Lecturer: . . . . . . . . . . . . . . . Date: . . . . . . . . . . . . . . .
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Approved by: . . . . . . . . . . . . . Date . . . . . . . . . . . . .
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Semester/Academic year
222 22-23
Date
14/05/2023
UNIVERSITY OF
Course title Calculus 1 (Advanced Program)
TECHNOLOGY
Course ID
006401
Faculty of AS
Duration
100 mins Question sheet code 2345
Notes: - This is a closed book exam. Only your calculator is allowed. Total available score: 10.
- Do not round between steps. Round your final answers to 4 decimal places.
- There are 16 multiple-choice questions and 2 essay questions.
Final Exam
Part 1. Multiple Choice Questions (8 points).
Z 5 2
x dx
mdx
√
√
=
Question 1. [L.O.2] Find the approximation of m which satisfies
x−1
81 − x2
1
0
A. 18.4858.
B. 17.4858.
C. 14.4858.
D. 15.4858.
E. 16.4858.
Z 9
Question 2. [L.O.2] A steady wind blows a kite due west. The kite’s height above the ground from
3
√
3.92 + 3x . Find the real number
horizontal position x = 0 to x = a > 0 meters is given by y =
a such that the distance travelled by the kite is 21.04 meters.
A. 4.8195.
B. 2.8195.
C. 1.8195.
D. 3.8195.
E. 5.8195.
Question 3. [L.O.2] Find the volume of the solid obtained by rotating the region bounded by the
curves x = 0, y = −9, y = x2 − 6x about the y−axis.
A. 27/4.
B. π.27/2.
C. 27/2.
D. π.27/4.
E. 27/8.
Question 4. [L.O.2] Evaluate the approximation of the displacement of one particle with velocity
v(t) = 7 arcsin(t) (m/s) from t = 0s to t = 0.9s.
A. 2.1058.
B. 5.1058.
C. 3.1058.
D. 4.1058.
E. 1.1058.
Question 5. [L.O.1] Find the antiderivative F (x) of the function f (x) = 7ex + 8x which satisfies
the condition F (0) = 15.
A. −7ex + 4x2 − 8. B. 7ex + 4x2 − 8. C. −7ex + x2 + 8. D. 7ex + 4x2 + 8. E. ex + 4x2 + 8.
Question 6. [L.O.1] The accumulated, or total, future value after T years of an income stream of
R(t) thousand dollars per year, earning interest at the rate of r per year compounded continuously,
Z T
rT
is given by A = e
R(t)e−rt dt. A company recently bought an automatic car-washing machine
0
that is expected to generate 3 thousand dollars in revenue per year, t years from now, for the next
9 years. If the income is reinvested in a business earning interest at the rate of r = 15% per year
compounded continuously, find the total accumulated value of this income stream at the end of 9
years.
A. 60.1485.
B. 59.1485.
C. 56.1485.
D. 58.1485.
E. 57.1485.
Question 7. [L.O.2] When a particle is located a distance x meter from the origin, a force of given
F (x) = ln(6 + 5x) (newton) acts on it. How much work is done in moving it from x = 3 to
x = 9.
A. 23.3176.
B. 20.3176.
C. 21.3176.
D. 19.3176.
E. 22.3176.
√
Question 8. [L.O.2] Let the region D be bounded by the curve y = 15x2 + 5, x−axis and the lines
x = 0, x = 3. Calculate the volume V of the solid, revolving the region D about x−axis?
A. 150.
B. 24.
C. 150π.
D. 195π.
E. 165π.
p
Question 9. [L.O.2] Rotating the curve x = − y 2 + 3, where 0 ⩽ y ⩽ 6 about the y−axis, evaluate the surface area of the obtained surface of revolution.
A. 179.5195.
B. 180.5195.
C. 176.5195.
D. 178.5195.
E. 181.5195.
Page 1/2 Code 2345
Question 10. [L.O.2] The interest rates charged by Madison Finance on auto loans for used cars
over a certain 12−month period in 2022 are approximated by the function
r(t) = −
1 3 1 2
t + t − 4t + 20,
18
5
(0 ⩽ t ⩽ 12)
where t is measured in months and r(t) is the annual percentage rate. What is the average rate on
auto loans extended by Madison over the 12−month period?
A. −18.40.
B. −15.40.
C. −21.40.
D. −17.40.
E. −16.40.
6
Question 11. [L.O.2] Which function is a solution of the following differential equation y ′ + y =
x
4x, where C is any constant?
A. y = 1/2.x2 + Cx6 .
B. y = 1/2.x2 + Cx−6 .
C. y = 4/9.x2 + Cx−6 .
D. y = 4/7.x2 + Cx6 .
E. y = 4/7.x2 + Cx−6 .
Question 12.Z [L.O.2] Using the midpoint rule with 5 equal interval to estimate the following
10
integral I =
f (x)dx where f (x) is given by the following table
4
x
4 4.6 5.2 5.8 6.4 7.0 7.6 8.2 8.8 9.4 10
f (x) 3.4 4.2 4.5 4.4 6.7 4.2 5.9 4.2 5.8 4.5 3.9
A. 28.8.
B. 29.8.
C. 26.8.
D. 27.8.
E. 25.8.
Question 13. [L.O.2]ZLet g(x) = f −1 (x) be the inverse function of function y = f (x) = ex/2 + 4.14.
9.21
Evaluate the integral
A. 11.321.
g(x)dx.
5.14
B. 8.3207.
C. 6.3207.
D. 9.3207.
E. 10.321.
Z x
Question 14. [L.O.2] Find the maximum and minimum values of f (x) =
(t2 − 13t − 30)dt on
−9
the interval [−9, 9].
A. fmax ≈ 529.8333; fmin ≈ −53.
B. fmax ≈ 530.8333; fmin ≈ −53.
C. fmax ≈ 531.8333; fmin ≈ −53.
D. fmax ≈ 529.8333; fmin ≈ −54.
E. fmax ≈ 530.8333; fmin ≈ −54.
Z 2
Z 2
Question 15. [L.O.1] If
f (x) dx = 8 and
−5
−5
i
9f (x) − 4g(x) dx.
A. −123.
B. −122.
C. −119.
Z 2h
20x +
g(x) dx = −4, then calculate I =
−5
D. −121.
E. −120.
1
Question 16. [L.O.2] Find the area of the region bounded by the graphs of y = √
; y = 0;
9 9 − x2
x = 0; and x = 3.
π
π
π
π
π
A. .
B. .
C. .
D. .
E. .
2
27
18
36
9
Part 2. Essay Questions (2 points).
Question 17. [L.O.2] Solve the differential equation y ′′ − 10y ′ + 21y = 77e14x
Question 18. [L.O.2] A tank initially contains 84 grams of salt dissolved in 136 liters of water. Pure
water flows into the tank at the rate of 14 liters per minute, and the well stirred mixture flows out
of the tank at the same rate. How much salt does the tank contain after 5 minutes?
- - - - - - - - - - END - - - - - - - - - Page 2/2 Code 2345
THE ANSWER
QUESTIONS SHEET
Question sheet 2345
1. B
11. B
2. C
12. E
3. B
13. B
4. C
14. E
5. D
15. B
6. E
16. C
1
7. C
8. C
9. D
10. A
THE DETAIL ANSWER FOR QUESTION SHEET 2345
Z 5 2
h
x dx
dx
x i9
π
√
√
= arcsin
Question 1. Let A =
= · Let B =
· Using substitu2
9 0
2
x − 1√
81 − x
1
0
Z 4
√
2
2(t2 + 1)2 dt =
tion method let t = x − 1 ⇒ t = x − 1 ⇒ 2tdt = dx, we have B =
0
√
√
ï 5
ò √4
√
2t
4t3
2( 4)5 4( 4)3
+
+ 2t
=
+
+ 2 4. Therefore m ≈ 17.4858
5
3
5
3
0
The correct answer is B
Z a»
Question 2. The arc length is L =
1 + [y ′ (x)]2 dx = 21.04 ⇒ a ≈ 1.8195.
Z 9
0
The correct answer is C
Question 3. We have
Z 3
Z 3
2
|x(−9)| − |x(x − 6x)| dx = 2π x(9 − 6x + x2 )dx = 27/2.π
VOy = 2π
0
0
The correct answer is B
Question 4. The displacement of one particle is
Z 0.9
Z 0.9
D=
v(t)dt =
7 arcsin(t)dt.
0
0


dt
du = √
u = arcsin t
Let
⇒
1 − t2
dv = dt

v = t
Therefore,
Z
Z 0.9
0.9
0.9
tdt
7 0.9
√
(1 − t2 )−1/2 d(1 − t2 ) =
D = 7t arcsin t
= 7t arcsin t
−7
+
2
2 0
0
0
1−t
0
ß
0.9
0.9
√
+ 7 1 − t2
≈ 3.1058.
0
0
= 7t arcsin t
The correct answer is C
Z
Question 5. We have F (x) =
f (x) dx = 7ex + 4x2 + C. Moreover,
F (0) = 7e0 + 4 × 02 + C = 15 ⇒ C = 8
The correct answer is D
Question 6. We have r = 15%, R(t) = 3, T = 9. Therefore, the total accumulated value of this
Z T
rT
income stream at the end of 9 years is A = e
R(t)e−rt dt ≈ 57.1485.
0
The correct answer is E
Z 9
Question 7. We have W =
Z 9
ln(6 + 5x)dx ≈ 21.3176.
F (x)dx =
3
3
The correct answer is C
2
Question 8. We have
Z 3 Ä√
Z 3
ä2
3
V =π
15x2 + 5 dx = π
15x2 + 5 dx = π 5x3 + 5x 0 = 150π.
0
0
The correct answer is C
Z 6
Question 9. The surface area is A = 2π
Z 6 p
»
|x| 1 + [x′ (y)]2 dy = 2π |− y 2 + 3| 1 +
0
178.5195.
The correct answer is D
0
y2
dy ≈
y2 + 3
Question 10. The average rate on auto loans extended by Madison over the 12−month period is
Z 12
Z Å
ã
1 12
1 3 1 2
1
r(t)dt =
− t + t − 4t + 20 dt = −18.40.
rave =
12 − 0 0
12 0
18
5
The correct answer is A
Question 11. This is the first order linear differential equation with p(x) =
Therefore
Z
−
y=e
Z
−
y=e
6
dx hZ
x
4xe
Z
6
and q(x) = 4x.
x
Z
p(x)dx hZ
p(x)dx
q(x)e
dx + C
i
6
i
h 8
i
dx
x dx + C = x−6 4 x + C = 1/2.x2 + Cx−6
8
The correct answer is B
10 − 4
= 1.2. Therefore, the subintervals
5
consist of [4, 5.2]; [5.2, 6.4]; [6.4, 7.6]; [7.6, 8.8]; [8.8, 10.0]. The midpoints of these subintervals are
4.6; 5.8; 7.0; 8.2; 9.4. Thus, the approximation of the integral using midpoint rule is
h
i
I ≈ ∆x × 4.2 + 4.4 + 4.2 + 4.2 + 4.5 = 25.8
Question 12. Each subinterval has length of ∆x =
The correct answer is E
x
Question 13. Since y = f (x) = e 2 + 4.14 ⇒ x = 2 ln(y − 4.14). So g(x) = f −1 (x) = 2 ln(x − 4.14).
Therefore,
Z
9.21
g(x)dx ≈ 8.3207.
5.14
The correct answer is B
Question 14. We have
Z x
ï
x = −2
2
′
2
f (x) =
(t − 13t − 30)dt ⇒ f (x) = x − 13x − 30 = 0 ⇔
x = 15
−9
Z −9
f (−9) =
(t2 − 13t − 30)dt = 0;
−9
Z −2
f (−2) =
(t2 − 13t − 30)dt ≈ 530.8333;
−9
Z 9
f (9) =
(t2 − 13t − 30)dt ≈ −54.
−9
The correct answer is E
3
Question 15. We have
Z 2
Z 2
Z 2
Z 2
[20x + 9f (x) − 4g(x)] dx = 20
x dx + 9
f (x) dx − 4
g(x) dx =
−5
−5
−5
−5
2
= 20
x2
+ 9 × 8 − 4 × (−4) = −122.
2 −5
The correct answer is B
Question 16. The required area is
Z 3
Å
ã
dx
x i3 1
π
1h
√
A=
arcsin
lim− arcsin b − arcsin 0 =
=
=
2
9
3 0 9 b→3
18
0 9 9−x
The correct answer is C
Question 17.
1. Step 1. Solve the homogeneous equation
y ′′ − 10y ′ + 21y = 0.
The characteristic equation k 2 − 10k + 21 = 0 has 2 real different roots
k1 = 7, k2 = 3
2. Step 2. The homogeneous solution is
yh = C1 e7x + C2 e3x
3. Step 3. Find a particular solution of nonhomogeneous equation
y ′′ − 10y ′ + 21y = 77e14x .
The particular solution has a form yp = xs .e14x .A. Since α = 14 is not the root of characteristic
equation then s = 0 and yp = A.e14x .
21×
−10×
1×
′′
yp − 10yp′ + 21yp
yp
=
Ae14x
yp′
=
14Ae14x
′′
yp
=
196Ae14x
= 77Ae14x = 77e14x
⇒A=1
4. Step 4. The general solution is
ygen = yh + yp = C1 e7x + C2 e3x + 1e14x .
Question 18.
1. Step 1. The rate of change of salt in the tank,
salt is flowing in minus the rate at which it is flowing out.
dQ
is equal to the rate at which
dt
dQ
= rate in − rate out
dt
2. Step 2. The rate at which salt enters the tank is the concentration 0 gr/l times the flow rate
14 l/min or 0 gr/min.
4
3. Step 3. To find the rate at which salt leaves the tank, we need to multiply the concentration
of salt in the tank by the rate of outflow, 14 l/min. The volume of water in the tank remains
constant at 136 liters, and since the mixture is well-stirred, the concentration throughout
Q
gr/l. Therefore, the rate at which salt leaves the tank is
the tank is the same, namely,
136
14Q
gr/min.
136
4. Step 4. Solve the differential equation using separable method
Z
Z
dQ
dQ
14Q
dQ
14
14
=0−
⇒
=−
dt ⇒
=
−
dt
dt
136
Q
136
Q
136
⇒ ln |Q| = −
14
t + ln C ⇒ Q = Ce−14t/136
136
5. Step 5. The initial condition is Q(0) = 84, so 84 = Ce0 = C. Thus Q = 84e−14t/136
⇒ Q(5) = 84e−14×5/136 = 50.2048.
5