Đề thi cuối kỳ môn Calculus 1 | Trường Đại học Quốc tế, Đại học Quốc gia Thành phố Hồ Chí Minh

Calculus 1
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CALCULUS 1 (MA001IU) – FINAL EXAMINATION
Semester 3, 2022-23 • Duration: 120 minutes • Date: August 7, 2023 SUBJECT: CALCULUS 1 Department of Mathematics Lecturer Nguyen Minh Quan
Nguyen Thi Thu Van, Nguyen Minh Quan INSTRUCTIONS:
• Use of calculator is allowed. Each student is allowed two double-sided sheets of notes (size A4
or similar). All other documents and electronic devices are forbidden.
• Write the steps you use to arrive at the answers to each question. No marks will be given for the answer alone.
• There are a total of 10 (ten) questions. Each one carries 10 points.
1. Evaluate the following limit √ 3 x − 2 lim . x→8 ln(x − 7) Ans We have √ √ 3 x − 2 ( 3 x − 2)′ 1 x−2/3 1 lim = lim = lim 3 = x→8 ln(x − 7) x→8 [ln(x − 7)]′ x→8 1 12 x−7 1
2. Find the point on the hyperbola y =
in the first quadrant that is closest to the point (0, 0). 2x 1
Ans 1. Let d be the distance between the points (0, 0) and (x, ). We minimize f (x) := d2 = 2x 1 1 1 x2 +
, for x > 0. One get f ′(x) = 2x −
. It implies that f (x) attains its minimum at x = √ . 4x2 2x3 2
3. A particle moves in a straight line and its velocity is given by v(t) = 3t + 2 and its initial position
is s(0) = 0. Find its position function when t = 2, i.e., find s(2).
Ans. We have s′(t) = v(t). Therefore Z 2 Z 2 Z 2 s′(t) dt = v(t) dt = (3t + 2) dt 0 0 0 =⇒ s(2) − s(0) = 10 =⇒ s(2) = 12. 4. Let Z π x2 √ F(x) = t + sint dt. 1 Find F′(1). Ans. We have q q F′(x) =
π x2 + sin(π x2) · (π x2)′x = 2πx π x2 + sin(π x2). √ Hence, F′(1) = 2π π . Calculus 1
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5. Evaluate the integral R x2e−xdx. 0 1 5
Ans. Use the integration by parts twice to obtain: R x2e−xdx = 2 − . 0 e 6. Evaluate the integral Z 1 x − 1 dx. 0 x2 + 4x + 3
Ans. We find A and B satisfying x − 1 A B = + x2 + 4x + 3 x + 1 x + 3 =⇒
x − 1 = A(x + 3) + B(x + 1) = (A + B)x + (3A + B) ( ( A + B = 1 A = −1 =⇒ =⇒ . 3A + B = −1 B = 2 Hence, Z 1 x − 1 Z 1 −1 2 8 dx = +
dx = (− ln |x + 1| + 2 ln |x + 3|) |10 = 3 ln 2 − 2 ln 3 = ln . 0 x2 + 4x + 3 0 x + 1 x + 3 9 Z ∞ 2x
7. Determine whether the improper integral √
dx is convergent or divergent. Explain. 1 1 + x2 Ans. We have Z ∞ 2x Z t 2x √ p p √ dx = lim √ dx = lim 2
1 + x2 |t1 = lim 2 1 + t2 − 2 2 = ∞. 1 1 + x2 t→∞ 1 1 + x2 t→∞ t→∞
8. Use the Trapezoidal rule with 6 sub equal intervals (i.e., n = 6) to approximate the value of the Z 2 1 integral dx 0 16 + x2 1 Ans. By putting f (x) = , we have 16 + x2 Z 2 1
dx ≈ (1/2)(1/3) [ f (0) + 2 f (1/3) + 2 f (2/3) + 2 f (1) + 2 f (4/3) + 2 f (5/3) + f (2)] = 0.116 0 16 + x2
9. Find the area of the region enclosed by the curves y = 6x − x2 and y = x. Z 5 125
Ans. The intersections are at x = 0 and x = 5. The area is A = (6x − x2 − x) dx = . 0 6
10. Use Newton’s method to approximate the positive root correct to six decimal places of the equation 2 − x2 = sin x.
Ans. Observe that 2 − x2 = sin x ⇔ f (x) ≡ 2 − x2 − sin x = 0.
Using Newton’s method, we have f (xn) 2 − x2 − sin xn x n n+1 = xn − = xn − f ′(xn) −2xn − cos xn
By choosing, for example, x1 = 1, we can deduce from the latter equation that x2 = 1.062405 x3 = 1.061549 x4 = 1.061549
Since x3 and x4 agree to six decimal places, the approximate positive root needed to find is 1.061549. Calculus 1
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—END OF THE QUESTION PAPER. GOOD LUCK!— Calculus 1
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CALCULUS 1 (MA001IU) – FINAL EXAMINATION
Semester 2, 2022-23 • Duration: 120 minutes • Date: August 7, 2023 SUBJECT: CALCULUS 1 Department of Mathematics Lecturer Nguyen Minh Quan
Nguyen Thi Thu Van, Pham Thanh Duong INSTRUCTIONS:
• Use of calculator is allowed. Each student is allowed two double-sided sheets of notes (size A4
or similar). All other documents and electronic devices are forbidden.
• Write the steps you use to arrive at the answers to each question. No marks will be given for the answer alone.
• There are a total of 10 (ten) questions. Each one carries 10 points. 1. Let f (x) = xsinx, x > 0. Find f ′(x). Ans 1. We have ln f (x) = sin x ln x f ′(x) sin x =⇒ = cos x ln x + f (x) x sin x =⇒ f ′(x) = cos x ln x + xsinx x
2. Evaluate the following limit √ 3 x − 2 lim . x→8 ln(x − 7) Ans We have √ √ 3 x − 2 ( 3 x − 2)′ 1 x−2/3 1 lim = lim = lim 3 = x→8 ln(x − 7) x→8 [ln(x − 7)]′ x→8 1 12 x−7 3. Let Z π x2 √ F(x) = t + sint dt. 1 Find F′(1). Ans. We have q q F′(x) =
π x2 + sin(π x2) · (π x2)′x = 2πx π x2 + sin(π x2). √ Hence, F′(1) = 2π π . 4. Evaluate the integral: Z 1 x − 1 dx. 0 x2 + 4x + 3 Calculus 1
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Ans. We find A and B satisfying x − 1 A B = + x2 + 4x + 3 x + 1 x + 3 =⇒
x − 1 = A(x + 3) + B(x + 1) = (A + B)x + (3A + B) ( ( A + B = 1 A = −1 =⇒ =⇒ . 3A + B = −1 B = 2 Hence, Z 1 x − 1 Z 1 −1 2 8 dx = +
dx = (− ln |x + 1| + 2 ln |x + 3|) |10 = 3 ln 2 − 2 ln 3 = ln . 0 x2 + 4x + 3 0 x + 1 x + 3 9 Z ∞ 2x
5. Determine whether the improper integral √
dx is convergent or divergent. Explain. 1 1 + x2 Ans. We have Z ∞ 2x Z t 2x √ p p √ dx = lim √ dx = lim 2
1 + x2 |t1 = lim 2 1 + t2 − 2 2 = ∞. 1 1 + x2 t→∞ 1 1 + x2 t→∞ t→∞
6. A particle moves in a straight line and its velocity is given by v(t) = 3t + 2 and its initial position
is s(0) = 0. Find its position function when t = 2, i.e., find s(2).
Ans. We have s′(t) = v(t). Therefore Z 2 Z 2 Z 2 s′(t) dt = v(t) dt = (3t + 2) dt 0 0 0 =⇒ s(2) − s(0) = 10 =⇒ s(2) = 12.
7. Use the Trapezoidal rule with 6 sub equal intervals (i.e., n = 6) to approximate the value of the Z 2 1 integral dx 0 16 + x2 1 Ans. By putting f (x) = , we have 16 + x2 Z 2 1
dx ≈ (1/2)(1/3) [ f (0) + 2 f (1/3) + 2 f (2/3) + 2 f (1) + 2 f (4/3) + 2 f (5/3) + f (2)] = 0.116 0 16 + x2
8. Determine the length of x = 2 (y − 1)3/2 between 1 ≤ y ≤ 4. 3
Ans. We have dx = (y − 1)1/2. The length of the curve corresponding to y ∈ [1, 4] is defined by dy s Z 4 dx 2 L = 1 + dy 1 dy Z 4 √ = y dy = 14/3. 1
9. Let f (x) be continuous on [0, 3] and differentiable over (0, 3). Suppose that f (0) = −3 and that
f ′(x) ≤ 2 for all x ∈ (0, 3). What is the largest possible value for f (3)?
Ans. Because f (x) is continuous on [0, 3] and differentiable over (0, 3), by the MVT, there exists c ∈ (0, 3) such that
f (3) − f (0) = f ′(c) × (3 − 0) =⇒
f (3) = 3 f ′(c) + f (0) ≤ 3 × 2 − 3 = 3. Calculus 1
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10. Use Newton’s method to approximate the positive root correct to six decimal places of the equation 2 − x2 = sin x.
Ans. Observe that 2 − x2 = sin x ⇔ f (x) ≡ 2 − x2 − sin x = 0.
Using Newton’s method, we have f (xn) 2 − x2 − sin xn x n n+1 = xn − = xn − f ′(xn) −2xn − cos xn
By choosing, for example, x1 = 1, we can deduce from the latter equation that x2 = 1.062405 x3 = 1.061549 x4 = 1.061549
Since x3 and x4 agree to six decimal places, the approximate positive root needed to find is 1.061549.
—END OF THE QUESTION PAPER. GOOD LUCK!—