班 微積分2 期考解答和評分標準

模組 06-12班
1111模

1. Evaluate the following integrals.

1 3 x
(a) (5%) ∫ 1 2 dx (b) (6%) ∫ ln(x2 + 1)dx. (c) (7%) ∫ √ dx
x +x
2 3 2 4x − x2

Solution:
(a) Let x = t6 (2 points). So dx/dt = 6t5 . The original integral becomes

1 6t2
∫ 6t5 dt = ∫ dt
t3 +t4 1+t
1
=6 ∫ t − 1 + dt
1+t
t2
=6( − t + ln ∣1 + t∣) + C
2
1 1 1
=3x 3 − 6x 6 + 6 ln ∣1 + x 6 ∣ + C.(3 points)

(b) Using integration by parts, we get

2 2 2x
∫ ln(x + 1)dx =x ln(x + 1) − ∫ x dx (3 points)
x2 + 1
2
=x ln(x2 + 1) − ∫ 2 − 2 dx
x +1
=x ln(x2 + 1) − 2x + 2 tan−1 x + C.(3 points)

(c) First we complete the square

3 x 3 x
∫ √ dx = ∫ √ dx.(2 points)
2 4x − x 2 2 4 − (x − 2)2

Then we use the substitution x − 2 = 2 sin θ and dx/dθ = 2 cos θ (2 points) . Therefore, the integral becomes
π
6 2 sin θ + 2
∫ 2 cos θdθ (because 0 ≤ θ ≤ π/6, cos θ is positive)
0 2∣ cos θ∣
π
6
=∫ 2 sin θ + 2dθ
0
π √ π
=2(− cos θ + θ)∣06 = 2 − 3+ .(3 points)
3

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 1
2. (12%) Let R be the region enclosed by the curve y = , 1 ≤ x ≤ 2 and the x-axis.
+ 2x + 2) x2 (x2
Find the volume of the solid obtained by rotating R about the y-axis.

Solution:
1. Set up integral (2 points in total):
By shell’s method (1 point), the volume is
2 1 2 1
V =∫ 2πx ⋅ dx = 2π ∫ dx. (1 point)
1 x2 (x2 + 2x + 2) 1 x(x2 + 2x + 2)

2. Partial fraction (4 points in total):

By partial fraction (1 point), we can assume

1 A Bx + C
= + . (1 point)
x(x2 + 2x + 2) x x2 + 2x + 2

Clear denominators, we have 1 = A(x2 + 2x + 2) + x(Bx + C). (1 point) Thus, by comparing the coefficients, we
1 −1
get A = , B = , and C = −1. (1 point)
2 2
*No matter what method is used, give all 4 points for the part of partial fraction if the final form is correct.
3. Evaluation (6 points in total): Hence, we have
2 1 x+2
V = π∫ ( − ) dx
1 x x2 + 2x + 2
2 1 2 x+2
= π (∫ dx − ∫ 2
dx) .
1 x 1 x + 2x + 2

2
2 1
Note that ∫ dx = ln ∣x∣∣ = ln 2 (1 point) and
1 x 1

2 x+2 2 x+2
∫ 2
dx = ∫ dx Let u = x + 1, du = dx
1 x + 2x + 2 1 (x + 1)2 + 1
3 u+1
=∫ du (1 point)
2 u2 + 1
3 u 3 1
=∫ 2
du + ∫ 2
du Let v = u2 + 1, dv = 2u du
2 u +1 2 u +1
1 10 dv 3
= ∫ (1 point) + arctan(u)∣2 (1 pint)
2 5 v
10
1
= ln ∣v∣∣ + arctan 3 − arctan 2
2 5
1
= ln 2 + arctan 3 − arctan 2. (1 point)
2

To sum up,
1
V = π (ln 2 − ( ln 2 + arctan 3 − arctan 2))
2
1
= π ( ln 2 − arctan 3 + arctan 2) . (1 point)
2

Page 2 of 9
 1−t
1+t tan
−1
x
3. For t ≠ −1, consider the function F (t) = ∫ dx.
t 1+x
√
(a) (1%) Evaluate F ( 2 − 1).
A
(b) (6%) Prove that F ′ (t) = with some constant A. Find the constant A.
1+t
1−t π
(Hint. You may use, without proof, the fact that tan−1 t + tan−1 ( ) = for t ≠ −1.)
1+t 4
1
2 tan
−1
x
(c) (4%) Use (a) and (b) to find F (0). Hence evaluate ∫ 1 dx.
3
1 + x

Solution:

√
√ 1−t 2− 2 √
(a) For t = 2 − 1, we have = √ = 2 − 1. Hence,
1+t 2
√
√ 2−1 tan−1 x
F ( 2 − 1) = ∫√ dx = 0 (1%).
2−1 1+x

(b) By the FTC (1% is allocated for the trial of computing the derivative via FTC, this point is given even if
the calculation is incorrect),
1−t
tan−1 ( 1+t ) 1 − t ′ tan−1 t
F ′ (t) = 1−t
( ) − (1% for the correct application of FTC)
1 + 1+t 1+t 1+t
tan−1 ( 1−t
1+t
) −2 tan−1 t 1−t ′ −2
= − (1% for the correct calculation of ( ) = )
1 + 1−t
1+t
(1 + t)2 1+t 1+t (1 + t)2
1−t
tan−1 ( 1+t ) tan−1 t
=− − (1% for the simplification (trial))
1+t 1+t
1 1−t
=− (tan−1 ( ) + tan−1 t) (1% for the correct simplification)
1+t 1+t
π 1
=− ⋅ (1% for the correct answer) .
4 1+t

(c) By (2), we have

π
F (t) = −
⋅ ln ∣1 + t∣ + C
4
with some constant C (1% for the determination of F (t) up to the constant). By (1),
√ π √
F ( 2 − 1) = − ⋅ ln( 2) + C = 0 (1% for setting up this equation),
4
π π
so C = ⋅ ln 2. Hence, F (0) = C = ⋅ ln 2 (1% for F (0)). Thus,
8 8
1
2 tan−1 x 1 π 9 π
∫1 dx = F ( ) = ⋅ ln = ⋅ (2 ln 3 − 3 ln 2) (1% for the correct answer).
3
1+x 3 8 8 8

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 ⎧
⎪x(t) = sec t
⎪ π √
4. Let C be the parametric curve defined by ⎨ , 0≤t< . Also we let P = (1, 0) and Q = ( 2, 1).
⎪
⎪y(t) = tan t 2
⎩

(a) (4%) Find the equation of tangent of C at Q.

(b) (3%) Express the arclength of the portion of C from P to Q as an integral. Do NOT evaluate the integral.
√
(c) (7%) Let R be the region bounded by C, the x-axis, and the line x = 2. Find the area of R.

Solution:

π
(a) (1%) The point Q corresponds to t = .
4
dx dy
(1%) = sec t tan t and = sec2 t.
dt dt
dy dy/dt sec t
Therefore, = = .
dx dx/dt tan t
´¹¹ ¹ ¸¹ ¹ ¹¶
(1%)
Hence, the equation of tangent is

sec π4 √ √ √
y−1= π (x − 2)(1%) ⇒ y − 1 = 2(x − 2)
tan 4

Marking scheme for 4a

1% - the value of t that corresponds to Q
1% - finding both x′ (t) and y ′ (t) correctly
1% - formula for dy/dx for a parametric curve
1% - correct equation of tangent line

(b) (1%) The point P corresponds to t = 0.

The arclength equals ¿
π Á 2 2 π √
À( dx ) + ( dy ) dt =
4 Á 4
sec2 t tan2 t + sec4 t dt
∫ ∫
0 dt dt 0
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹(1%)
¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶
(1%)

Marking scheme for 4b

1% - the value of t that corresponds to P
1% - correct arclength element ds in parametric form
1% - correct answer (correct integrand AND integration limits)

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(c) The area of R equals
π π
4 4
∫ y(t) ⋅ x′ (t)dt = ∫ tan t ⋅ sec t tan t dt (1 + 1%)
0 0
π
4
=∫ tan2 t ⋅ sec t dt
0
π
4
=∫ sec3 t − sec t dt
0
π
⎡ ⎤4
⎢ ⎥
⎢ ⎥
⎢ ln ∣ sec t + tan t∣ + sec t tan t ⎥
= ⎢⎢ − ln ∣ sec t + tan t∣⎥⎥
⎢ 2 ¹ ¹ ¹ ¸¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶⎥⎥
⎢´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹(1%)
⎢ (3%)
⎥
⎣ ⎦
√ √ √ √ 0
ln( 2 + 1) + 2 √ 2 − ln( 2 + 1)
= − ln( 2 + 1) = (1%)
2 2

Marking scheme for 4c

1% - formula for area : ∫ y dx
1% - setting up the correct integral
3% - (*) integral of sec3 t
1% - integral of sec t
1% - answer
Remark for (*). No derivation is required. 1M or 2M can be awarded to a candidate with an
incorrect evaluation who (1) attempts reasonably to compute ∫ sec3 θdθ or (2) makes minor typos.

Page 5 of 9
5. The following diagram shows the graph of two polar curves r = 1 + cos θ and r = 2 − cos θ.

(a) (4%) Find, in polar coordinates, the intersection points of the two curves.
(b) (1%) Shade clearly in the diagram above the region that lies inside r = 1 + cos θ and outside r = 2 − cos θ.
(c) (6%) Find the area of the region in (b).

Solution:
(a)
Since r ≥ 0 for both polar curves and both are periodic over 2π, the intersections can only happen when the r
values are the same for some θ value.
1 π
Set 1 + cos θ = 2 − cos θ, then cos θ = and θ = ± + 2kπ, k ∈ Z.
2 3
π 3 π 3
When θ = , r = . When θ = − , r = .
3 2 3 2
3 π
The intersection points are ( , ± ) .
2 3 (r,θ)
(b)
1 π π
2 − cos θ ≤ 1 + cos θ ⇒ cos θ ≥ ⇒ − + 2kπ ≤ θ ≤ + 2kπ.
2 3 3
Shade in the right-most region.
(c)
The area is
1π/3 π/3 1
∫ (1 + cos θ)2 dθ − ∫ (2 − cos θ)2 dθ
−π/3 2 −π/3 2

1 π/3 3 π/3
√
= ∫ (6 cos θ − 3) dθ = [2 sin θ − θ]−π/3 = 3 3 − π
2 −π/3 2

Grading:
(a) 2% for solving for θ and 2% for final answer. The student does not need to write 2kπ and the final answer
could be any equivalent point. Any incorrect answer here still needs to be used in (c).
(b) No partial credit.
(c) 4% for the correct setup (2% for using answer in (a) and 2% for integrand) and 2% for evaluating the integral.
An extra -1% if the answer is negative and the student just added absolute value for no reason.

Page 6 of 9
6. An object of mass 1 kg falls near the surface of the earth experiences air resistance that is proportional to the square
of its velocity. Therefore, its equation of motion is given by
dv 1
= 9.8 − v 2 .
dt 5
where v = v(t) is the velocity of the object at time t. It is known that 0 ≤ v < 7 and v(0) = 0.
(a) (9%) Find v(t).
(b) (1%) Find lim v(t).
t→∞

Solution:
(a)

dv 1 dv 1
= (49 − v 2 ) ⇒ ∫ 2
= ∫ dt 2 pts
dt 5 49 − v 5
1 1 1 1
⇒ ∫ + dv = t + C 2 pts for correct partial fractions
14 7−v 7+v 5
7+v 1 1
⇒ ln ∣ ∣ = 2.8t + C ′ 2 pts for integrating and
7−v 7−v 7+v
7+v
Because 0 ≤ v < 7, we conclude that = Ae2.8t , where A is a constant. 1 pt
7−v
Because v(0) = 0, we have 1 = A ⋅ e0 = A 1 pt
2.8t
7+v 14 7(e − 1)
Hence = e2.8t ⇒ v(t) = 7 − = 1 pt
7−v 1 + e2.8t e2.8t + 1
14
(b) lim v(t) = lim 7 − =7 1 pt
t→∞ t→∞ 1 + e2.8t

t
7. Initially a tank contains 30 L of pure water. At time t min, brine solution of concentration c(t) = e− 15 (2 + sin t) kg/L
enters the tank at a rate of 2 L/min. The solution is kept mixed thoroughly and drains from the tank at a rate of 2
L/min. Let A(t) (in kg) be the amount of salt in the tank after t minutes.
(a) (4%) Derive a differential equation satisfied by A(t).
(b) (8%) Hence solve for A(t).

Solution:
dA t A(t) t 1
(a) = rate in − rate out = 2 × e− 15 (2 + sin t) − 2 × = 2 ⋅ e− 15 (2 + sin t) − A(t)
dt 30 15
t
2 pts for rate in = 2 × e− 15 (2 + sin t)
A
2 pts for rate out = 2 ×
30
dA 1 t
(b) + A(t) = 2 × e− 15 (2 + sin t)
dt 15 t
Choose the integrating factor I(x) = e 15 2 pts
t dA 1 t ′
Then e (15 + A(t)) = 4 + 2 sin t ⇒ (e ⋅ A(t)) = 4 + 2 sin t
15 2 pts
t
dt 15
And e 15 A(t) = 4t − 2 cos t + C 2 pts
Because A(0) = 0, we have e0 ⋅ A(0) = 0 = −2 cos 0 + C. Hence C = 2. 1 pt
t t t
Therefore, A(t) = 4te− 15 − 2e− 15 cos t + 2e− 15 1 pt

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8. Munch-Munch Restaurant in Taipei displays the poster in Figure 1 that indicates every customer should receive their
orders within 90 seconds.

It is known that the waiting time for an order is a continuous random variable X whose density is given by
⎧
⎪0
⎪ if x < 0
f (x) = ⎨ (x in seconds).
⎪
⎪c ⋅ 2−0.1x if x ≥ 0
⎩
b
Recall that P(a ≤ X ≤ b) = ∫ f (x) dx.
a

(a) (3%) Find the value of the constant c.

(b) (3%) A customer receives a gift card as a compensation if his/her order arrives after 90 seconds. Find the
probability that a customer will receive a gift card.
(c) In order to shorten the serving time, the manager of the restaurant has purchased a few food serving robots
√
X
(See Figure 2). Having implemented these robots, the serving time becomes a new random variable Y = .
2
(i) (3%) Write down the distribution function F (y) = P(Y ≤ y) as an integral.
(ii) (3%) Find the probability density function fY (y) of Y . (Hint. fY (y) = F ′ (y))

Solution:

∞
(a) (1%) Since ∫ c ⋅ 2−0.1x dx = 1,
0

t
LHS = lim ∫ c ⋅ 2−0.1x dx See below
t→∞ 0
t
2−0.1x
= lim [c ⋅ ] (1%)
t→∞ −0.1 ln 2 0
2−0.1t 1
= lim c ( + )
t→∞ −0.1 ln 2 0.1 ln 2
c
=
0.1 ln 2
Hence c = 0.1 ln 2.(1%)

(b)
∞ t
P(X > 90) = ∫ 0.1 ln 2 ⋅ 2−0.1x dx = lim ∫ 0.1 ln 2 ⋅ 2−0.1x dx
90 t→∞ 90
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶
(1%) (1%)
x=t
= lim [−2−0.1x ]x=90
t→∞
= lim (2−9 − 2−0.1t )
t→∞
= 2−9 (1%)

Page 8 of 9
 Marking scheme for 8ab
1% - knowing that the total probability equals 1
1% - anti-derivative of 2−0.1x
1% - correct value of c
1% - setting up the correct integral for P(X > 90)
1% - (*) definition of improper integral
1% - correct answer
Remark for (*). The definition of improper integrals need to appear at least once in either 8a or
8b. Otherwise, this 1% will be taken off.

(c) (a) For y ≥ 0 (1%), we have

√ 2
X 4y
F (y) = P(Y ≤ y) = P( ≤ y) = P (X ≤ 4y 2 ) = ∫ f (x) dx
2 ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ 0¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶
(1%) (1%)

and for y < 0, we have F (y) = 0.

(b) Let fY (y) be the density of Y . For y ≥ 0 (See above), by FTC, we have
2
fY (y) = F ′ (y) = f (4y 2 ) ⋅ 8y = 0.1 ln 2 ⋅ 2−0.4y ⋅ 8y .
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶
(1%) (2%)

and for y < 0, we have fY (y) = 0.

Marking scheme for 8c
1% - (*) distinguish the cases y > 0 and y ≤ 0
1% - transforming P(Y ≤ y) into P(X ≤ 4y 2 )
1% - correct distribution function (both integrand and integration limits need to be correct)
1% - differentiating F (y) by FTC
2% - correct density fY (y) (1% if a candidate obtains incorrect value for c)
Remark for (*). Candidates need to demonstrate the differences of the cases when y < 0 and y ≥ 0
in either (c) (i) or (c) (ii). Otherwise, this 1% will not be awarded.

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