2M for the term by replacing t with 2x2 ˆ 2M for the term coming from the Chain rule (b) By chain rule, f′(x

1101模模模組組組13-16班班班 微微微積積積分分分2 期期期考考考解解解答答答和和和評評評分分分標標標準準準 1. Let g(x) =∫

2x² 4

t

1 + t³dt and f (x) = ln(2 + g(x)).

(a) (4%) Find g^′(x). (b) (4%) Find f^′(

√ 2).

Solution:

(a) By FTC, g^′(x) = 2x² 1 + 8x⁶

´¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¶

(2M)

⋅ 4x

(2M)¯ Marking scheme :

ˆ 2M for the term by replacing t with 2x²

ˆ 2M for the term coming from the Chain rule (b) By chain rule, f^′(x) = g^′(x)

2 + g(x)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(2M)

Since g(√ 2) = 0

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

, we have f^′(

√ 2) =g^′(

√2)

2 =

8√ 2 65

²(1M)

Marking scheme :

ˆ 2M for differentiating f(x) correctly

ˆ 1M for knowing g(√ 2) = 0

ˆ 1M for answer

2. Evaluate the following integrals.

(a) (8%)∫ (1 − x

2)

3

2⋅x³dx (b) (10%) ∫

e^x

(e^x+1)²(e^2x+1)dx

Solution:

(a) Method1:

Let x = sin(θ).

Then dx = cos(θ)dθ and (1 − x²)

3

2x³= (1 − sin(θ))³²sin³(θ) = cos³(θ) sin³(θ).

Thus

∫ (1 − x

2)

3

2x³dx =∫ cos³(θ) sin³(θ) cos(θ)dθ

= ∫ cos⁴(θ) sin³(θ)dθ =∫ cos⁴(θ) sin²(θ) sin(θ)dθ =∫ cos⁴(θ)(1 − cos²(θ)) sin(θ)dθ Let u = cos(θ). Then du = − sin(θ)dθ and sin(θ)dθ = −du.

So

∫ cos⁴(θ)(1 − cos²(θ)) sin(θ)dθ

= ∫ u⁴(1 − u²)(−du) =∫ −u⁴+u⁶du = −u⁵ 5 +

u⁷

7 +C = −cos⁵(θ)

5 +

cos⁷(θ)

7 +C

Recall that sin(θ) = x. We have cos(θ) =√

1 − x². Thus −cos⁵(θ)

5 +

cos⁷(θ)

7 = −

(1 − x²)

5 2

5 +

(1 − x²)

7 2

7 and

∫ (1 − x

2)

3

2x³dx = −(1 − x²)

5 2

5 +

(1 − x²)

7 2

7 +C.

1 point for the correct substitution x = sin(θ) and dx = cos(θ)dθ, 1 point for getting∫ (1 − x

2)

3

2x³dx =∫ cos⁴(θ) sin³(θ)dθ 1 point for the substitution u = cos(θ) and du = − sin(θ)dθ,

1 point for getting∫ cos⁴(θ)(1 − cos²(θ)) sin(θ)dθ =∫ −u⁴+u⁶du 1 point for ∫ −u⁴+u⁶du = −u⁵

5 + u⁷

7 +C 1 point for getting −cos⁵(θ)

5 +

cos⁷(θ)

7 +C

1 point for getting cos(θ) =

√ 1 − x² 1 point for the final answer∫ (1 − x

2)

3

2x³dx = −(1 − x²)

5 2

5 +

(1 − x²)

7 2

7 +C.

Method2:

Let u = 1 − x².

Then du = −2xdx, xdx = −1

2du, x²=1 − u and (1 − x²)

3

2x³dx = (1 − x²)

3

2x²xdx =∫ u³²(1 − u)(−1

2)du = −u³² +u⁵²

2 du.

∫ (1 − x

2)

3

2x³dx =∫

−u³²+u⁵²

2 du

= − u⁵²

5 + u⁷²

7 +C = −(1 − x²)

5 2

5 +

(1 − x²)

7 2

7 +C.

1 point for the correct substitution u = 1 − x² 1 point for du = −2xdx,

1 point for getting x²=1 − u 2 point for getting∫ (1 − x

2)

3

2x³dx =∫ u³²(1 − u)(−1 2)du

2 point for getting∫

−u³²+u⁵²

2 du = −u⁵² 5 +

u⁷² 7 +C 1 point for the final answer∫ (1 − x

2)

3

2x³dx = −(1 − x²)

5 2

5 +

(1 − x²)

7 2

7 +C.

Method3:

Let u = x²and dv = (1 − x²)

3

2x. Then du = 2xdx, v =∫ (1 − x

2)

3

2xdx = −1

5(1 − x²)

5 2.

∫ (1 − x

2)

3

2x³dx =∫ x²(1 − x²)

3 2xdx

=x²(−

1

5(1 − x²)

5 2) +

1

5∫ (1 − x

2)

5 22xdx

= − x²

5 (1 − x²)

5 2−

1 5⋅

2

7⋅ (1 − x²)

7 2 +C

= − x²

5 (1 − x²)

5 2−

2(1 − x²)

7 2

35 +C.

Remark: We can simplify the answer

− x²

5 (1 − x²)

5 2−

2(1 − x²)

7 2

35 = (1 − x²)

5 2(−

x² 5 −

2(1 − x²)

35 )

= (1 − x²)

5 2(−

1 5+

1 − x²

5 −

2(1 − x²)

35 )

= −

(1 − x²)

5 2

5 +

(1 − x²)

7 2

7 1 point for u = x², du = 2xdx,

1 point for getting dv = (1 − x²)

3 2x 2 point for getting v = −1

5(1 − x²)

5 2

2 point for getting∫ (1 − x

2)

3

2x³dx = x²(−1

5(1 − x²)

5 2) +1

5∫ (1 − x

2)

5 22xdx 2 point for the final answer∫ (1 − x

2)

3

2x³dx = −x²

5 (1 − x²)

5 2 −

2(1 − x²)

7 2

35 +C..

(b) Let u = e^x. Then du = e^xdx and∫

e^x

(e^x+1)²(e^2x+1)dx =∫

1

(u + 1)²(u²+1)du.

The function 1

(u + 1)²(u²+1) has a partial fraction decomposition of the form 1

(u + 1)²(u²+1) = a u + 1+

b (u + 1)² +

cu + d u²+1. Multiplying by (u + 1)²(u²+1), we have

1 =a(u + 1)(u²+1) + b(u²+1) + (cu + d)(u + 1)²

=a(u³+u + u²+1) + bu²+b + (cu + d)(u²+2u + 1)

=au³+au²+au + a + bu²+b + cu³+2cu²+cu + du²+2du + d

=(a + c)u³+ (a + b + 2c + d)u²+ (a + c + d)u + a + b + d

Comparing the coefficient, we have a + c = 0, a + b + 2c + d = 0, a + c + d = 0, a + b + d = 1. From a + c = 0, we have c = −a. From a + c + d = 0 and a + c = 0, we have d = 0. From a + b + d = 1 and d = 0, we have b = 1 − a.

From a + b + 2c + d = 0, we have a + 1 − a − 2a + 0 = 0, 2a = 1 and a = 1

2. Thus b = 1 − a = 1 −1 2 =

1

2, c = −a = −1 2. 1

(u + 1)²(u²+1)= 1 2

1 u + 1+

1 2

1 (u + 1)²−

1 2

u u²+1. Remark: one can also plug in u = −1 first to get b = 1

2 first and then determine other coefficients accordingly.

Thus

∫

1

(u + 1)²(u²+1)du =∫ 1 2

1 u + 1+

1 2

1 (u + 1)²−

1 2

u u²+1du

= 1

2ln ∣u + 1∣ −1 2

1 u + 1−

1

4ln ∣u²+1∣ + C and

∫

e^x

(e^x+1)²(e^2x+1)dx = 1

2ln ∣e^x+1∣ −1 2

1 e^x+1 −

1

4ln ∣e^2x+1∣ + C.

1 point for the correct substitution u = e^x and du = e^xdx, 1 point for getting∫

e^x

(e^x+1)²(e^2x+1)dx =∫

1

(u + 1)²(u²+1)du 1 point for setting up 1

(u + 1)²(u²+1)= a u + 1+

b (u + 1)² +

cu + d u²+1 4 points for getting 1

(u + 1)²(u²+1)= 1 2

1 u + 1+

1 2

1 (u + 1)² −

1 2

u

u²+1 (each coefficient 1 point) 2 points for getting∫

1

(u + 1)²(u²+1)du = 1

2ln ∣u + 1∣ −1 2

1 u + 1−

1

4ln ∣u²+1∣ + C 1 point for the final answer∫

e^x

(e^x+1)²(e^2x+1)dx = 1

2ln ∣e^x+1∣ −1 2

1 e^x+1 −

1

4ln ∣e^2x+1∣ + C

3. Find the volume of the solid obtained by rotating the following regions about the specific axes.

(a) (8%) Region bounded by y =

√x

2 and y = x²

4 ; rotate about the x-axis.

(b) (8%) Region bounded by y = x²e^x2, x = 0, x = 1 and y = 0 (See Figure below); rotate about the y-axis.

Solution:

(a) The curves y = f (x) =

√x

2 and y = g(x) = x²

4 intersect when

√x 2 =

x²

4 . This implies x 2 =

x⁴

16, x⁴−8x = 0 and x(x³−8) = 0. So x = 0 and x = 2. We can plug in x = 1

2 and get f (1 2) =

√ 1 4 =

1

2 and g(1 2) =

1 16. So we know that

√x 2 ≥

x²

4 on [0, 2]. The cross section has the shape of a washer (an annular ring) with inner radius x²

4 and outer radius

√x

2. The area of the cross-sectional is A(x) = π(

√x

2)²−π(x²

4 )²=π(x 2 −

x⁴ 16).

So the volume is∫

2 0

A(x)dx =∫

2 0

π(x 2 −

x⁴

16)dx = π(x² 4 −

x⁵ 80)∣

2 0=π(4

4− 32 80) =

3π 5 . 2 points for finding the intersecting x-coordinate x = 0 and x = 2 ,

1 point for showing or explaining

√x 2 ≥

x²

4 on [0, 2]

2 points for finding area of cross section A(x) = π(x 2 −

x⁴ 16) 1 points for setting up volume ∫

2 0

A(x)dx =∫

2 0

π(x 2 −

x⁴ 16)dx 2 points for the definite integral final answer∫

2 0

π(x 2−

x⁴

16)dx = π(x² 4 −

x⁵ 80)∣

2 0=π(4

4− 32 80) =

3π 5 (b) We can use the shell method to find the volume. So the volume is∫

1 0

2πx ⋅ x²e^x2dx = 2π∫

1 0

x³e^x2dx.

We find ∫ x³e^x2dx first. We write ∫ x³e^x2dx as ∫ x²xe^x2dx. Let u = x² and dv = xe^x2dx. Then du = 2x and v =∫ xe^x2dx = e^x2

2 . Using integration by parts, we get∫ x²xe^x2dx = x²e^x2 2 − ∫

e^x2

2 ⋅2xdx = x²e^x2

2 − ∫ xe^x

2dx =x²e^x2

2 −

e^x2 2 +C.

Thus 2π∫

1 0

x³e^x2dx = 2π(x²e^x2

2 −

e^x2

2 )∣¹₀=π(x²e^x2−e^x2)∣¹₀=π(e − e − (0 − 1)) = π ⋅ 1 = π.

So the volume is π.

4 points for setting up the volume formula 2π∫

1 0

x³e^x2dx , 2 points for indefinite integral∫ x³e^x2dx = x²e^x2

2 −

e^x2 2 +C 2 points for finding the definite integral Volume= 2π∫

1 0

x³e^x2dx = 2π(x²e^x2

2 −

e^x2

2 )∣¹₀=π(x²e^x2−e^x2)∣¹₀= π(e − e − (0 − 1)) = π ⋅ 1 = π

4. The latest model of XPhone has just released. It is very popular that there is a shortage in supply.

(a) The XPhone is priced at p = D(q) = 64(4 + 2^−(q−220)) dollars when q-thousand units are demanded. While the supply function is given by p = S(q) = 256 + 4^10q , only q = 10-thousand units can be supplied.

(i) (6%) Find the total surplus in this case.

(ii) (2%) How many additional units of XPhone need to be supplied to maximize the total surplus ? Reminder. The total surplus (TS) is defined by TS(¯q) =∫

¯ q

0 [D(q) − S(q)] dq.

(b) The waiting time for customers to receive their new XPhone is a continuous random variable whose probability density function equals

f (x) =

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

0 if x < 0

ln 2

5 ⋅2^−x/5 if x ≥ 0 (x in days).

(i) (8%) What is probability for a customer to wait for more than a week ?

(ii) (6%) The newly released earphone ‘XPod’ is also highly demanded and its waiting time is a random variable given by Y = 0.5X + 1. Find the probability density function for Y .

Solution:

(a) (i) The total surplus is

∫

10

0

±(1M)

64 ⋅ 2^−(q−220)−4^10q

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(2M)

dq =

⎡⎢

⎢⎢

⎢⎢

⎢⎢

⎢⎣

−1280

ln 2 ⋅2^−(q−220)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

− 10 ln 4⋅4^10q

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

⎤⎥

⎥⎥

⎥⎥

⎥⎥

⎥⎦

q=10

q=0

= −1280 ⋅ 2⁻²⁵ ln 2 − 40

ln 4+1280 ⋅ 2¹⁰¹ ln 2 + 10

ln 4

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

=

1280(2¹⁰¹ −2⁻²⁵) −15 ln 2

Marking scheme :

ˆ 1M for correct integration limits

ˆ 2M for correct integrand

ˆ 1M for correct anti-derivative of 2^−q−2)20

ˆ 1M for correct anti-derivative of 4^q/10

ˆ 1M for correct answer

(ii) The total surplus is maximized at equilibrium quantity q^⋆at which the quantity demanded and supplied agree. Equate

64(4 + 2⁻⁽

q⋆−2 20 )

) =256 + 4^q⋆10

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

⇒ q^⋆= 24.4

±(1M)

So 24.4 − 10 = 14.4-thousand units need to be produced additionally. Marking scheme :

ˆ 1M for setting D(q) = S(q)

ˆ 1M for correct answer

(b) (i) The required probability equals to

∫

∞ 7

±(2M)

ln 2 5 ⋅2^−x/5

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

dx^def= lim

t→∞∫

t 7

ln 2

5 ⋅2^−x/5dx

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

= lim

t→∞

⎡⎢

⎢⎢

⎢⎢

⎣

−2^−x/5

´¹¹¹¹¹¹¹¸¹¹¹¹¹¹¶

(2M)

⎤⎥

⎥⎥

⎥⎥

⎦

t

7

= lim

t→∞(−2^−t/5+2^−7/5) = 1 2^7/5

±(2M)

Marking scheme :

ˆ 2M for correct integration limits

ˆ 1M for correct integrand

ˆ 1M for definition of improper integral

ˆ 2M for correct antiderivative of f(x)

ˆ 2M for correct answer

(ii) The distribution function of Y is (for y ≥ 1) FY(y) = P(Y ≤ y)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

=P(X ≤ 2y − 2) = ∫

2y−2 0

ln 2

5 ⋅2^−x/5dx

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

By differentiating the distribution function, we obtain the density functiin of Y :

fY(y) =

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

0 if y < 1 (1M)

2 ln 2

5 ⋅2^{−(2y−2)/5} if y ≥ 1 (3M) (y in days).

Marking scheme :

ˆ 1M for definition of probability distribution function

ˆ 1M for transforming into P(X ≤ 2y − 2)

ˆ 1M for specifying fY(y) = 0 for y < 1 (or y ≤ 1)

ˆ 3M for correct density function when y ≥ 1.

5. Solve, for y = y(x), the following differential equations.

(a) (10%) dy

dx =x²y²−x²with y(0) = 2. (b) (10%) dy

dx=x − 2y + 1 with y(0) =1 8.

Solution:

(a) We rewrite the differential equation as y^′= (y²−1)x². Case 1: y = ±1. y^′=0. Since y(0) = 2, it is impossible that the case occurs.

Case 2: y ≠ ±1. It implies that

∫ dy

y²−1 = ∫ x²dx(1 point)

⇒ 1 2∫

1 y − 1−

1

y + 1dy =1

3x³(1 point)

⇒ 1

2ln ∣y − 1 y + 1∣ =

1

3x³+C(2 points)

⇒ y − 1

y + 1 =Ae^2x33 (2 points)

⇒ y(x) =1 + Ae^2x33

1 − Ae^2x33 where A ∈ R.(1 point) Since y(0) = 2, 1 + A

1 − A=2 ⇒ A =1

3.(2 points)

Hence the general solution of the differential equation is

y(x) =1 +¹₃e^2x33 1 −¹₃e^2x33

=

3 + e^2x33 3 − e^2x33

.(1 point)

(b) The integration factor I(x) is

I(x) = e^{∫ 2dx}=e^2x.(2 points) Multiply both sides of y^′+2y = x + 1 by e^2x. We have that

(e^2xy)^′=e^2x(x + 1)(1 point)

⇒ e^2xy =∫ e^2x(x + 1)dx(1 point)

⇒ e^2xy =∫ xe^2xdx +∫ e^2xdx

⇒ e^2xy =x 2e^2x− ∫

1

2e^2xdx +∫ e^2xdx(2 points)

⇒ e^2xy =x 2e^2x+

1

4e^2x+C(1 point)

⇒ y =x 2+

1

4+Ce^−2x.(1 point) Since y(0) = 1

4+C =1

8, C =−1

8 (1 point). Hence y = x 2 +

1 4−

1

8e^−2x(1 point).

6. The parametric equations of a trochoid (see figure) are given by

x = θ − 2 sin(θ) and y = 1 − 2 cos(θ) where 0 ≤ θ ≤ 2π.

(a) (8%) Find the equation of the tangent to the curve at θ =π 4. (b) (2%) Find the values of θ at which the curve passes the x-axis.

(c) (6%) Find the area of the region under the curve and above the x−axis.

Solution:

(a) We have that dx

dθ =1 − 2 cos(θ) (1 point) and dy

dθ =2 sin(θ) (1 point). We have that dy

dx =

2 sin(θ)

1 − 2 cos(θ).(2 points)

Thus dy

dx∣_θ=π/4 =

2 sin(θ)

1 − 2 cos(θ)∣_θ=π/4 =

√2 1 −√

2 = −(

√

2 + 2)(2 points). Hence the tangent line is y − (1 − 2 cos(π/4)) = −(√

2 + 2)(x − π/4 + 2 sin(π/4) ⇒ y − 1 +√

2 = −(√

2 + 2)(x − π/4 +√

2)(2 points).

(b) The values of θ at which the curve passes the x-axis are 1 − 2 cos(θ) = 0(1 point). It implies that θ = π

3(0.5 point) and θ =5π

3 (0.5 point).

(c) The area is

A = ∫ ydx

= ∫

5π 3 π 3

(1 − 2 cos θ)d(θ − 2 sin(θ))(1 point)

= ∫

5π 3 π 3

(1 − 2 cos θ)(1 − 2 cos θ)dθ(1 point)

= ∫

5π 3 π 3

(1 − 4 cos θ + 4 cos²θ)dθ(1 point)

= [θ − 4 sin θ]

5π π3 3 + ∫

5π 3 π 3

2 cos(2θ) + 2dθ(1 point)

= 4π

3 +4√

3 + [2θ + sin(2θ)]

5π π3

3 (1 point)

= 4π

3 +4√ 3 +8π

3 −

√

3 = 4π + 3√

3(1 point).