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） =========================== 對 齊 這 裡 ============================= 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){(1/4) ∫ 1/(1-u) du + (1/4) ∫ 1/(1+u) du ０ ０ -1/2 + (-1/2) ∫ 1/[(1+u)^2] du} ０ = (-1){(1/4)[-ln(3/2)] + (1/4)[ln(1/2)] + (1/2)(2-1)} = (-1/4)ln(1/3) － 0.5 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 Consider Σ ────── n=2 n^p + (-1)^n 1/[n^p + (-1)^n] Using limit comparison test we get lim ──────── = 1 n→∞ 1/n^p (1) If p＞1, then the series is convergent. (2) If 0＜p≦1, then the series is divergent. Therefore the original series is absolutely convergent when p＞1. Now we try to find out whether the original series conv./div. when 0＜p≦1. Here Leibniz's Test fails since a_n≧a_(n+1) is not always true for any n. Let A = the original series. ∞ (-1)^n Consider B = Σ ──── is convergent (Leibniz's Test). n=2 n^p - 1 ∞ (-1)^n ∞ (-1)^n ∞ 1 1 B－A = Σ ──── － Σ ─────── = Σ ───── － ───── n=2 n^p - 1 n=2 n^p + (-1)^n n=1 (2n)^p - 1 (2n)^p + 1 ∞ 2 = Σ ─────── n=2 (2n)^(2p) - 1 2/[(2n)^(2p) -1] Using limit comparison, we get lim ───────── = 2 n→∞ 1/[(2n)^(2p)] ∞ 1 When 0＜p≦1/2, Σ ───── diverges n=2 (2n)^(2p) => B－A diverges => A = B － (B－A) diverges. conv. div. ∞ 1 When 1/2＜p≦1, Σ ───── converges n=2 (2n)^(2p) => B－A converges => A = B － (B－A) converges. ※ Conclusion: When p＞1, the original series is absolutely convergent. When 1/2＜p≦1, the original series is conditionally convergent. When 0＜p≦1/2, the original series is divergent. 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 (有省略一些推論) 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. ※ 編輯: liltwnboiz 來自: 114.24.179.35 (01/08 20:01) ※ 編輯: liltwnboiz 來自: 114.24.179.35 (01/08 20:02) ※ 編輯: liltwnboiz 來自: 114.24.144.40 (01/10 20:20)