Calculus II Midterm Practice problems Time: 4/30(M), 3:10-5:00; Place:
數學系 3173, 3174 (1F)Chap 4: Sec. 4.10.
1. Evaluate the given integral
i)∫−∞∞ (1+ee−x−x)2dx ∫−∞∞ x3dx iii) limR→∞∫−RR x3dx Ans: (i) 1 (ii) DIV (iii) 0
2. Determine whether the integral converges or diverges:
i)∫01x−1/3dx ii)∫01x−4/3dx
iii)∫1∞x−1/3dx iv)∫−11 x−1/3dx Ans: (i) 3/2
(ii) DIV (iii) DIV (iv) 0 Chap 5: Sec. 5.3-Sec. 5.4.
1. Set up a definite integral for the arc length of an ellipse x2+ 4y2 = 4. Ans: 4∫02
√
1 + (− x
4√
1−x2/4)2dx or 4∫0π/2√(−2 sin θ)2+ (cos θ)2dθ
2. Set up the integral for the surface area of the surface of revolution. y = ex, 0 ≤ x ≤ 1, revolved about x-axis. Ans: S =∫012πex√1 + (ex)2dx
3. An object is released from a height of 10m with an upward velocity of 5m/s. Let y(t) be the height of the object. Identify the initial conditions y(0) and y0(0). Find y(t). Ans: y(0) = 10;y0(0) = 5;y(t) =
−12gt2+ 5t + 10;
Chap 6: Sec. 6.1-Sec. 6.3.
1. Two years ago, there were 4 grams of a radioactive substance . Now there are 3 grams. How much was there 10 years ago?
Ans: 4354
2. Find the size of permanent endowment needed to generate an annual $2,000 forever at 10% (annual) interest compounded continuously. Ans: 2000· e−0.1
3. Select the differential equation which corresponds to the direction field below.
(a) y0 =−2y1 (b) y0 =−2yx,
(c) y0 =−2xy (d) y0 =−y12
4. Select the differential equation which corresponds to the direction field below.
(a) y0 =−xy (b) y0 = y + x2,
(c) y0 = y2+ x3 (d) y0 =−x − y2
5. Solve the IVP:
i) y0 = xy−12 , y(0) = 2; ii) y0 = x−1y , y(0) =−2.
Ans: (i) y = (32x2− 3x + 8)1/3;(ii) y(t) =−√
x2− 2x + 4 6. Solve the initial value problem
i) y0(t) = y(t), y(0) = 0; ii) y0(t) = (y(t)− 4), y(0) = 4.
Ans: (i) y(t0 = 0;(ii) y(t) = 4 Chap 7. Sec. 7.1-Sec. 7.8.
1. Determine the convergence of a sequence. If it converges, find the limit of the sequence.
A) an= 3n4n2+n2−n−2 B) an= cos nn2
C) an= 1 D) an= (−1)n
Ans: A) 34; B) 0; C) 1; D) Div
2. Determine if the series converges or diverges. If it converges, find the sum of the series.
i)∑∞k=0(13)k ii) ∑∞k=0 kk+12−4
iii)∑∞k=2 k(k1−1)
Hint/Ans: i) Geometric Series; ii) DIV; Limit comparison Test (Compared with Harmonic Series); iii) Telescope Series; example 2.3.
3. Determine if the series converges or diverges.
i)∑∞k=0 √33
k ii) ∑∞k=2 k(k1−1)
iii)∑∞k=0((−1)k− (13)k) iv)∑∞k=0 kk+13−4
v)∑∞k=2cos k+1k2
Hint/Ans:i) Div (p-series) ;ii) Conv (Telescope Series); iii)Div (Thm 2.3); iv) Conv (Limit Comparison Test, p-series) ;v)Conv (Comparison Test)
4. Use the Integral Test to test the convergence of the series ∑∞k=2k(ln k)1 3
Hint: Substitution s = ln x 5. If ak=
{ 1/k if k is odd
1/k2 if k is even , determine the convergence of the series,∑∞k=1(−1)kak. Hint: Div (Thm 2.3)
6. Determine if the series is absolutely convergent, conditionally convergent or divergent.
i)∑∞k=1(−1)kk3k+1 ii) ∑∞k=1(−1)kk+1√k iii)∑∞k=1(−1)k 3(2k)!k iv)∑∞k=1(−1)k(2k+1)1
Hint/Ans: i)Abs Conv (Limit Comparison Test, p-series);ii) Conditionally Conv (Limit Comparison Test, p-series, Alternating Series Test);
iii) Abs Conv (Ratio Test);iv) Conditionally Conv (Limit Comparison Test, p-series, Alternating Series Test)
7. Determine the radius and interval of convergence of the power series.
i)∑∞k=12kk(x− 2)k ii) ∑∞k=1(x− 3)k iii)∑∞k=1k!(x− 2)k iv)∑∞k=1 2k!k(x− 3)k Ans: i) (0, 4), r = 2;ii) (2, 4), r = 1;iii){2},r = 0;iv) (−∞, ∞),r = ∞
8. For f (x) = ex, find the Taylor polynomial of degree 2 expanded about c = 1.
Ans: P2(x) = e +1e(x− 1) + e22(x− 1)2
9. Find the Taylor series of ex, e−x2 and x2e−x2 about c = 0. Determine the corresponding radius and interval of convergence.
Ans: ex=∑∞k=0 k!1xk, for x∈ (−∞, ∞), r = ∞;
e−x2 =∑∞k=0(−1)k 1k!x2k, for x∈ (−∞, ∞), r = ∞;
x2e−x2 =∑∞k=0(−1)k 1k!x2k+2, for x∈ (−∞, ∞), r = ∞
10. Given that 1+x1 =∑∞k=0(−1)kxk, for − 1 < x < 1, find the Taylor series of
i) ln (1 + x) ii) 1+x1 2.
iii) x ln (1 + x) iv) 1+xx22.
Determine the corresponding radius and interval of convergence.
Ans: i)ln (1 + x) =∑∞k=0(−1)k+1kxk+1, for − 1 < x ≤ 1, r = 1;
ii)1+x1 2 =∑∞k=0(−1)kx2k, for − 1 < x < 1,, r = 1;
iii)x ln (1 + x) =∑∞k=0 (−1)k+1kxk+2, for − 1 < x ≤ 1, r = 1 iv)1+xx22 =∑∞k=0(−1)kx2k+2, for − 1 < x < 1,, r = 1;
11. ∑∞k=0(−1)kk+11 =?
Ans: ln 2