Re: [試題] 97上微積分乙王振男期末考詳解

看板NTU-Exam作者liltwnboiz (TCL)時間13年前 (2011/01/10 22:09)推噓3(3推 0噓 5→)

留言8則, 4人參與討論串1/1

※ 引述《ayuiop ([限]2008最後醫夜)》之銘言：課程名稱︰微積分乙課程性質︰系必帶課程教師︰王振男開課學院：醫學院開課系所︰醫學系考試日期（年月日）︰12/13 考試時限（分鐘）：140min 1.判斷以下命題，如為真須證明、否則需舉反例 (a) ΣAn converges absolutely => ΣAn^2 converges (b) ΣAn^2 converges => ΣAn converges absolutely (c) ΣAn converges => ΣAn^2 converges sol> (a) ΣA_n coverges absolutely => Σ│A_n│ converges and │A_n│→0 Therefore there exist a M＞0 such that │A_n│＜1 for any n≧M => 0≦(A_n)^2≦│A_n│＜1 for any n≧M Now let the first interger larger than or equal to M be N ∞ N-1 ∞ N-1 ∞ Σ (A_n)^2 = Σ (A_n)^2 + Σ ≦ Σ (A_n)^2 + Σ │A_n│ converges n=1 n=1 n=N n=1 n=N => TRUE (b) Let (A_n)^2 = 1/(n^2) Then Σ(A_n)^2 converges absolutely but Σ(A_n) = Σ(1/n) diverges => FALSE (c) Let A_n = (-1)^n/√(n) Then ΣA_n converges by the Leibniz's Test But Σ(A_n)^2 = Σ(1/n) diverges => FALSE 2.判斷以下級數是否收斂，需註明以何種方式檢驗 (a)Σ(-1)^(n+1)*sin(π/n) (b)Σ(-1)^(n+1)*nsin(π/n) (c)Σ1/(㏑n)^p , for all p > 1 sol> (a) First, lim(n→∞) (a_n) = 0 Second, for any n≧2, sin(π/n)≧sin[π/(n+1)] Third, for any positive n, sin(π/n)＞0 Therefore the series passes the Leibniz's Test (b) lim(n→∞) nsin(π/n) = lim(n→∞) π[sin(π/n)]/(π/n) = π The series diverges 3.決定此瑕積分是否收斂，如收斂需計算其值 1 ∫ (㏑(1/x))^2 dx 0 (hint: set t = ㏑(1/x)) sol> Let t = ln(1/x), dt = x/(-x^2) = -1/x dx => dx = -x dt = -e^(-t) dt ∫(㏑(1/x))^2 dx = ∫[-(t)^2][e^(-t)] dt = ∫t^2 d[e^(-t)] = (t^2)e^(-t)－∫(2t)e^(-t) dt = (t^2)e^(-t) + ∫2t d[e^(-t)]} = (t^2)e^(-t) + (2t)e^(-t)－∫2 e^(-t) dt = (t^2)e^(-t) + (2t)e^(-t) + 2 e^(-t) + C = x[(lnx)^2 － 2lnx + 2] + C Let the above result be F(x) 所求 = F(1) - F(0) = F(1) - lim(b→0)F(b) = 2 - lim(b→0)[b(lnb)^2 － 2blnb] (省略羅畢達) = 2 4. x^3 ∫──── dx √(1-x^2) sol> Let x = sinu , dx = cosu du x^3 [(sinu)^3](cosu) du ∫──── dx = ∫────────── = ∫1-(cosu)^2 d(-cosu) √(1-x^2) cosu = (1/3)(cosu)^3 － cosu + C = (1/3)(1-x^2)^(3/2) － (1-x^2)^(1/2) + C 5. ∫x㏑(x+2) dx sol> Let u = x+2 , du = dx ∫x㏑(x+2) dx = ∫(u-2)ln(u) du = (1/2)∫ln(u) d[(u-2)^2] = (1/2)ln(u)(u-2)^2 － (1/2)∫(u-2)^2/u du = (1/2)ln(u)(u-2)^2 － (1/2)∫[u-4+(4/u)] du = (1/2)ln(u)(u-2)^2 － (1/2)[(1/2)u^2 - 4u + 4ln(u)] + C 代回 u = x+2 後即為答案 6. A 200-gal tank is half full of distilled water. At time t=0, a solution containing 0.5 pound/gal of concentrate enters the tank at the rate of 5gal/min, and the well-strred mixture is withdrawn at the rate of 3gal/min a. At what time will the tank be full? b. At the time the tank is full, how many pounds of concentrate will it contain? sol> 請見98上期末考 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 221.169.194.146

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iodmepobole

01/17 01:34,

01/17 01:34

-- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.24.144.40 ※ liltwnboiz:轉錄至某隱形看板 01/10 22:10

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ALegmontnick

01/10 23:06, , 1^F

01/10 23:06, 1^F

※ 編輯: liltwnboiz 來自: 114.24.144.40 (01/11 00:52)

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PaulErdos