The areas of these three parts are represented by the following integrals:

1031微微微甲甲甲01-04班班班期期期末末末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. (8%) Evaluate lim

x→0

∫

tan x x

√ 1 + t³dt

x³ . (You may use the Mean Value Theorem for Integrals.) Solution:

Method1.

By the Mean Value Theorem, there is c between x and tan x such that

∫

tan x x

√1 + t³dx tan x − x =

√

1 + c³ (3pts)

By the Squeeze Theorem,

xlim→0x = 0, lim

x→0tan x = 0 imply

xlim→0c = 0 (1pt)

Now we evaluate

xlim→0

∫

tan x x

√1 + t³dx

x³ = lim

x→0

∫

tan x x

√1 + t³dx tan x − x

tan x − x x³

= lim

x→0

√

1 + c³⋅tan x − x x³

= lim

x→0

√

1 + c³⋅sec²x − 1 3x²

= lim

x→0

√

1 + c³⋅tan²x 3x²

= lim

x→0

√

1 + c³(sin x x )² 1

3 cos²x

=

√

1 + 0³⋅1²⋅ 1 3⋅ 1²

= 1

3 (4pt)

Method2.

By the Fundamental Theorem of Calculus,

xlim→0

∫

tan x x

√1 + t³dx

x³ = lim

x→0

√

1 + tan³x sec²x −√ 1 + x³

3x² (3pts)

= lim

x→0

3 tan²x sec²x 2√

1+tan³x sec²x +√

1 + tan³x⋅ 2 tan x sec²x − ^3x2

2√ 1+x³

6x (2pts)

= lim

x→0(1 4

tan x sec⁴x

√

1 + tan³x tan x

x +1 3

√

1 + tan³xtan x

x sec²x − x 4√

1 + x³)

= 1

3 (3pts)

Remark.

No point if you are with intent to evaluate∫

√

1 + t³dt.

Calculation error: (−1pt) to (−3pts) each part.

2. (8%) Let F (x) =∫

x 0 (∫

u³

0 f (t)dt) du and G(x) =∫

x³

0 f (u)(x −√³

u)du, x ≥ 0. Show that F (x) = G(x) for x ≥ 0.

Solution:

From

dF (x)

dx = d

dx∫

x 0 (∫

u³

0 f (t)dt) du

= ∫

x³

0 f (t)dt (2pts) dG(x)

dx = d

dx∫

x³

0 f (u)(x −√³ u)du

= d

dx(x∫

x³

0 f (u)du) − d dx∫

x³ 0 f (u)√³

udu (2pts)

= (∫

x³

0 f (u)du + x⋅ f (x³)⋅3x²) −f (x³)

√3

x³⋅3x²

= ∫

x³

0 f (u)du (3pts) d

dx(F (x) − G(x)) = d

dxF (x) − d dxG(x)

= ∫

x³

0 f (t)dt −∫

x³

0 f (u)du = 0 we have

F (x) − G(x) = C for some constant C.

Since

F (0) = ∫

0 0 (∫

u³

0 f (t)dt) du = 0 G(0) = ∫

0³

0 f (u)(0 −√³

u)du = 0 we have C = 0. (1pt)

Therefore, F (x) = G(x).

Remark.

Wrong calculation dG(x) dx = d

dx∫

x³

0 f (u)(x −√³

u)du = f (x³)(x −√³

x³)= 0 : (−5pts).

Calculation error: (−1pt) or (−3pts) each part.

3. (16%) Three of these six antiderivatives are elementary. Compute them.

(a)∫ x cos xdx (b)∫

cos x x dx (c)∫

x

ln xdx (d)∫

ln(x²) x dx (e)∫

√x − 1√ x√

x + 1dx (f) ∫

√x − 1√

x + 1xdx

Solution:

(a)

∫ x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x + C.

(d) Put u = ln (x²), du =2

xdx. So,

∫

ln (x²) x dx =∫

u

2 du =u²

4 +C =(ln (x²))

2

4 +C.

(f) Put u = x²−1, du = 2x dx. So,

∫

√x − 1√

x + 1x dx =∫ x

√

x²−1 dx =∫

√u

2 du =u³²

3 +C =(x²−1)³² 3 +C.

評分標準：

(1) 僅完成能夠得到答案的變數變換、分部積分或三角代換之步驟，得2分。

(2) 題目(f)使用三角代換但未將tan (sec⁻¹x)(或其他類似表示式)換成根式，扣2分。

(3) 每次出現少寫積分常數、未將變數代回成x、化簡對數未加絕對值等零星錯誤，但其餘答案皆無誤，扣1分。

(4) 題目(b)、(c)、(e)的任何作答過程與此題計分無關。

4. (16%) (a) Evaluate the integral ∫ dx x√

x⁶−1. (b) Evaluate the integral∫

(e^3x+1) (e^2x+1)²dx.

Solution:

(a) [6pts]∫ dx x√

x⁶−1 Solution.

∫ dx x√

x⁶−1

= ∫

x⁵dx x⁶√

x⁶−1 (letu =√

x⁶−1du = 3(x⁶−1)⁻¹²x⁵dx)

= 1 3∫

udu (u²+1)u

= 1 3∫

du

(u²+1) [+4pts]

= 1

3tan⁻¹√

x⁶−1 + C [+2pts]

note:

∫ dx x√

x⁶−1 (let x³=sect3x²=sec t tan tdt)

= ∫

sec t tan tdt sec t∣ tan t∣ (

√ x⁶−1 =

√

tan²t = ∣ tan t∣) [+5pts]

(b) [10pts]∫

(e^3x+1)dx (e^2x+1)² Solution.

∫

(e^3x+1)dx

(e^2x+1)² (letu = e^xdu = e^xdx)

= ∫

(u³+1)du

u(u²+1)² [+2pts]

= ∫ 1

u− u − 1

u²+1− u + 1

(u²+1)²du [+3pts]

= ln ∣u∣ + tan⁻¹u − 1

2ln(u²+1) +1 2

1

u²+1 − ∫ 1

(u²+1)²du [+2pts]

(∫ 1

(u²+1)²du =1

2tan⁻¹e^x+1 2

e^x

e^2x+1+C) [+3pts]

= ln ∣e^x∣ +tan⁻¹e^x−1

2ln(e^2x+1) +1 2

1 e^2x+1

−1

2tan⁻¹e^x−1 2

e^x e^2x+1+C

= x − 1

2ln(e^2x+1) +1

2tan⁻¹e^x+1 2

1 − e^x e^2x+1+C 不定積分沒有 +C 扣一分

5. (8%) Find the values of s such that F (s) =∫

∞

0 sin t e^−stdt converges and evaluate the integrals.

Solution:

”s ≠ 0” ∫ sin t e^−stdt = −sin t s +1

s∫ cos t e^−stdt

= −sin t s − 1

s²cos t e^−st− 1

s²∫ sin t e^−stdt Ô⇒ ∫ sin t e^−stdt = −1

1 + s²(s sin t e^−st+cos t e^−st) (3%) Ô⇒ lim

b→∞∫

b

0 sin t e^−stdt = −1 1 + s² lim

b→∞(s sin b e^−bs+cos t e^−bs) + 1 1 + s² Ô⇒ ∫

∞

0 sin t e^−stdt =

⎧⎪

⎪

⎨

⎪⎪

⎩ 1

1 + s² s > 0 (2%)

diverge s < 0 (1%)

”s=0” F (0) =∫

∞

0 sin t dt = lim

b→∞(−cos b + 1) diverges (2%)

6. (8%) Find the length of the loop of the curve 3ay²=x(a − x)², a > 0.

Solution:

3ay²=x(a − x)², a > 0

Ô⇒ 6ayy^′= (a − x)²−2x(a − x) = (a − x)(a − 3x) Ô⇒ y^′= (a − x)(a − 3x)

6ay = (a − 3x)

√12ax

Length = 2∫

a 0

√ 1 + (dy

dx)²dx = 2∫

a 0

√

1 +(a − 3x)²

12ax dx (2%)

= 2∫

a 0

a + 3x

√12axdx = 1

√3a∫

a

0 (ax^−1/2+3x^1/2)dx (3%)

= 4

√3a (3%)

7. (20%) Let C be the curve y = ln x, 0 < x ≤ 1 and R be the region bounded by x-axis, y-axis and the curve C.

(a) Compute the area of R if it is finite.

(b) Compute the arc length of C if it is finite.

(c) Rotate C about the y-axis. Compute the area of the generated surface if it is finite.

(d) Rotate R about the line y = x. Compute the volume of the generated solid if it is finite.

Solution:

(5%)(a) Area(R) = − ∫

1

0 ln x dx (3%)

= −lim

t→0⁺(x ln x − x)∣¹_t

=1 + lim

t→0⁺t ln t (1%)

=1 (1%)

(5%)(b) Length(C) = ∫

1 0

√

1 + (y^′)² dx (3%)

= ∫

1 0

√ 1 + (1

x)² dx

≥ ∫

1 0

1

x dx (1%)

= ∞ (1%)

(5%)(c) S =2π∫

1

0 x ds

=2π∫

1 0 x

√ 1 + (1

x)²dx let x = tan θ, dx = sec²θdθ (2%)

=2π∫

π 4

0 sec³θ dθ (1%)

=2π

2 (sec θ tan θ + ln ∣sec θ + tan θ∣)∣

π 4

0

=π(√

2 + ln(√

2 + 1)) (2%)

(5%)(d) x¯ = − ∫

1

0 x ln x dx/Area(R)

= −(lim

t→0⁺(x²

2 ln x)∣¹_t− ∫

1 0

x 2 dx)/1

=1

4 (2%)

¯

y = −1 2∫

1

0 (ln x)²dx/Area(R)

= −1 2(lim

t→0⁺(x(ln x)²)∣¹_t−2∫

1

0 ln x dx)/1

= −1 (2%)

The distance from (¯x, ¯y) to the line y = x is 5 4√

2 By Puppus Theorem, volume = 2π 5

4√

2Area(R) =5√ 2

4 π (1%)

8. (8%) (a) Sketch the curve with polar equation r = 1 + 2 cos 2θ.

(b) Find the area of the region inside both curves r = 1 and r = 1 + 2 cos 2θ.

Solution:

(a) First find some values of r for different θ :

θ 0 π

4 π 3

π 2

2 3π 3

4π π

r 3 1 0 -1 0 1 3

Since the graph is symmetric about the x-axis, so it looks like:(2 pts)

(b)The region to be integrated is as follows :(1pt)

Divide the required region into 3 parts:

(a) I (b) II (c) III

The areas of these three parts are represented by the following integrals:

(1pt for each) I. 1

· π ·

=

II. R

π 4

1

2

(1 + 2 cos 2θ)

dθ III. R

π3

1

2

(1 + 2 cos 2θ)

dθ =

(3θ + 2 sin 2θ +

sin 4θ   

π2

π 3

) =

−

− 1 Hence the area is 4( I + II + III)

=

+2 R

π4

(1 + 2 cos 2θ)

dθ+ 2 R

π3

(1 + 2 cos 2θ)

dθ

=

+2 R

π

4

(1 + 2 cos 2θ)

dθ

= 2(3θ + 2 sin 2θ +

sin 4θ   

π 2 π4

) = 2(

π −

π − 2) +

= 2π − 4. (2pts)

The areas of these three parts are represented by the following integrals: (1pt for each)

I. 1²⋅π ⋅ 45^○ 360^○ =π

8 II. ∫

π 3 π 4

1

2(1 + 2 cos 2θ)²dθ III.∫

π 2 π 3

1

2(1 + 2 cos 2θ)²dθ =1

2(3θ + 2 sin 2θ +1 2sin 4θ∣

π 2 π 3

)= π 4 −3√

3 8 −1 Hence the area is 4( I + II + III)

= π 2 +2∫

π 3 π 4

(1 + 2 cos 2θ)²dθ+ 2∫

π 2 π 3

(1 + 2 cos 2θ)²dθ

= π 2 +2∫

π 2 π 4

(1 + 2 cos 2θ)²dθ

Page 6 of 8

= 2(3θ + 2 sin 2θ +1 2sin 4θ∣

π 2 π 4

)= 2(3 2π −3

4π − 2) +π

2 = 2π − 4. (2pts)

9. (10%) An inverted cycloid is defined by the parametried equations

x(θ) = r(θ − sin θ), y(θ) = −r(1 − cos θ), 0 ≤ θ ≤ 2π.

Consider the motion of a particle without friction and rolling down the inverted cycloid released from (x(α), y(α)), where α ∈ [0, π]. By the conservation of energy, the velocity of the particle at (x(θ), y(θ)) is given by

v =ds dt =

√

2gr(cos α − cos θ) (∗) where s is the arc length function and α ≤ θ ≤ π.

(a) Derive a separable differential equation for dθ

dt from (*).

(b) Compute the time T =∫

π

θ=αdt for the particle to get to the lowest point (x(π), y(π)).

Solution:

(a) (4 points) We found dx

dφ and dy dφ first:

dx

dφ was r − r cos φ, and dy

dφ was −r sin φ.

S(θ) =∫

θ α

√ (dx

dφ)²+ (dy dφ)² dφ

=

√ 2r∫

θ α

√

1 − cos φ dφ.

v =ds dt = ds

dθ dθ dt

⇒ dθ dt = ds

dt dθ ds Therefore, dθ

dt =

¿ ÁÁ

À g(cos α − cos θ) r(1 − cos θ) (b) (6 ponits)

T =∫

π θ=α dt

= ∫

π θ=α

dt dθ dθ

= ∫

π θ=α

¿ ÁÁ

À r(1 − cos θ) g(cos α − cos θ) dθ Consider∫

¿ ÁÁ

À r(1 − cos θ) g(cos α − cos θ) dθ , Let A = cos α, u = cosθ

2 and du = −1 2 sinθ

2 dθ The above equation = −2√

2∫

1

√A + 1 − 2u²du Then, Let sin t =

√ 2

A + 1u, and cos tdt =

√ 2 A + 1du . The above equation = −2√

2

√A + 1

√ A + 1

2 ∫ cos t cos tdt

= −2t

= −2 sin⁻¹( 1

cos^α₂ ⋅cosθ 2) Consequently,

T =

√r

g(−2 sin⁻¹( 1

cos^α ⋅cosα 2)) ∣^π_α

=π

√r g.

Other proper methods are permitted to solve these problems.

10. (8%) Solve the differential equation y^′+2 xy = y³

x². (Hint: let y²= 1 u.)

Solution:

y^′+P (x)y = Q(x)yⁿ, where P (x) = 2

x, Q(x) = 1

x² and n = 3.

We found that this equation is Bernoulli equation.

Consequently, we mutiplied y⁻³at both sides.

The original equation ⇒ y⁻³y^′+y⁻³(2 xy) = 1

x²

⇒y⁻³y^′+ 2

xy⁻²= 1 x². Let y²= 1

u, that is, u = y⁻².

After that, we calculated and found that u^′= −2y⁻³y^′. Therefore, the above equation ⇒ −u^′+4

xu = 2

x². (3 ponits ) The integral factor: I(x) = e^{∫ x4udx}=e^{−4 ln x}=x⁻⁴. (2 points)

We mutiplied x⁻⁴on both sides of above equation, and we got: −x⁻⁴u^′+4x⁻⁵u = 2x⁻⁶.

⇒ d

dx(−x⁻⁴u) = 2x⁻⁶

We integrated both sides, and we had: −x⁻⁴u =∫ 2x⁻⁶dx + C, where C is a constant.

⇒ −x⁻⁴u = 2

5x⁻⁵+C, where C is a constant.

⇒ −u = 2

5x+Cx⁴. We known that u = 1

y².

Terefore, the solution of this problem is that y⁻²= 2

5x+Cx⁴, where C is a constant.

⇒y²= 1

2

5x+cx⁴, where C is a constant. (3 points)

Other proper methods are permitted to solve this problem.