1031微
微甲07-11班
班期 末 考 解 答 和 評 分 標 準
1. (10%) Solve the initial value problem:
dy
ey
, y(0) = −1.
=
dx
1 + x2
Solution:
dy
ey
(separable)
=
dx
1 + x2
1
dx (2%)
⇒ e−y dy =
1 + x2
⇒ −e−y = tan−1 x + C (3% for each side)
With y(0) = −1, we have −e = tan−1 0 + C = C.
⇒ C = −e (2%)
∴ The solution is e−y = e − tan−1 x
2. (a) (10%) Solve the initial value problem:

 2x(x + 3)y 0 + (4x + 3) y = 2x 21 (x + 3) 21 ,
1
 y(1) = , x > 0.
2
(b) (3%) Find lim y(x) and lim+ y(x).
x→∞
x→0
Solution:
(a)
Divide the equation by 2x (x + 3)
4x + 3
1
We get y 0 +
y=p
2x(x + 3)
x(x + 3)
Multiply integration factor I(x) on both side, then
1
4x + 3
y = Ip
Iy 0 + I
2x(x + 3)
x(x + 3)
´ 4x+3
4x + 3
Solve that I 0 = I
, we let I = e 2x(x+3) dx
2x(x + 3) ˆ
ˆ
4x + 3
1
1
3
1
2
Where
dx =
( +
)dx = ln x + ln (x + 3) + C, when x > 0
2x(x +p
3)
2
x x+3
2 p 3
Hence I(x) = eC x(x + 3)3 , Let C = 0 ⇒ I(x) = x(x + 3)3
4x + 3
1
0
Iy 0 + I
=x+3
y = Iy 0 + I 0 y = (Iy) = I p
2x(x + 3)
x(x + 3)
ˆ
p
1
1
Iy = (x + 3)dx = x3 + 3x + D ⇒ x(x + 3)3 y = x3 + 3x + D
2
2
1
1
Bring y(1) = into the equation, we find D =
2
2
x2 + 6x + 1
Hence y = p
2 x(x + 3)3
(b)
s
r
(1 + x6 + x12 )2
1
1
lim y = lim
=
=
x→∞
x→∞
4
2
4(1 + x3 )3
p
2
Because lim+ x + 6x + 1 = 1, and lim+ 2 x(x + 3)3 = 0
x→0
So lim+ y → +∞
x→0
x→0
評分標準如下，
(a)
解出 I(x) +5分
解出y 的不定積分 +3分
代入初始條件求出完整解答 +2分
Page 1 of 7
計算錯誤該部分加分折半
(b)
(a)解出的 y 錯誤不管答案一律0分
(有些同學計算錯誤會影響答案，有些不會，覺得不應該同樣因為計算錯誤而有差別)
此外對一題2分，兩題3分
3. (a) (7%) Find the length of the curve in polar coordinates: r =
(b) (7%) Find the area enclosed by the curve given in (a).
√
1 + sin 2θ, 0 ≤ θ ≤ 2π.
Solution:
(a)
dr
cos 2θ
d√
1 + sin 2θ = √
=
dθ
dθ
1 + sin 2θ
The length of the curve L is
ˆ
2π
r2
L=
0
ˆ
2π
=
+
s
dr
dθ
2
1 + sin 2θ +
0
ˆ
2π
=
0
ˆ
2π
=
0
ˆ
=
s
2π
s
r
√
dθ
(2 points)
(2 points)
cos2 2θ
dθ
1 + sin 2θ
1 + 2 sin 2θ + sin2 2θ + cos2 2θ
dθ
1 + sin 2θ
2 + 2 sin 2θ
dθ
1 + sin 2θ
2 dθ
0
√
= 2 2π
(3 points)
(b) Because r ≥ 0 for 0 ≤ θ ≤ 2π (or due to observation from the figure below), the area A of the enclosed
region is
ˆ 2π
1 2
A=
r dθ
(4 points)
2
0
ˆ
1 2π
=
(1 + sin 2θ) dθ
2 0
2π
1
1
=
θ − cos 2θ
2
2
0
1
1
1
=
2π − − 0 +
2
2
2
=π
Page 2 of 7
(3 points)
4. (a) (7%) Find the length of the parametric curve C: x = ln(sec t + tan t) − sin t, y = cos t, 0 ≤ t ≤
π
.
3
(b) (7%) Rotate the curve C about x-axis. Find the surface area.
√
(c) (7%) Find the volume of the solid bounded by the surface given in (b) and the planes x = 0, x = ln(2+ 3)−
√
3
.
2
Solution:
(a).
d
(x(t)) = sec(t) − cos(t),
dt
d
(y(t)) = −sin(t).
dt
ˆ π
ˆ π
3 ds = 3 tan(t)dt
0
. . . (1 points)
. . . (2 points)
0
π/3
= −ln|cos(t)|0
. . . (2 points)
= ln(2) ]
. . . (2 points)
, where ds =
s
d
(x(t))
dt
2
2
d
+
(y(t)) dt = tan(t)dt.
dt
(b).
ˆ π
ˆ π
3 2πy(t)ds = 3 2πcon(t)tan(t)dt
0
. . . (2 points)
0
π/3
= −2πcos(t)|0
=π
]
. . . (3 points)
. . . (2 points)
(c).
ˆ
ˆ π
πy (t)d(x(t)) = 3 πcon2 (t)[sec(t) − cos(t)]dt
2
. . . (2 points)
0
ˆ π
= 3 πsin2 (t)con(t)dt
0
π
π/3
sin3 (t)|0
. . . (3 points)
3
√
3
π ]
. . . (2 points)
=
8
=
, where d(x(t)) = [sec(t) − cos(t)]dt.
ˆ
5. (10%) Find
1
dx.
x2 (x2 + x + 1)
Solution:
1
a
b
cx + d
= + 2 + 2
, where a, b, c, d are determined. (1 point) Then solve a = −1,b =
x2 (x2 + x + 1)
x
x
x +x+1
1,c = 1,d = 0. (2 points, if solve partially 1 point)
so we have
ˆ
ˆ
ˆ
ˆ
1
−1
1
x
dx =
dx +
dx +
dx
x2 (x2 + x + 1)
x
x2
x2 + x + 1
Let
Page 3 of 7
1
+
x
ˆ
x
dx (2 points)
x2 + x + 1
ˆ
ˆ
ˆ
x + 21
x
1
1
1
dx
=
dx
−
dx = ln(x2 + x + 1) (2 points)
x2 + x + 1
x2 + x + 1
2
x2 + x + 1
2
ˆ
2
2x + 1
1
√
= √ arctan √ (3 points)
3 2
1 2
3
3
(x + 2 ) + ( 2 )
= − ln |x| −
1 1
1
2x + 1
+ ln(x2 + x + 1) − √ arctan √
+C
x 2
3
3
If you don’t wirte +C, you will lose 1 point.
so Answer is − ln |x| −
ˆ
6. (10%) Find
1
dx.
sec x − 1
Solution:
Solution 1.
ˆ
ˆ
1
sec x + 1
sec x + 1
·
dx =
dx (2分)
sec x − 1 sec x + 1
sec2 x − 1
ˆ
ˆ
cos2 x
cos x
sec x + 1
dx (3分) =
+
=
dx
tan2 x
sin2 x
sin2 x
ˆ
ˆ
1 − sin2 x
1
d
sin
x
+
dx
=
2
sin2 x
ˆ
ˆ
ˆ sin x
1
2
d sin x + csc x dx − 1 dx
=
sin2 x
1
=−
− cot x − x + C.
sin x
1
dx =
sec x − 1
後面的三項積分個別計分，第一項積出 −
Solution 2. Let t = tan
cos x =
ˆ
1
再得 4 分，其它兩項及常數 C 一共佔 3 分。
sin x
x
(佔 1 分), then we have
2
2t
2
1 − t2
(佔 1 分), sin x =
(沒有用到), and dx =
dt. (佔 1 分)
1 + t2
1 + t2
1 + t2
So
ˆ
ˆ 1−t2
ˆ
2
cos x
1 − t2
1+t2 · 1+t2
dx (2分) =
dt (5分)
dt
=
2
2
1−t
1 − cos x
t (1 + t2 )
1 − 1+t2
ˆ 1
2
1
=
−
dt (7分) = − − 2 tan−1 t + C (9分)
2
2
t
t +1
t
1
x
x
=−
− 2 tan−1 tan
+ C = − cot − x + C. (10分)
tan x2
2
2
1
dx =
sec x − 1
ˆ
前面兩種解法是大部分人採用的積分策略，所以在此列出詳細配分標準，以下還有數種解法，處理方式都很漂亮，在此也
完整呈現供大家參考及學習，配分以個案處理， 但原則上和前兩種配分方式雷同。
Solution 3.
ˆ
Solution 4.
ˆ
ˆ
1 − 2 sin2 x2
cos x
dx =
dx
1 − cos x
2 sin2 x2
ˆ
ˆ
x x
x
= csc2 d
− 1 dx = − cot − x + C.
2
2
2
1
dx =
sec x − 1
1
dx =
sec x − 1
ˆ
ˆ
1
sec x + 1
·
dx =
sec x − 1 sec x + 1
Page 4 of 7
ˆ
sec x + 1
dx =
sec2 x − 1
ˆ
sec x + 1
dx.
tan2 x
Since
ˆ
and
ˆ
ˆ
sec x
sec x tan x
1
dx
=
d
sec
x
=
−
sec x d cot3 x
tan3 x
tan3 x
tan3 x
ˆ
ˆ
sec x
sec x
cos x
2
2
dx
+ 3 sec x cot x csc x dx =
+3
=
3
3
tan x
tan x
sin4 x
ˆ
sec x
sec x
1
1
=
+3
d sin x =
−
+ C,
tan3 x
tan3 x sin3 x
sin4 x
sec x
dx =
tan2 x
ˆ
1
dx =
tan2 x
we get
Solution 5.
ˆ
ˆ
ˆ
1
dx =
sec x − 1
Since
and
ˆ
cot2 x dx =
(csc2 x − 1) dx = − cot x − x + C,
1
1
sec x
− cot x − x + C.
−
dx =
sec x − 1
tan3 x sin3 x
ˆ
ˆ
ˆ
ˆ
sec x + 1
1
·
dx =
sec x − 1 sec x + 1
sec x
dx =
tan2 x
ˆ
sec x + 1
dx =
sec2 x − 1
ˆ
sec x + 1
dx.
tan2 x
ˆ
cot x csc x dx = − csc x + C,
ˆ
ˆ
cos2 x
2
2
dx
=
cos
x
csc
x
dx
=
−
cos2 x d cot x
sin2 x
ˆ
= − cos2 x cot x + cot x d cos2 x
ˆ
= − cos2 x cot x − 2 cot x cos x sin x dx
ˆ
ˆ
= − cos2 x cot x − 2 cos2 x dx = − cos2 x cot x − (1 + cos 2x) dx
1
dx =
tan2 x
ˆ
sin 2x
+C
2
− cos2 x cot x − x − sin x cos x + C
= − cos2 x cot x − x −
we get
ˆ
1
dx = − csc x − cos2 x cot x − x − sin x cos x + C.
sec x − 1
x
Solution 6. Let t = cot , then we have
2
cos x =
t2 − 1
,
1 + t2
sin x =
2t
,
1 + t2
and
dx = −
2
dt.
1 + t2
So
ˆ
ˆ t2 −1 −2
ˆ
cos x
1 − t2
1+t2 · 1+t2
dt
=
dx =
dt
2 −1
t
1 − cos x
1 + t2
1 − 1+t2
ˆ 2
=
−1 +
dt = −t + 2 tan−1 t + C
1 + t2
x
x
+ C.
= − cot + 2 tan−1 cot
2
2
1
dx =
sec x − 1
ˆ
1
Solution 7. Let t = sec x+tan x. Since sec2 x−tan2 x = (sec x+tan x)(sec x−tan x) = 1, we have = sec x−tan x
t
and
sec x =
t2 + 1
,
2t
tan x =
t2 − 1
,
2t
Page 5 of 7
dt =
t2 + 1
dx.
2
(b) Discuss the asymptotic behavior lim y(x) and lim+ y(x).
x→∞
x→0
Problem 3. (%+%)
(a) Find the length of the polar curve: r =
√
1 + sin 2θ, 0 ≤ θ ≤ 2π.
(b) Find the area enclosed by the curve given in (a).
So
ˆ 4. (%+%+%)
Problem
ˆ
ˆ
1
2
1
4t
dx =
· 2
dt =
dt
2 (t
t2 +1
(a) Find
curve
C:
x
=
ln(sec
t
+
tan
t)
−
sin2 t,+y 1)
= cos t, 0 ≤ t ≤ π3 .
secthe
x −length
1 of the parametric
t
+
1
(t
−
1)
2t − 1
ˆ (b) Rotate the curve C about the 2x-axis. Find2the surface area. 2
−
=
dt = −
− 2 tan−1 t + C
(t − 1)2
t2 + 1
t−1
(c) Find the volume of the solid bounded by the surface given in (b) and the planes x = 0, x =
√
2
ln(2 + 3).
=−
− 2 tan−1 (sec x + tan x) + C.
Z
sec1x + tan x − 1
Problem 5. (%) Find
x2 (x2 + x + 1)
Z
dx.
1
Problem
(%) Find
dx. with radius R and all axes of cylinders lie in a plane with polar
7. Consider the intersection
of6.three
circular
cylinders
sec x − 1
π
2π
Problem
the intersection
three circular
cylinders
with radius
and alllooks like the union
equations θ = 0, θ =
, and7.θ (%+%)
=
. Consider
The cross-section
is of
a hexagon,
and
the shape
of theR solid
3
3
π
2π
axes
of
cylinders
lie
in
a
plane
with
polar
equations
θ
=
0,
θ
=
,
and
θ
=
.
The
cross-section
of two umbrellas.
3
3
is a hexagon, and the shape of the solid looks like the union of two umbrellas.
θ=
2π
3
θ=
π
3
y
R
θ=0
(a)
(b)
Figure 1: Intersection of three cylinders. (a) Vertical view. (b) The solid.
(a) Find
the cross
area ofsection
the crossof
section
of the when
solid when
heightisisy,
y, y
y∈
R].R].
(a) (6%) Find the area
of the
the solid
the the
height
∈[−R,
[−R,
(b) (6%) Find the volume
solid.
(b) Findof
thethe
volume
of the solid.
Problem 8. (%) Evaluate the improper integral
Solution:
(a)Observe the figures below.
Z
∞
2
x3 e−x dx.
0
1
(a)Observe the figures below.
d
d
θ=
π
6
y
R
When the height is y, the area of the cross section(hexagon) is decided by
p When the height is y, the area of the cross section(hexagon)
d = R2 − y 2 .
Hence,
Hence,
is decided by d =
p
1
2
Area = 6 × Area of1equilateral
triangles = 6 × × d × √ d
2
2
3
Area = 6 × Area of equilateral triangles = 6 × × √
d× √ d
2
√
=
2 3(R32 − y 2 ) (3%)
2
2
= 2 3(R − y )
ˆ R
√
Z R
Z R √
√
1
8 3 3
(b)
R
=
2 3(R2 − y 2 )dy = 2 3(R2 y − y 3 )|R
(b)VV== A(y)dy
A(y)dy(3%)
−R =
3
3
−R
−R
√
ˆ−R
R √
√
1 3 R
8 3 3
2
2
2
=
2 3(R − y )dy = 2 3(R y − y )|−R (2%) =
R (1%)
3
3
−R
ˆ
8. (10%) Evaluate the improper integral
∞
2
x3 e−x dx.
0
Page 6 of 7
R2 − y 2 .(3%)
Solution:
ˆ
0
∞
3 −x2
x e
ˆ
ˆ 2
1 t
x e
dx = lim
x e
dx = lim
u e−u du (3分)
t→∞ 0
t→∞ 2 0
0
!
ˆ t2
ˆ 2
2
1 t
1 h −u iu=t
−u
−u
−
e du (6分)
= lim −
ue
u de = lim −
t→∞
t→∞
2 0
2
u=0
0
2
2
1
1 h −u iu=t
= lim − t2 e−t −
e
t→∞
2
2
u=0
1 2 −t2 1 −t2 1
1
= lim − t e
− e
+
(7分) = ,
t→∞
2
2
2
2
t
3 −x2
1
dx = lim
t→∞ 2
ˆ
t
2 −x2
2
where we use the l’Hospital Rule to get find the limit:
2
t2
2
t→∞ et
lim t2 e−t = lim
t→∞
∞
(∞
,L0 )
=
lim
t→∞
1
2t
= 0.
2 = lim
t
t→∞
2t e
et2
ˆ
∞
這題重點在於是否具有「瑕積分的觀念」，而判定的依據是以有寫出「極限」為準；縱使前面所有的式子都是寫
或
0
2
2 −t
是上、下限寫 |∞
= 0，就認定你知道「瑕積分的意義」；這個極限值不是顯然
0 ， 但最後一定要用極限判斷 lim t e
t→∞
2
的，一定要討論，佔 3 分， 也就是說，最後有完整討論這個極限就可以拿到滿分 10 分。 另一個極限式 lim e−t = 0
t→∞
可接受直接寫答案，沒有解釋不會扣分。
Page 7 of 7