作者liltwnboiz (TCL)
看板NTU-Exam
標題Re: [試題] 97上 微積分乙 王振男 期末考
時間Mon Jan 10 22:09:32 2011
※ 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5gal/min, and the well-strred mixture is withdrawn at the rate of 3gal/min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
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1F：推 iodmepobole :ayuoip真強者竟然把題目記下來了 01/17 01:34
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※ liltwnboiz:轉錄至某隱形看板 01/10 22:10
2F：推 ALegmontnick:done 01/10 23:06
※ 編輯: liltwnboiz 來自: 114.24.144.40 (01/11 00:52)
3F：推 PaulErdos :2. (c) 跟Σ 1/n 作comparison 01/11 00:57
4F：推 kennyli :2.(c)用integral test ，let x=e^t去做 01/11 02:41
5F：→ kennyli :把積分範圍從t=0~無限，拆成t=0~1和t=1~無限 01/11 02:42
6F：→ kennyli :0~1的那個，和∫t^(-p)dt做comparison，發現div 01/11 02:43
7F：→ kennyli :t=1~無限的那個，必大於0，加起來還是發散 01/11 02:44
8F：→ kennyli :所以積分後發散，原級數就發散 01/11 02:45
9F：→ liltwnboiz :喔 有強者 Orz 01/11 23:01