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課程名稱︰微積分乙上
課程性質︰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）︰2010／1／12
考試時限（分鐘）：140min(?)
是否需發放獎勵金：yes please
（如未明確表示，則不予發放）
1. Find ∫sin(ln x) dx (10%)
sol> Let ln(x) = u, (1/x) dx = du => dx = e^u du
∫sin(ln x) dx = ∫sin(u)(e^u) du = ∫sin(u) d(e^u)
= (e^u)sin(u)﹣∫(e^u)cos(u) du = (e^u)sin(u)﹣(e^u)cos(u)﹣∫(e^u)sin(u) du
=> ∫sin(u)(e^u) du = (1/2)[(e^u)sin(u)﹣(e^u)cos(u)]
= (1/2)[xsin(ln x)﹣xcos(ln x)] + C
x^2 dx
2. Evaluate∫───────, x＞1 (10%)
(x^2-1)^(5/2)
sol> Let x = secu, dx = (secu)(tanu) du
x^2 dx (secu)^3 du 1
∫─────── = ∫────── = ∫──── d(sinu)
(x^2-1)^(5/2) (tanu)^4 (sinu)^4
= (-1/3)(sinu)^(-3) + C = (-1/3)(√(x^2-1)/x)^(-3), x＞1
2π/3 cosθ dθ
3. Find ∫ ───────── (10%)
π/2 sinθcosθ+sinθ
2π/3 sinθcosθ dθ
sol> 原式上下乘 sinθ = ∫ ──────────────
π/2 cosθ(sinθ)^2 + (sinθ)^2
2π/3 cosθ d(﹣cosθ) 2π/3 cosθd(-cosθ)
= ∫ ─────────── = ∫ ───────────
π/2 (cosθ+1)[1-(cosθ)^2] π/2 [(1+cosθ)^2](1-cosθ)
cosθ = A(1+cosθ)^2 + B[1-(cosθ)^2] + C(1-cosθ)
=> A = 1/4, B = 1/4, C = ﹣1/2 (Let u = cosθ)
-1/2 -1/2
原式 = (1/4) ∫ 1/(1-u) du + (1/4) ∫ 1/(1+u) du
０ ０
-1/2
+ (-1/2) ∫ 1/[(1+u)^2] du (Let u = tanψ, du = (secψ)^2 dψ)
０
tan^(-1)(-1/2)
= (1/4)[-ln(3/2)] + (1/4)[ln(1/2)] + (-1/2) ∫ dψ
tan^(-1)(0)
= (1/4)ln(1/3) + (-1/2)[tan^(-1)(-1/2)]
4. Test for conditional and absolute convergence of
∞ (-1)^n
Σ ───────, p＞0 (20%)
n=2 n^p + (-1)^n
(-1)^n
sol> First, lim ────── = 0
n→∞ n^p + (-1)^n
[(-1)^n]/[n^p]
Using limit comparison test, we get lim ─────────── = 1
n→∞ [(-1)^n]/[n^p +(-1)^n]
∞ (-1)^n
Since Σ ──── is an alternating series and satisfies the Leibniz's
n^p
test ( 1/n^p＞0 for any n≧2, 1/n^p≧1/(n+1)^p for any n≧2, 1/n^p→0),
it is convergent for any p＞0
∞ 1
Now consider Σ ───
n=2 n^p
(1) If p＞1, then the series is convergent.
(2) If 0＜p≦1, then the series is divergent.
Therefore we know that the original series is absolutely convergent when
p＞1 and is conditionally convergent when 0＜p≦1.
5. Test for the convergence fo the following series. (20%)
∞ 1×3×...×(2n-1)
(a) Σ ─────────────
n=1 [2×4×...×(2n)](3^n+1)
1×3×...×(2n-1)
sol> First lim ──────────── = 0
n→∞ [2×4×...×(2n)](3^n+1)
[a_(n+1)] (2n+1)(3^(n) + 1)
Using ratio test, we get lim ───── = lim ──────────
n→∞ [a_n] n→∞ (2n+2)(3^(n+1) + 1)
= 1/3＜1. Therefore the series is convergent.
∞ n(ln n)
(b) Σ ─────
n=1 1+n-n^2
n(ln n)
sol> First, lim ───── = 0
n→∞ 1+n-n^2
[n(ln n)]/(1+n-n^2)
Using limit comparison test, we get lim ────────── = 1
n→∞ (ln n)/n
∞ ∞ ∞
∫(ln x)/x dx = ∫ (ln x) d(ln x) = (1/2)(ln x)^2│ diverges
１ １ １
Therefore the original series diverges (有省略一個推論)
6. A 200-gal tank is half full of distilled water. At time t = 0, a solution
containing 0.5 lb/gal of concentrate enters the tank at the rate of
5 gal/min, and the well-stirred mixture is withdrawn at the rate of
3 gal/min. (10%)
(a) At what time will the tank be full?
sol> enter rate － exit rate = 5－3 = 2 gal/min
100/2 = 50 min for the tank to be full.
(b) At the time the tank is full, how many pounds of concentrate will
it contain?
sol> Let y(t) be the amount of concentrate in the tank at time t (lb)
The volume of the solution in the tank at time t is 100 + 2t (gal)
So the y(t) satisfies the differential equation
dy y(t)
── = (5)(0.5)－ ────(3)
dt 100 + 2t
=> dy/dt + [3/(100+2t)]y = 2.5
The integrating factor I(t) = e^[∫3/(100+2t) dt] = (100+2t)^(3/2)
=> [(100+2t)^(3/2)](dy/dt) + 3y[(100+2t)^(1/2)] = d[y(100+2t)^(3/2)]
= 2.5(100+2t)^(3/2)
=> y = (100+2t)^(-3/2) ×∫2.5(100+2t)^(3/2) dt
= (100+2t)^(-3/2) ×[(1/2)(100+2t)^(5/2) + C]
= (1/2)(100+2t) + C(100+2t)^(-3/2)
When t = 0, y = 0 = 50 + C/1000 => C = -5×10^4
When t = 50 , y = 100 + (-5×10^4)(200)^(-3/2) ≒ 82.32 lb
7. Test the convergence of the improper integral (20%)
∞ cos x
∫ ──── dx
π/2 x
(3/2+n)π
∞ cos x ∞ ∫ cos x
Hint: Consider ∫ ──── dx = Σ ∫ ──── dx
π/2 x n=0 ∫ x
(1/2+n)π
and identify the series you get.
∞ cos x ∞ d(sin x) ∞ ∞ sinx dx
sol> ∫ ──── dx = ∫ ──── = (sinx/x)│ － ∫ ────
π/2 x π/2 x π/2 π/2 ﹣x^2
(converges) (converges)
(這邊稍微偷懶一下...) Therefore the improper integral converges.
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※ liltwnboiz:轉錄至某隱形看板 01/07 21:42