1091!D06-12í ®M2 ãTU
1. (8 pts) Let f be a continuous function on R such that
∫
x3 0
f (t) dt = x3⋅cos(πx) for all x.
Find f (1).
Solution:
Since f is continuous on R, by Fundamental Theorem of Calculus (2%), we have d
dx∫
x3 0
f (t) dt = d
dx(x3⋅cos(πx))
⇒ f (x3) ⋅3x2 (2%) = 3x2⋅cos(πx) − πx3⋅sin(πx) (1%) To find f (1), we solve x3=1 to obtain that x = 1 (1%). For x = 1, we have
f (1) ⋅ 3 = 3 ⋅ cos(π) − π ⋅ sin(π) = −3 ⇒ f (1) = −1 (2%).
2. (16 pts) Evaluate the following definite integrals.
(a) ∫
1 0
√x dx (1 +√
x)4. (b) ∫
1 0
sin−1(
√x) dx.
Solution:
Answer: Must show clearly the steps of substitution and integration by parts (a) Method-1:
– (2%) Set y =√
x ⇒ dx = 2ydy ⇒∫
1 0
√x (1+√
x)4dx =∫
1 0
2y2 (1+y)4dy (1%) = 2∫
1 0
1−(1−y2)
(1+y)4 dy = 2∫
1 0[ 1
(1+y)4 − (1−y)
(1+y)3]dy (2%) = 2∫
1 0[ 1
(1+y)4 −2−(1+y)
(1+y)3 ]dy = 2∫
1 0[ 1
(1+y)4 − 2
(1+y)3 + 1
(1+y)2]dy (2%) = −2[3(1+y)1 3 − 1
(1+y)2 + 1
(1+y)]10 = 1
12 ⋯ Method-2:
– (2%) Set y = 1 +√
x ⇒ dx = 2(y − 1)dy (3%) ⇒∫
1 0
√x (1+√
x)4dx =∫
2 1
2(y−1)2
y4 dy = 2∫
2
1[y12 −y23 +y14]dy (2%) = −2[1y − 1
y2 + 1
3y3]21 = 1
12 ⋯ (b) Method-1:
– (2%) Set y =√
x ⇒ dx = 2ydy ⇒∫
1 0 sin−1(
√x)dx =∫
1
0 sin−1(y) 2ydy (2%) =∫
1
0 sin−1(y) (y2)′dy = [y2sin−1(y)]10− ∫01 y
2
√ 1−y2dy (2%) = π2 − ∫011−(1−y
2)
√
1−y2 dy = π2 − ∫01[√1
1−y2 +
√
1 − y2]dy (2%) = π2 − [sin−1(y)]10+ ∫0π/2cos2(θ)dθ where y = sin θ (1%) = 12[θ +sin(2θ)2 ]π/20 = π
4 ⋯ Method-2:
– (4%) Set y = sin−1(
√x) ⇒ x = sin2(y) and dx = sin(2y)dy (2%) ⇒∫
1 0 sin−1(
√x)dx =∫
π/2
0 y sin(2y)dy (2%) =∫
π/2
0 y (−cos(2y)2 )′dy =−12 [y cos(2y)]π/20 +1
2∫
π/2
0 cos(2y)dy (1%) = π4 +14[sin(2y)]π/20 = π4 ⋯
Page 2 of 10
3. (13 pts)
(a) Decompose x2+4x + 5
(x + 1)2(x2+2x + 3) into partial fractions.
(b) Evaluate the indefinite integral ∫
x2+4x + 5
(x + 1)2(x2+2x + 3)dx.
Solution:
(a) Let
x2+4x + 5
(x + 1)2(x2+2x + 3) = A x + 1 +
B (x + 1)2 +
Cx + D
x2+2x + 3. (2 pts) By clearing the denominator we have
x2+4x + 5 = A(x + 1)(x2+2x + 3) + B(x2+2x + 3) + (Cx + D)(x + 1)2. (1 pt) By comparing the coefficients, we have
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
A +C =0,
3A +B +2C +D =1, 5A +2B +C +2D = 4,
3A +3B +D =5,
(4 pts) ⇒
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
A =1, B =1, C = −1, D = −1.
(1 pts)
Therefore, we get
x2+4x + 5
(x + 1)2(x2+2x + 3) = 1
x + 1+ 1 (x + 1)2 +
−x − 1
x2+2x + 3. (1 pt) (b) By (a) we have the indefinite integral
x2+4x + 5
(x + 1)2(x2+2x + 3)dx =∫ [ 1
x + 1+ 1
(x + 1)2 − x + 1
x2+2x + 3]dx (1 pt)
=ln ∣x + 1∣ − 1 x + 1 −
1
2ln(x2+2x + 3) + C. (3 pts)
4. (16 pts)
(a) Find the orthogonal trajectories of the family of curves y = K ⋅ tan2x, where K is an arbitrary constant.
(b) Solve the initial value problem : x(x + 1)dy
dx+y = (x + 1)2sin x cos x, y (π 2) =0.
Solution:
(a) (i) The slope of each point on the family of curves is dy
dx =K ⋅ 2 tan x ⋅ sec2x (2 pts) = y
tan2x⋅2 tan x ⋅ sec2x
=2y ⋅ sec2x ⋅ cot x = 2y
sin x ⋅ cos x. (2 pts) (ii) Solve dy
dx =
−sin x ⋅ cos x
2y . (2 pts) We have
dy dx =
−sin x ⋅ cos x 2y
⇒ ∫ 2ydy = ∫ (− sin x ⋅ cos x)dx = −1
2 ∫ sin(2x)dx
⇒y2= cos(2x)
4 +C. (2 pts)
The orthogonal trajectories of the family of curves y2 =
cos(2x) 4 +C.
(b) The standard form of the differential equation is dy
dx + 1
x(x + 1)⋅y = x + 1
x ⋅sin x ⋅ cos x.
Since
e∫ x(x+1)1 dx=e∫ (1x−x+11 )dx=eln ∣x+1x ∣= ∣ x x + 1∣, we can take the integration factor as I(x) = x+1x . (3 pts)
Then we have
I(x) ⋅ [dy dx +
1
x(x + 1) ⋅y] = I(x) ⋅ [x + 1
x ⋅sin x ⋅ cos x]
⇒ [ x
x + 1⋅y]′= [I(x) ⋅ y]′= 1
2sin(2x)
⇒ x
x + 1 ⋅y = I(x) ⋅ y =∫ 1
2sin(2x)dx = −1
4 cos(2x) + C
⇒y = x + 1 x ⋅ [
−1
4 cos(2x) + C]. (3 pts) Since y(π2) =0, we get −14 =C. So the solution is y = x + 1
x ⋅ [
−1
4 cos(2x) −1
4]. (2 pts)
Page 4 of 10
5. (14 pts) Let C be a curve whose parametrisation is given by
⎧⎪
⎪
⎨
⎪⎪
⎩
x = etcos t
y = etsin t with 0 ≤ t ≤ π.
(a) Find the arclength of C.
(b) Find dy
dx in terms of t and find the point Q in x-y coordinates at which the tangent to C is perpendicular to the x-axis.
Solution:
(a)
dx
dt =et(cos t − sin t) dy
dt =et(sin t + cos t).
Arc length is then computed by the formula
∫
π 0
√ (dx
dt)2+ (dx
dt)2 dt =∫
π 0
√
2et dt =
√
2(eπ−1).
Grading Guideline. Arc length formula (including lower and upper limits) 3pt differentiation computation 2pt
integral computation 2pt.
(b) With the computation above
dy
dx =sin t + cos t cos t − sin t.
Vertical tangents occur, if exists, when cos t − sin t = 0, that is t = π4. We need to check if it is indeed a vertical tangent. We can either compute
±
lim
t→π4
dy dx = ±∞
or just say the numerator is non-zero. Finally, converting it into cartesian coordinates, we get Q = eπ4(√1
2,√1
2).
Grading Guideline. dydx (2pt= chain rule (1pt) + computation (1pt)) vertical tangent (3pt)=denominator=0 (1pt)+t-value(1pt)+limit(1pt) converting to cartesian coordinates (1pt).
6. (10 pts) Consider two polar curves C1 and C2 defined by
C1 ∶r2 =4 sin(4θ), C2 ∶r2 =4 cos(2θ).
Find the area of the region that lies inside both C1 and C2.
Solution:
By symmetry, we compute the intersection of curves C1 and C2 when 0 ≤ θ ≤ π4.
{ r2=4 sin(4θ)
r2 =4 cos(2θ), ⇒4 cos(2θ) ⋅ (sin(2θ) − 1) = 0 (2 pts)
⇒cos(2θ) = 0 or sin(2θ) = 1 2
⇒2θ = π 2 or π
6 ⇒θ = π 4 or π
12. (3 pts) So we have
Area of the region inside C1 and C2
=2 ⋅ [∫
π 12
0
1
2⋅4 sin(4θ)dθ +∫
π 4 π 12
1
2⋅4 cos(2θ)dθ] (3 pts)
= (−cos(4θ)) ∣
π 12
0
+(2 sin(2θ)) ∣
π 4
π 12
= (−1
2+1) + 2(1 − 1 2) = 3
2. (2 pts)
Page 6 of 10
7. (10 pts) The ellipse
x2+y2 5 =1
is rotated about the x-axis to form a surface called an oblate spheroid. Find the surface area of this oblate spheroid.
Solution:
Marking scheme.
(2M) *Evaluation of ds (2M) ** Integrand (1M) Integration limits
(2M) ***Convert the integral into C ⋅∫ sec3θdθ by a suitable substitution (2M) ****Correct evaluation of ∫ sec3θdθ
(1M) Correct answer Partial credits.
(*) 1M is awarded as long as the candidate attempts to compute ds (**) Here a candidate will receive at most 1M for
making mistakes up to a constant factor
mentioning∫ 2πy ⋅ ds (or any equivalent form)
(***) 1M is taken away if a student makes mistakes in the integration limits after conducting a substitution
(****) No derivation is required. At most 1M can be awarded to a candidate with an incorrect evaluation who (1) attempts to derive ∫ sec3θdθ or (2) makes minor/obvious typos.
Sample Solution 1.
One can obtain an ellipsoid by using the upper half of the ellipse y =
√
5(1 − x2)for which dy
dx =
√
5 ⋅ −x
√
1 − x2 and
√
1 + (dy dx)
2
=
√
1 + 5x2 1 − x2 =
√
1 + 4x2 1 − x2
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(2M )
By symmetry, the surface area is given by
2 ⋅∫
1 0
2π
√
5(1 − x2)
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
y
√
1 + 4x2 1 − x2 dx
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
ds
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(3M )
=4
√ 5π∫
1 0
√
1 + 4x2dx
To evaluate ∫
√
1 + 4x2dx, we substitute x = 12tan θ, then
∫
√
1 + 4x2dx = 1
2∫ sec3θdθ (2M)
= 1
4(sec θ tan θ + ln ∣ sec θ + tan θ∣) + C (2M)
1 2x
√
1 + 4x2 ln(2x +
√
1 + 4x2 C
Hence, the surface area equals to √
5π(2√
5 + ln(2 +√
5)). (1M) Sample Solution 2 (Parametric way).
The ellipse can be parametrised by x = cos t and y =√ 5 sin t.
Then
√ (
dx dt)
2
+ ( dy dt)
2
=
√
sin2t + 5 cos2t =√
1 + 4 cos2t
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(2M )
.
By symmetry, the surface area thus equals to 2 ⋅∫
π/2 0
2π√ 5 sin t
´¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¶
y
√
1 + 4 cos2tdt
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
ds
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(3M )
=4√ 5π∫
π/2 0
sin t
√
1 + 4 cos2tdt
To evaluate ∫ sin t
√
1 + 4 cos2tdt, we let 2 cos(t) = tan θ
∫ sin t
√
1 + 4 cos2tdt = −1
2∫ sec3θdθ (2M)
= − 1
4(sec θ tan θ + ln ∣ sec θ + tan θ∣) + C (2M)
= − 1
4(2 cos t
√
1 + 4 cos2t + ln(2 cos t +
√
1 + 4 cos2t)) + C
Hence, the surface area equals to √
5π(2√
5 + ln(2 +√
5)). (1M) Alternative form of the final answer : 10π +√
5π ln(2 +√ 5).
Page 8 of 10
8. (13 pts) Consider the region bounded by the curve y = 1
√
e2x+1, the x-axis and x = 1.
(a) Find the volume of the solid obtained by revolving the region about the x-axis.
(b) Student A claims that ‘The volume of the solid obtained by revolving the region about the y-axis is also finite.’. Give a proof to Student A’s claim by using the comparison test for improper integrals.
Solution:
(a)
Marking scheme.
(2M) *Integrand
(1M) Integration limits
(1M) Writing down the definition of improper integrals (3M) **Using a suitable substitution to evaluate the integral (1M) Correct answer
Partial credits.
(*) At most 1M is taken away for any missing/extra scalar factors.
(**) A candidate will receive
2M for having done a correct substitution but incomplete evaluation of the integral
3M as long as he/she integrates ∫
b a
1
e2x+1dx correctly - does not matter his/her choice of integration limits.
Sample solution.
By disk method, the volume of the solid equals to
∫
∞ 1
π ⋅ 1
e2x+1dx
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(2+1M )
= lim
t→∞∫
t 1
π ⋅ 1
e2x+1dx
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
.
To evaluate ∫ 1
e2x+1dx, we let u = e2x. Then
∫ 1
e2x+1dx = 1 2∫
du u(u + 1) =
1 2∫
1 u −
1
u + 1du = 1
2ln ∣ u
u + 1∣ +C = 1
2ln ∣ e2x e2x+1∣ +C
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(3M )
.
Hence, the volume equals to
π e2t e2 1
Alternative form of the final answer : −π
2ln e2 e2+1 =
π
2lne2+1 e2 =π ln
√ e2+1
e2 . (b)
Marking scheme.
(2M) *Setting up the correct integral (1M) Any correct upper bound for √ x
e2x+1
(2M) **Correct argument using the comparison test Partial credits.
(*) At most 1M is taken away for any missing/extra scalar factors.
(**) 1M can be awarded to candidates with an incorrect upper bound, but with some attempts of an argument using the comparison test.
In an extreme situation that a candidate didn’t set up a correct integral (not of the form C ⋅∫ √e2xx+1dx) but demonstrated some understandings of the comparison test, at most 1M will be awarded.
Sample solution.
By shell method, the volume of the solid equals to ∫
∞ 1
2πx ⋅ 1
√
e2x+1dx. (2M)
Since 0 ≤ x
√
e2x+1 ≤ x
ex (1M)
and ∫
∞ 1
x
exdx = lim
t→∞∫
t 1
x
exdx = lim
t→∞(−te−t−e−t+2e−1) =2e−1 is convergent, (2M) the comparison test implies that ∫
∞ 1
x
√
e2x+1dx is convergent. This verifies the claim of Student A.
Alternative argument.
By shell method, the volume of the solid equals to ∫
∞ 1
2πx ⋅ 1
√
e2x+1dx. (2M)
Since 0 ≤ x
√
e2x+1 ≤ 1
x2 for x sufficiently large (1M)
and ∫
∞ 1
1
x2dx converges as a p-integral with p > 1, (2M)
the comparison test implies that ∫
∞ 1
x
√
e2x+1dx is convergent. This verifies the claim of Student A.
Remark. 1/x2 can be replaced by 1/xp for any p > 1 in the above argument.
Page 10 of 10
