MATH4007 Calculus 2 (M05-09) Final Examination Solution
Date : 20th December, 2025
NTU 114-1
Part A. Short Questions (60%)
A1 Evaluate.
π/2
(a) (12%) ∫
π/4
cos3 x
dx
1 + sin2 x
Solution:
Use u = sin x. Then du = cos x dx and cos2 x = 1 − sin2 x.
π/2
∫
π/4
Hence
π/2
∫
1/2
(b) (12%) ∫
π/4
1
1
cos3 x
1 − u2
2
du = ∫ √ −1 +
du
dx
=
√
∫
2
2
1 + u2
1/ 2
1/ 2 1 + u
1 + sin x
√
cos3 x
1
π
1
dx = [−u + 2 tan−1 u]1/√2 = − 1 + √ − 2 tan−1 (1/ 2)
2
2
1 + sin x
2
√
x 1 − x2 sin−1 x dx
0
Solution:
√
Use integration by parts with u = sin−1 x and dv = x 1 − x2 dx.
dx
−1
Then du = √
and v =
(1 − x2 )3/2 .
2
3
1−x
1/2
∫
With sin−1 (1/2) =
0
1/2
√
1 1/2
−1
1 − x2 dx
x 1 − x2 sin−1 x dx = [ (1 − x2 )3/2 sin−1 x] + ∫
3
3 0
0
π
. Hence
6
1/2
∫
√
x
1 − x2
−1
sin
0
√
√
3π
−π 3 3 1 1
11
x dx =
⋅
+ −
=
−
18
8
6 72 72
48
Grading:
Correctly applying integration techniques can earn up to 4 points. This is for any work that is too far from
the correct answer.
The final answer do not need to be simplified. But the numbers still need to be plugged in.
Other than the 2 cases above, it will be -1 for each algebra mistake and -2 for each calculus mistake.
A2 (8%) Find the limit
cos x
1
f (t) dt
∫
4
x→0 x
1
lim
where f is a smooth function with f (1) = 0 and f ′ (1) = 3. Recall lim
x→0
sin x
= 1.
x
Solution:
Since lim x4 = lim ∫
x→0
x→0
cos x
f (t) dt = 0, we can apply the L’Hôpital’s rule and the fundamental theorem of calculus
1
(with the chain rule) to get
cos x
1
f (cos x) ⋅ − sin x
1
sin x
f (cos x)
1
f (cos x)
f (t) dt = lim
= − lim
⋅ lim
= − lim
.
∫
4
x→0 x
x→0
4x3
4 x→0 x x→0
x2
4 x→0
x2
1
lim
Since lim x2 = 0 and lim f (cos x) = f (1) = 0, we can apply the L’Hôpital’s rule again to get
x→0
x→0
cos x
1
1
1
f (cos x)
f ′ (cos x) ⋅ − sin x 1
sin x 1 ′
3
f (t) dt = − lim
= − lim
= lim f ′ (cos x) ⋅ lim
= f (1) ⋅ 1 = .
∫
4
2
x→0 x
x→0 x
4 x→0
x
4 x→0
2x
8 x→0
8
8
1
lim
Page 1 of 8
Grading policy:
ˆ Students may apply the Fundamental Theorem of Calculus together with the chain rule without explicitly
′
cos x
stating it. If they differentiate correctly, they earn 2 points. If they write (∫
′
cos x
(∫
f (t) dt) = f (cos x) or
1
f (t) dt) = f (x) ⋅ − sin x, then they earn 1 point. Any other result earns 0 points.
1
0
∞
or . Each
0
∞
application is worth 2 points: 1 point for identifying the form, and 1 point for differentiating correctly.
ˆ Students must indicate where they use L’Hôpital’s rule and identify the indeterminate form
ˆ Using the continuity of f ′ to conclude lim f ′ (cos x) = f ′ (1) = 3 is worth 1 point.
x→0
ˆ Using the limit lim
x→0
sin x
= 1 to simplify the limit is worth 1 point.
x
ˆ Another method: If a student applys L’Hôpital’s rule four times and correctly derives
cos x
−f (cos x) sin x
f ′ (cos x) sin2 x − f (cos x) cos x
1
f
(t)
dt
=
lim
=
lim
∫
x→0
x→0
x→0 x4 1
4x3
12x2
3
′′
′
−f (cos x) sin x + 3f (cos x) sin x cos x + f (cos x) sin x
= lim
x→0
24x
f ′′′ (cos x) sin4 x − 6f ′′ (cos x) sin2 x cos x + (7 cos2 x − 4)f ′ (cos x) + f (cos x) cos x
= lim
x→0
24
3⋅3+0⋅1 3
= ,
=
24
8
lim
then they earn 8 points in total.
A3 (10%) Find the arc length of the graph of the function
x√
f (x) = ∫
for 0 ≤ x ≤
64 cos6 θ − 1 dθ
0
b√
π
. Recall L = ∫
1 + [f ′ (x)]2 dx.
3
a
Solution:
Note that, by the Fundamental Theorem of Calculus,
√
f ′ (x) = 64 cos6 x − 1 .
(correct use of FTC: 3 points)
π
Also notice that 64 cos6 x − 1 ≥ 0 for any x ∈ [0, ].
3
The arc length is then given by
π
3
∫
0
√
1 + [f ′ (x)]2 dx = ∫
π
3
√
64 cos6 x dx
0
= 8∫
= 8∫
π
3
cos3 x dx
(removal of the radical: 3 points)
0
x= π
3
cos2 x d(sin x)
x=0
√
= 8∫
3
2
(1 − u2 ) du
(u ∶= sin x , cos2 x = 1 − sin2 x)
0
√
3
3
u 2
= 8 [u − ]
3 0
√ √
√
=4 3− 3=3 3.
(rest of the computation: 4 points)
(-0.5 points: for each minor mistake)
Page 2 of 8
A4 A fish population N (t) (in thousands) grows exponentially with intrinsic growth rate r > 0 per year when left alone.
Suppose 20 thousand is harvested per year, then we have a differential equation for the population.
N ′ (t) = r ⋅ N (t) − 20
(a) (2%) If we want the fish population to stay a constant, what should N (0) be in terms of r?
(b) (8%) If the initial value N (0) = N0 is a number between 0 and your answer in (a), then the fish population will
reach zero (extinction) at time t∗ . Find t∗ in terms of r and N0 .
Solution:
(a) Since the fish population stays a constant, we have N (t) = N (0) for all t. (1 pt) Then we have
0=
dN
= rN (0) − 20
dt
⇒
N (0) =
20
. (1 pt)
r
(b) The integrating factor is e∫ −r dt = e−rt . (2 pts) Then
dN
d −rt
− re−rt N (t) = −20e−rt ⇒
[e N (t)] = −20e−rt
dt
dt
20
e−rt N (t) = e−rt + C (3 pts)
r
e−rt
⇒
∗
20
20
20
Consider N (0) = N0 , we have N0 =
+C ⇒ C = N0 − . (1 pt) Since N (t∗ ) = 0, we have 0 = (e−rt −1)+N0 .
r
r
r
1
rN0
∗
So t = − ln (1 −
). (2 pts)
r
20
A5 (8%) Solve the initial value problem by first considering a substitution u = g(y),
du
dy
= g ′ (y) .
dx
dx
⎧
dy
⎪
⎪
+ y 2026 = 4ex
⎪y 2025
⎨
dx
⎪
⎪
⎪
⎩y(0) = 1
Solution:
Apply the substitution u = y 2026 to the differential equation and note that
du
dy
= 2026 y 2025 .
dx
dx
It follows that the differential equation becomes
1 du
+ u = 4ex
2026 dx
⇐⇒
du
+ 2026 u = 4 ⋅ 2026 ex .
dx
(eqn transformation: 1 point)
To obtain an integrating factor I(x),
(Method 1: Find I(x) by computation)
take
I(x) ∶= e∫ 2026 dx = e2026x .
(Method 2: Find I(x) by observation)
observe that
(I(x) from formula / by solving DE: 2 points)
d 2026x
du
(e
u) = e2026x
+ 2026 e2026x u
dx
dx
du
= e2026x (
+ 2026 u) .
dx
(choice of I(x): 1 point)
(verification: 1 point)
d 2026x 2026
(e
) = 2026 e2026x (y 2025 y ′ + y 2026 ) with verification is also acceptable.)
y
dx
Thus, multiplying I(x) ∶= e2026x to both sides of the differential equation, we obtain
(Observing directly
d 2026x
(e
u) = 4 ⋅ 2026 e(2026+1)x
dx
Ô⇒ e2026x y 2026 = e2026x u = 4 ⋅ 2026 ∫ e2027x dx
(differential or integral equation of I(x)u: 2 points)
4 ⋅ 2026 2027x
=
e
+C .
2027
Page 3 of 8
(integration: 1 point)
The initial condition y(0) = 1 implies that
1=
4 ⋅ 2026
+C
2027
⇐⇒
C=−
4 ⋅ 2026 − 2027
6077
(= −
).
2027
2027
(value of C: 1 point)
(When definite integrals are used instead, the 1 point is given to correct upper and lower limits of integration
and suitable use of the initial condition.)
As a result, y 2026 =
4 ⋅ 2026ex −(4 ⋅ 2026 − 2027) e−2026x
8104ex − 6077e−2026x
(=
).
2027
2027
(conclusion: 1 point)
(-0.5 points: for each minor mistake, e.g. wrong sign, wrong coef., missing constant C, ...)
Page 4 of 8
Part B. Structured Questions (40%)
B1 (16%) (a) (6%) Use the method of partial fractions to evaluate
√
3
1
dx
√
1/ 3 (x + 1)(x2 + 1)
∫
Solution:
According to the partial fractions, we wrtie
1
A
Bx + C
=
+ 2
,
(x + 1)(x2 + 1) x + 1
x +1
where A, B and C are constants to be determined. Multiplying (x + 1)(x2 + 1) on both sides yields
1 = A(x2 + 1) + (Bx + C)(x + 1) = (A + B)x2 + (B + C)x + (A + C).
1
1
By comparing the coefficients, we get A = C = , B = − . Hence, the integral can be evaluated as follows.
2
2
√
√
3
3 1 1
1
1 x−1
dx
=
(
−
) dx
√
∫
∫ √
2 x2 + 1
1/ 3 (x + 1)(x2 + 1)
1/ 3 2 x + 1
√
3
1
1
1
π
1 π π
= ln ∣x + 1∣ − ln(x2 + 1) + tan−1 x∣ √ = ( − ) = .
2
4
2
2 3 6
12
1/ 3
Grading policy:
ˆ Getting the correct coefficients earns 2 points. If they get only one or two coefficients correct, then
they earn 1 point.
1
x
1
dx, ∫ 2
dx, and ∫ 2
dx is worth 1 point. (Even if students obtain the
x+1
x +1
x +1
wrong coefficients, they can still earn points if they correctly calculate the integrals.)
ˆ Each integral ∫
ˆ The final answer is worth 1 point. If there is a minor typo, then they earn 0.5 points.
(b) (6%) Evaluate
√
∫
3
1
dx
√
1/ 3 (x2 + 1)(x2 + 1)
Solution:
√
√
Let x = tan θ. Then dx = sec2 θ dθ. When x = 1/ 3, we have θ = π/6; when x = 3, we have θ = π/3.
Thus, the integral can be transformed as
√
∫
3
π/3
1
dx = ∫
√
1/ 3 (1 + x2 )(1 + x2 )
π/6
=
π/3
1
⋅ sec2 θ dθ = ∫
cos2 θ dθ
4
sec θ
π/6
1 π/3
1
sin 2θ π/3 1 π
π
(1 + cos 2θ) dθ = (θ +
)∣
= ⋅ = .
∫
2 π/6
2
2
2
6
12
π/6
Grading policy:
ˆ Making a suitable substitution and changing the limits of integration is worth 3 points.
ˆ Using the double angle formula is worth 1 point.
ˆ Evaluating the integral and obtaining the final answer is worth 2 points.
ˆ It is acceptable to evaluate the indefinite integral. If students obtain the antiderivative correctly in
terms of x, then they earn 4 points. Evaluating the definite integral (by applying the limits) is worth
2 points.
ˆ It is acceptable to apply the result of (c) to earn full points.
Page 5 of 8
(c) (4%) Let a be a positive constant. By considering the substitution x =
√
∫
Solution: √
Let I = ∫
3
1
√
1/ 3 (xa + 1)(x2 + 1)
3
1
, we can evaluate
u
1
dx
√
1/ 3 (xa + 1)(x2 + 1)
√
√
√
√
dx. When x = 1/ 3, we have u = 3; when x = 3, we have u = 1/ 3. In
1
addition, dx = − 2 du. Then
u
√
1/ 3
I = ∫√
3
√
3
1
ua
1
⋅
−
du
=
du.
√
∫
(u−a + 1)(u−2 + 1) u2
1/ 3 (ua + 1)(u2 + 1)
By renaming the dummy variable, we replace u by x, i.e.,
√
I =∫
xa
3
dx.
√
1/ 3 (xa + 1)(x2 + 1)
Therefore, we find
√
√
√
√
3
3
3
1
xa
1
π
2I = I + I = ∫ √
dx
+
dx = ∫ √ 2
dx = tan−1 x∣ √ = ,
√
∫
a
2
a
2
1/ 3
6
1/ 3 (x + 1)(x + 1)
1/ 3 (x + 1)(x + 1)
1/ 3 x + 1
which implies I =
3
π
.
12
Grading policy:
ˆ Correctly applying the substitution x =
1
(including changing the limits) is worth 2 points.
u
ˆ Renaming the dummy variable and adding the two expressions for I to obtain the final answer is worth
2 point.
B2 (14%) From Calculus 1, we know that ln x ≤ x − 1 < x for all x > 0.
1
(a) (2%) Show that, for any number p > 0, the inequality ln x < xp is true for all x > 0.
p
(b) (5%) Show that
∞
dx
∫
q
x (ln x)2
2
is divergent for all q < 1. (Hint: Choose a suitable p-value for (a) to set up comparison with
(c) (5%) By comparing with the function
1
, show that the improper integral
x(ln x)2
∞
∫
2
dx
xq (ln x)2
is convergent for all q ≥ 1.
(d) (2%) Darken the correct circle: T True / F False
∞
dx
For any constant q, ∫
is always a divergent improper integral.
xq (ln x)2
0
Solution:
(a) Consider x = tp , t > 0 in the given information.
1−q
(b) Choose p =
or any smaller p.
2
1
(c) They are all smaller than
and
x(ln x)2
∞
∫
2
t
dx
dx
1
=
lim
=
∫
x(ln x)2 t→∞ 2 x(ln x)2 ln 2
Page 6 of 8
1
.)
x
Then use comparison theorem.
(d) True.
Grading:
No partial credits for (a) and (d).
For (b) and (c), 2 points for the comparison target’s behavior and the statement of comparison theorem. 3 points
for the inequality that allows the comparison theorem.
If student evaluates a special case in (b) or (c), they can get 1 point for that part (if everything else is bad).
Page 7 of 8
0—'S
B3 (10%) Elio and Erofen heard that there is a magic lamp appearing at a random place on the Royal Palm Boulevard
(
). Suppose the Royal Palm Boulevard is L meters long. Let X be the random variable denoting the position
of the magic lamp in [0, L] with probability density function
⎧
⎪
⎪C(L − x), 0 ≤ x ≤ L
fX (x) = ⎨
⎪
0,
x ∉ [0, L]
⎪
⎩
∞
Recall a probability density function satisfies ∫
And the expected value is given by ∫
∞
−∞
b
fX (x) dx = 1. The probability of X ∈ [a, b] is given by ∫
−∞
a
fX (x) dx.
x ⋅ fX (x) dx.
(a) (2%) Find the value of the constant C in terms of L.
(b) (3%) Find the expected position of the magic lamp.
(c) (2%) Suppose they decide to search for the magic lamp separately. Elio searches the first L/3 meters and Erofen
searches the last 2L/3 meters. Who is more likely to find the magic lamp?
√
(d) (3%) Suppose a magic tool is used to shorten the Royal Palm Boulevard, with a new random variable Y = X
for the position. Find the expected value of Y in terms of L.
Solution:
(a)
1=∫
∞
−∞
L
fX (x) dx = ∫
0
L
⎛
2
x2 ⎞ CL2
(1 pt) ⇒ C = 2 . (1 pt)
C(L − x) dx = C Lx − ∣ =
2 0⎠
2
L
⎝
(b) The expected position is
∞
∫
=
−∞
L
x ⋅ fX (x) dx = ∫
0
L
⎛ x2 x3 ⎞
(2 pts)
C(Lx − x2 ) dx = C L − ∣
3 0⎠
⎝ 2
CL3
2 L3 L
= 2
= . (1 pt)
6
L 6
3
(c) For Elio, the probability is
L/3
P (0 ≤ x ≤ L/3) = ∫
0
L/3
⎛ 2x x2
⎞
2
2 1
5
= ( − ) = (1 pt)
(L
−
x)
dx
=
−
∣
L2
3 9
9
⎝ L L2 0 ⎠
For Erofen, the probability is
L
⎛ 2x x2
⎞
2
2 1
4
P (L/3 ≤ x ≤ L) = ∫
(L
−
x)
dx
=
− 2∣
= (2 − 1 − + ) = .
L
L
3
9
9
1/3 L2
⎝
L/3 ⎠
L
4
(or P (L/3 ≤ x ≤ L) = 1 − P (0 ≤ x ≤ L/3) = .) (1 pt)
9
So Elio is more likely to find the magic lamp.
(d) The expected of Y is
∞√
L
√
E[Y ] = E[ X] = ∫
x ⋅ fX (x) dx(1 pt) = ∫ C(Lx1/2 − x3/2 ) dx
−∞
3/2
=
2 ⎛ 2Lx
L2 ⎝ 3
−2
0
5/2 L ⎞
2Lx
5
∣
(1 pt) =
0⎠
Page 8 of 8
8√
L. (1 pt)
15