2(\pi/4-x/2)} \,dx =\sqrt 2 \int_0^{\pi/2} \sin(\pi/4-x/2)\,dx =\left. \left[ 2\sqrt 2 \cos (\pi/4-x/2)\right] \right|_0^{\pi/2} \\= 2\sqrt 2(1-{\sqrt 2\over 2})=\bbox[red, 2pt]{2\sqrt 2-2}$$

解答： $$\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \iint_R \sqrt{3-x^2-y^2}\,dA =\int_0^{2\pi} \int_0^\sqrt 3 \sqrt {3-r^2}r\,drd\theta =\int_0^{2\pi} \int_3^{0} -{1\over 2} \sqrt u\,dud\theta (u=3-r^2) \\=\int_0^{2\pi} \left.\left[ -{1\over 6}u^{3/2}\right] \right|_3^0 \,d\theta =\int_0^{2\pi} ={1\over 6}3^{3/2} \times 2\pi =  \bbox[red, 2pt]{2\sqrt 3 \pi}$$

解答： $$\lim_{n \to \infty}\left|{a_{n+1}\over a_n} \right| =\lim_{n \to \infty}\left|{2(n+1)(x-3)^{n+1}\over (n+2)!} \cdot {(n+1)!\over 2n(x-3)^n}\right| = \lim_{n \to \infty}\left|{n+1\over n} \cdot {x-3\over  n+2}\right| =0 \\ \Rightarrow 收斂半徑為 \bbox[red, 2pt]{\infty}$$

解答： $$\int_1^3 \int_0^{2x} \ln x\,dydx =\int_1^3   2x\ln x\, dx = \left.\left[ x^2\ln x-{1\over 2}x^2 \right]\right|_1^3 =9\ln 3-{9\over 2}+{1\over 2} =\bbox[red, 2pt]{9\ln 3-4}$$

解答： $$z=f(x,y)= \ln(xy)^{1/2} ={1\over 2}\ln(xy) \Rightarrow \cases{f_x= {y\over 2} \cdot{1\over xy} ={1\over 2x} \\ f_y= {x\over 2}\cdot {1\over xy} = {1 \over 2y}} \\ (5,10) \to (5.03,9.96) \Rightarrow \cases{dx=0.03 \\ dy =-0.04} \Rightarrow dz = f_x(5,10)dx +f_y(5,10)dy \\= {1\over 10}\cdot 0.03 +{1\over 20}\cdot (-0.04) = \bbox[red, 2pt]{0.001}$$

解答： $${dy\over dt} ={k\over v}(10-y) \Rightarrow \int {1\over 10-y}\,dy = \int {k\over v}\,dt \Rightarrow -\ln(10-y)= {k\over v}t+C \Rightarrow 10-y = e^{-(kt/v+C)} \\ \Rightarrow y=10-e^{-(kt/v+C)} ＝10-C^*e^{-(kt/v)}\\再將初始值y(0)= y_0代入\Rightarrow y_0= 10-C^* \Rightarrow C^* =10-y_0 \Rightarrow y=10-(10-y_0)e^{-(kt/v)} \\ \Rightarrow \lim_{t\to \infty} y = \bbox[red, 2pt]{10}$$

解答： $$利用\text{ Lagrange's multiplier }求極值：\cases{f(x,y,z)= xy+2yz+2xz\\ g(x,y,z)= xyz-108} \Rightarrow \cases{f_x =\lambda g_x\\ f_y =\lambda g_y\\ f_z= \lambda g_z\\ g=0} \\ \Rightarrow \cases{y+2z = \lambda yz \cdots(1) \\ x+2z = \lambda xz \cdots(2)\\ 2y+2x = \lambda xy \cdots(3) \\ xyz=108 \cdots(4)} \Rightarrow \cases{(1)\div (2) \Rightarrow {y+2z\over x+2z}={y\over x} \\ (2)\div (3) \Rightarrow {x+2z\over 2(x+y)}={z\over y}} \Rightarrow \cases{x=y\\ y=2z} 代入（4）\Rightarrow 4z^3=108\\ \Rightarrow z=3 \Rightarrow x=y= 6 \Rightarrow f(6,6,3)= 36+ 36+ 36= \bbox[red, 2pt]{108}$$

乙、計算、證明題：共3題，每題12分，共36分

解答： $$本題少了飛機高度的數據，無法計算！$$

解答： $$\mathbf{a.}\;積分檢定法\text{ integral test: }\int_1^\infty {e^{2/x}\over x^2}\,dx = \left.\left[ -{1\over 2} e^{2/x}\right]\right|_1^\infty =-{1\over 2}+{1\over 2}e^2 \lt \infty \Rightarrow 級數\bbox[red, 2pt]{收斂} \\\mathbf{b.}\; \lim_{n\to \infty}{n\over \sqrt{3n^2+5}} = \lim_{n\to \infty}{n\over n\sqrt{3+5/n^2}} = {1\over \sqrt 3} \ne 0 \Rightarrow 級數\bbox[red, 2pt]{發散}$$

解答： $$f為機率密度函數\Rightarrow \iint f(x,y)\,dydx=1 \Rightarrow \int_0^\infty \int_0^\infty ke^{-(x+y)/a}\,dydx= \int_0^\infty \left.\left[ -ake^{-(x+y)/a} \right] \right|_0^\infty \,dx \\=\int_0^\infty  ake^{-x/a}\,dx =\left.\left[ -a^2ke^{-x/a} \right]\right|_0^\infty =a^2k = 1 \Rightarrow \bbox[red, 2pt]{a^2k =1}$$

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