FACULTY OF SCIENCE
DEPARTMENT OF MATHEMATICS AND APPLIED MATHEMATICS
MODULE
MATENB1
APPLICATIONS OF CALCULUS FOR ENGINEERS
CAMPUS
ASSESSMENT
APK
SEMESTER TEST 1 MEMORANDUM
DATE 28/08/2021
TIME 09:00
ASSESSOR(S)
MR. S. MAFUNDA
DR. K. SEBOGODI
DURATION 90 MINUTES
MARKS 47
SURNAME AND INITIALS MEMORANDUM
STUDENT NUMBER MEMORANDUM
CONTACT NUMBER MEMORANDUM
NUMBER OF PAGES: 1 + 9 PAGES
INSTRUCTIONS: 1. ANSWER ALL THE QUESTIONS ON THE PAPER IN PEN.
2. NO CALCULATORS ARE ALLOWED.
3. SHOW ALL CALCULATIONS AND MOTIVATE ALL ANSWERS.
4. IF YOU REQUIRE EXTRA SPACE, CONTINUE ON THE
ADJACENT BLANK PAGE AND INDICATE THIS CLEARLY.
MATENB1 SEMESTER TEST 1 MEMORANDUM– 28 AUGUST 2021
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Question 1 [12 marks]
Answer the following True or False questions and give a short justification/counterexample:
Z
Z
n x
n x
(a)
x e dx = x e − n xn−1 ex dx.
(2)
TRUE
FALSE
X
Justification: Let u = xn , dv = ex dx =⇒ du = nxn−1 dx, v = ex . Then the solution follows.
.
Z
(b)
sec2 (3x) dx = 3 tan(3x) + C.
TRUE
FALSE
(2)
X
d
[3 tan(3x) + C] = 9 sec2 (3x) ̸= sec2 (3x).
dx
Justification:
.
Z a
(c)
x
2
√
a2 − x2 dx = a4
0
TRUE
FALSE
Z a
sin2 θ cos2 θ dθ
(2)
0
X
Justification: The limits of integration must change in accordance with the substitution
used.
.
(d) The partial fraction decomposition of
TRUE
FALSE
Justification:
.
2x + 8
A
B
C
is
+
+
.
2
(x − 1)(x + 4)
x−1 x+2 x−2
X
A
Bx + C
+ 2
x−1
x +4
(2)
MATENB1 SEMESTER TEST 1 MEMORANDUM– 28 AUGUST 2021
e
Z e
ln(x − 1)dx = (x − 1) ln(x − 1) − x
(e)
2/9
1
TRUE
FALSE
(2)
1
X
Justification: We have that lim+ ln(x − 1) = −∞ so x = 1 is a discontinuity of the intex→1
grand in the interval [1, e] and the integral is improper and cannot be calculated by directly
substituting the limits of integration.
.
(f) Suppose f is a differentiable function such that f ′ (x) ≤ 3 for all x ∈ [−3, 4]. If f (−3) = 4,
the Mean Value Theorem says that f (4) ≤ 25.
(2)
TRUE
X
FALSE
Justification:Since f is differentiable on (−3, 4), it is also continuous on the interval [−3, 4].
By the MVT, there exists c ∈ (−3, 4) such that f (4) − f (−3) = f ′ (c)(4 − (−3)), i.e.,
f (4) = f (−3) + 7f ′ (c) ≤ 4 + 7(3) = 25.
.
MATENB1 SEMESTER TEST 1 MEMORANDUM– 28 AUGUST 2021
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Question 2 [6 marks]
Evaluate the following integrals using the method of integration by parts:
Z
(a)
sin x ln(cos x) dx
(3)
Solution:
Let
u = ln(cos(x))
du = − tan(x) dx
dv = sin(x)dx
v = − cos(x)
Hence
Z
Z
sin x ln(cos x) dx = [ln(cos(x)) · (− cos(x))] − tan(x) cos(x) dx
Z
= [− cos(x) ln(cos(x))] − sin(x) dx
= cos(x) [1 − ln(cos(x))] + C
Z π/4
(b)
x cos(2x) dx
(3)
0
Solution:
Let
u =x
du =dx
dv = cos(2x)dx
1
v = sin(2x)
2
Hence
Z π/4
0
Z
π/4
1 π/4
x
x cos(2x) dx = sin(2x)
−
sin(2x) dx
2
2 0
0
π/4
π
π 1 1
=
sin
−0 −
− cos(2x)
4
2
2
2
0
π 1
= + (0 − 1)
8 4
π−2
=
8
MATENB1 SEMESTER TEST 1 MEMORANDUM– 28 AUGUST 2021
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Question 3 [6 marks]
Evaluate the following integrals:
Z
sin(2x)
(a)
dx
1 + cos2 (x)
(3)
Solution:
Let
u =1 + cos2 (x)
du = − 2 sin(x) cos(x)dx
Hence
Z
sin(2x)
dx =
1 + cos2 (x)
Z
2 sin(x) cos(x)
dx
1 + cos2 (x)
Z
1
=−
du
u
= − ln |u| + C
= − ln |1 + cos2 (x)| + C
= − ln 1 + cos2 (x) + C
Z π√
2 sin x. cos3 x dx
(b)
(3)
0
Solution:
Let
u = sin(x)
=⇒ u(0) = 0 and u
π 2
=1
du = cos(x) dx
Z π√
Z π√
2 sin x. cos3 x dx = 2 sin x. cos2 x cos x dx
0
0
Z π√
= 2 sin x(1 − sin2 x) cos x dx
Z0 1
√
u(1 − u2 ) du
0
1
2 3 2 7
= u2 − u2
3
7
0
8
=
21
=
MATENB1 SEMESTER TEST 1 MEMORANDUM– 28 AUGUST 2021
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Question 4 [6 marks]
Evaluate the following integrals using the method of trigonometric or hyperbolic substitution:
Z
1
√
(a)
dx
(4)
x4 x2 − 2
Solution:
Use trigonometric
substitution:
√
√
Let x = 2 sec θ, where 0 ≤ θ < π2 or π ≤ θ < 3π
. Whence we obtain dx = 2 sec θ tan θdθ;
2
√
√
x4 = 4 sec2 θ; x2 − 2 = 2 tan θ and sec θ = √x2
Then
Z
Z
√
1
1
1
√
√
dx =
2 sec θ tan θ dθ
4 θ 2 tan θ
4
x4 x2 − 2
sec
Z
1
1
=
dθ
4
sec3 θ
Z
1
=
cos3 θ dθ
4
Z
1
cos2 θ cos θ dθ
=
4
Z
1
(1 − sin2 θ) cos θ dθ
=
4
1
1
=
(sin θ − sin3 θ) + C
4√
3p
2
(x2 − 2)3
x −2
−
+C
=
4x
12x3
√
The last line follows from sec θ = √x2 , which implies cos θ = x2 and sin θ =
Z
2
(b)
dx
4 + x2
√
x2 −2
.
x
(2)
Solution:
Let x = 2 tan θ, where − π2 < θ < π2 . Then dx = 2 sec2 θdθ and 4 + x2 = 4 sec2 θ. Whence
Z
Z
2
2
dx =
2 sec2 θ dθ
2
4+x
4 sec2 θ
Z
= 2 dθ
= 2θ + C
x
= tan−1 + C
2
MATENB1 SEMESTER TEST 1 MEMORANDUM– 28 AUGUST 2021
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Question 5 [5 marks]
Z
1
dx
x2 − 4
(a) Use partial fractions to determine
(3)
Solution:
1
A
B
+
x+2 x−2
1 =A(x − 2) + B(x + 2)
x2 − 4
Hence A = −
1
4
and B =
Z
=
1
4
Z
Z
1
1
dx
1
dx
dx = −
+
2
x −4
4
x+2 4
x−2
x−2
1
+C
= ln
4
x+2
Z
(b) Use you answer in (a) to compute
1
√
dx
(x − 2) x + 2
Solution:
Let u =
√
x + 2 → u2 = x + 2 → x − 2 = u2 − 4 and 2udu = dx
Z
1
√
dx =
(x − 2) x + 2
Z
2u
du
u(u2 − 4)
1
u−2
= ln
+C
2
u+2
√
1
x+2−2
+C
= ln √
2
x+2+2
(2)
MATENB1 SEMESTER TEST 1 MEMORANDUM– 28 AUGUST 2021
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Question 6 [6 marks]
Z ∞
(a) Evaluate
e
1
dx.
x (ln(x))2
(3)
Solution:
Letting u = ln(x), we have that du = x1 dx and
Z
Z
1
1
1
u−2 du = − = −
+ C.
2 dx =
u
ln(x)
x (ln(x))
It follows that
Z ∞
e
Z t
1
2 dx
e x (ln(x))
t
1
1
1
= lim −
= − lim
−
t→∞
t→∞
ln(x) e
ln(t) ln(e)
= −(0 − 1) = 1 (Since ln t → ∞ as t → ∞)
1
dx = lim
t→∞
x (ln(x))2
Z π/2
ln(x) dx.
(b) Evaluate
(3)
0
Solution:
We have
π/2
Z π/2
ln(x) dx =
0
=
=
=
=
=
lim (x ln(x) − x)
π π πt
lim+
ln
− − t ln(t) + t
t→0
2
2
2
π
π
π
ln(t)
ln
− − lim+
+ lim+ t
t→0
2
2
2 t→0 1/t
π
π
π
1/t
ln
− − lim+
+ lim t
2
2
2 t→0 −1/t2 t→0+
π π π
ln
− − lim −t + lim+ t
t→0
2 h 2 2 i t→0+
π
π
ln
−1 .
2
2
t→0+
MATENB1 SEMESTER TEST 1 MEMORANDUM– 28 AUGUST 2021
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Question 7 [6 marks]
(a) Suppose f (x) = x3 + 3x + 1. Show that the equation f (x) = 0 has exactly one real root in
the interval [−1, 0].
(3)
Solution:
f (0) = 1 ≥ 0 and f (−1) = −3 ≤ 0. Since f is continuous on [−1, 0], by the Intermediate Value Theorem, there exists c ∈ [−1, 0] such that f (c) = 0. So, the given equation
has a root c.
Assume that there is another root c′ , then f (c′ ) = f (c) = 0. Since f is a polynomial, it
is continuous on the interval [−1, 0] and and differentiable on (−1, 0). By Rolle’s Theorem,
there exists a ∈ (c′ , c) such that f ′ (a) = 0. i.e., 3a2 + 3 = 0 and there is no real number a
satisfying this equation. This gives a contradiction. Therefore, the equation can’t have two
real roots.
(b) Sketch the graph of a function that satisfies the following conditions:
i) f ′ (−2) = f ′ (2) = 0, f ′ (x) < 0 for |x| < 2.
ii) f ′ (x) > 0 for 2 < x < 4.
iii) f ′′ (x) < 0 for x ∈ (−4, 0).
iv) f ′ (x) = 0 for x > 4.
v) f has an inflection point at (0, 2).
vi) f (−4) = 1; f (−2) = 4; f (2) = 12 ; f (4) = 3.
Solution:
(3)
MATENB1 SEMESTER TEST 1 MEMORANDUM– 28 AUGUST 2021
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