THE INTERNATIONAL UNIVERSITY (IU) - VIETNAM NATIONAL UNIVERSITY - HCMC
FINAL EXAMINATION
Semester 1, 2016-17 • 10th January 2016 • 120 minutes
Head of Dept. of Mathematics:
CALCULUS I
Lecturers:
Assoc. Prof. Nguyen Dinh
TT Duong, JC Harris, HB Minh, NM Quan, MD Thanh
INSTRUCTIONS: Each student is allowed a scientific calculator and a maximum of two
double-sided sheets of reference material (size A4 or similar), stapled together and marked
with their name and ID. All other documents and electronic devices are forbidden.
Question 1.
(15 marks) Use L’Hôpital’s rule to evaluate the limit
lim x(ln x)2
x→0+
Question 2. (15 marks) A population of animals is infected with a disease. After t days,
t
the percentage of the population infected is modelled by the function p(t) = 8te− 12 for
0 ≤ t ≤ 60. Find the maximum value of p and the time at which it occurs.
Question 3.
(10 marks) Show that eu ≥ 1 + u for all u ≥ 0.
Question 4.
Evaluate the integrals:
Z
x
√
a) (10 marks)
dx,
1 + x2
Z1
b) (10 marks)
(1 + |t3 |) dt.
−1
1
from x = 0 to x = ∞.
Question 5. (10 marks) Find the area under the curve y =
(x + 2)2
Question 6. (15 marks) A cylinder is generated by rotating a rectangle about one of its
edges. The perimeter of the rectangle is 30 cm. To obtain a cylinder of maximum volume,
what dimensions should the rectangle have?
Question 7. (15 marks) A torus (a donut-shaped solid) is generated by rotating a circle
of radius 2 cm about an axis 5 cm from the center of the circle, as shown in theR figures below.
b
Find the volume of the torus. [Hint: Use the method of cylindrical shells, V = a 2πxf (x)dx.]
—– END —–
SOLUTIONS
Question 1.
Using L’Hôpital’s rule for the indeterminate form
lim+ xln2 x = lim+
x→0
x→0
∞
∞
twice, we have
2 ln x
1/x
ln2 x
ln x
= lim+ x 2 = −2 lim+
= −2 lim+
=0
x→0 −1/x
x→0 −1/x2
x→0 1/x
1/x
Question 2.
t −t
e 12 .
so p0 (t) = 8 1 −
12
On [0, 60], p has one critical value p0 (t) = 0 at t = 12. Comparing p(0) = 0, p(12) = 96e−1 ≈
≈ 35.3% after 12 days.
35.3 and p(60) = 480e−5 ≈ 3.23, the maximum value is p(12) = 96
e
t
p(t) = 8te− 12
Question 3.
Let f (u) = eu − u − 1. Then f is differentiable with f (0) = 0 and f 0 (u) = eu − 1 ≥ 0 for all
u ≥ 0. Hence f (u) ≥ 0 for all u ≥ 0.
Question 4.
√
a) Set u = 1 + x2 , so u2 = 1 + x2 , udu = xdx. Thus
Z
Z
Z
√
x
u
√
du = du = u + C = 1 + x2 + C.
dx =
u
1 + x2
b) Because the function is even,
Z1
(1 + |t3 |) dt = 2
−1
Z1
(1 + |t3 |) dt = 2
0
Z1
(1 + t3 ) dt = 2[t +
t4 1 5
] =
4 0 2
0
Question 5.
Z
Area =
0
∞
dx
= lim
(x + 2)2 t→∞
Z
0
t
dx
(x + 2)2
= lim
t→∞
−1
x+2
t
!
= lim
0
t→∞
−1
1
+
t+2 2
=
1
2
Question 6.
Let r be the length of the base and h the height of the rectangle. We need to maximize
the volume of the resulting cylinder, where V = πr2 h. The perimeter of the rectangle is
30 = 2r + 2h. So, h = 15 − r. Therefore, V = πr2 (15 − r).
Differentiate w.r.t r: V 0 (r) = 3πr(10 − r). We get critical points at r = 0 and r = 10. By
using a derivative test, we find the maximum volume is V (10) = 500π (cm3 ) when r = 10
cm and h = 5 cm.
Question 7.
A circle of radius 2 centred at (5,0) has equation (x − 5)2 + y 2 = 22 . Consider rotating this
Rb
around the y-axis. Using the method of cylindrical shells, V = a 2πxf (x)dx, we get
8
Z
p
2πx 22 − (x − 5)2 dx
Volume = 2
3
(here letting t = x − 5)
Z 2
√
= 4π
(t + 5) 4 − t2 dt
−2
Z 2√
Z 2 √
2
t 4 − t dt + 20π
4 − t2 dt
= 4π
−2
−2
|
{z
}
= 0 (odd function)
Z
2
= 20π
√
4 − t2 dt
−2
(here letting t = 2 sin θ)
Z π/2
= 20π
22 cos2 θdθ
−π/2
= 40π
2