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Calculus II Midterm Practice problems Time: 4/30(M), 3:10-5:00; Place:

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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)1x−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: 402

1 + ( x

4

1−x2/4)2dx or 40π/2(−2 sin θ)2+ (cos θ)2

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 =012πex1 + (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

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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+1k 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 = ∞

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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

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