1011微
微乙 01-05班
班期 末 考 解 答 和 評 分 標 準
sin x − x cos x
.
x→0 tan−1 x − sin x
1. (10%) 求 lim
Solution:
由定理4.3(l’Hopital 法則):
( sinx x ) + cos x
sin x − x cos x
cos x − cos x + x sin x
sin x + x cos x
lim
=
lim
=
lim
=
lim
=
1
−2
x→0 tan−1 x − sin x
x→0
x→0 −2x
x→0
+ sin x
( 1+x
+ ( sinx x )
2 ) − cos x
(1+x2 )2
(1+x2 )2
1+1
= −2
−2 + 1
評分標準:
(1)使用l’Hopital法則:第一次微對3分，第二次對3分，接下來的運算及答案4分。
(2)使用泰勒展式:一個展式對3分共9分，接下來的運算及答案1分。
Page 1 of 8
2. (10%) 計算
Z
x2
1
dx.
+ 3x + 2
Solution:
Z
Z x + 1
dx
1
1
+C[5%] = ln |x+1|−ln |x+2|+C.
2.
=
−
dx[5%] = ln x2 + 3x + 2
x+1 x+2
x + 2
Note.
1. No constant C or absolute value for ln ⇒ −2%
2. Other methods: partial grades, 3% or 5%.
Page 2 of 8
3. (10%) 計算
Z
x2 sin x dx.
Solution:
Z
Z
Z
2
2
2
x sin xdx = − x d(cos x) = −[x cos x − 2 cos x(dx2 )].......................(6 pts)
Z
Z
2
2
= −x cos x + 2 x cos xdx = −x cos x + 2 x(d sin x)
Z
2
= −x cos x + 2[x sin x − sin xdx]...........................................(3 pts)
= −x2 cos x + 2x sin x + 2 cos x + C, C ∈ R...............................(1 pt)
Page 3 of 8
4. (10%) 求函數 y = sin x, y = cos x 的曲線在 x = 0 到 x = π 之間所夾的面積大小.
Solution:
The area is equal to the following integration.
Z π
| sin x − cos x|dx
0
π
π
and sin x ≥ cos x if ≤ x ≤ π, the area is equal to
4
4
Z π
Z π
4
(cos x − sin x)dx +
(sin x − cos x)dx
Because cos x ≥ sin x if 0 ≤ x ≤
(1)
π
4
0
Then
Z
π
π
4
Z
| sin x − cos x|dx =
0
Z
π
(cos x − sin x)dx +
(sin x − cos x)dx
π
4
0
π
= (sin x + cos x) |04 +(− cos x − sin x) |ππ
4
√
√
= ( 2 − 1) + (1 + 2)
√
= 2 2
Grading evaluation:
1. If you write down the equation (1), you will get 2 points.
2. If one of the two integrations in the equation (1) is right, you will get 4 points, respectively.
Page 4 of 8
1
5. (15%) 求函數 f (x) = x2 由 x = 0 到 x = 1 的曲線長度.
2
Solution:
f 0 (x) = x Z
1√
length l =
1 + x2 dx (7 pts)
0
Z πp
4
π π
1 + tan2 θ sec2 θdθ (if we let x = tan θ,θ ∈ (− , ), then dx = sec2 θdθ)
=
2 2
0
Z π
4
=
sec3 θdθ (3 pts)
0
Z π
4
sec θd tan θ
=
0
Z π
π
4
4
= sec θ tan θ|0 −
sec θ tan2 θdθ
0
Z π
π
4
4
= sec θ tan θ|0 −
sec θ(sec2 θ − 1)dθ
0
Z π
Z π
π
4
4
3
4
= sec θ tan θ|0 −
sec θdθ +
sec θdθ
0
0
Z π
π
4
4
= (sec θ tan θ + ln | sec θ + tan θ|)|0 −
sec3 θdθ
0
Z π
π
4
1
⇒l=
sec3 θdθ = (sec θ tan θ + ln | sec θ + tan θ|)|04
2
0
√
√
2 ln ( 2 + 1)
+
(5 pts)
=
2
2
Page 5 of 8
√
6. (15%) 求 x =
2y
, y = 1 及 x = 0 所圍成的區域繞 y 軸旋轉之體積.
1 + y2
Solution:
Z
y=1
x2 dy
2
Zy=0
1 √
2y
π
dy (10 points)
1 + y2
Z0 1
2y
π
dy
(1 + y 2 )2
0
Z 1
du
(3 points)
π
2
0 (1+ u)
1
π −
1 + u 0
π
(2 points)
2
The volume = π
=
=
=
=
=
Page 6 of 8
7. (10%) 求 y =
tan x
4
, y = 及 x = 0 所圍成的區域繞 y 軸旋轉之體積.
x
π
Solution:
We have the following simple observations:
tan x
=1
x→0
x
0
tan x
2x − sin(2x)
π
(ii )
=
> 0, ∀x ∈ (0, ) =⇒
2
2
x
2x cos x
2
(i ) lim
tan x
π
% on (0, )
x
2
4
π
tan x
= has a unique solution which is x = clearly.
x
π
4
4
π
tan x
≤ on (0, ]
(iv )
x
π
4
(iii ) From (ii ),
Hence,
Area =
=
=
=
=
π/4
4 tan x
2π
x
−
dx
π
x
0
Z π/4 4
2π
x − tan x dx
π
0
h
iπ/4
2 h iπ/4
2π x2
+ 2π ln | cos x|
π
0
0
2
π
− π ln 2
4
1
π2
+ 2π ln √
4
2
Z
Grade:
• 10 points if you have the right answer like (2) or (3) for example.
• 5 points if you have a mistake(s) with +/− in the coefficients.
• 0 points otherwise.
Remark.
tan x
4
π
π
= on (0, ).
You don’t have to claim that x = is the unique solution to
4
x
π
2
Page 7 of 8
(2)
(3)
2
8. 令函數f (x) = e−x .
(a) (10%) 求 f (x) 對 x = 0 的泰勒展開式, 並寫出一般項.
Z 1
2
(b) (10%) 以(a)中非零的前三項估計積分
f (x) dx, 誤差忽略不計.
0
Solution:
a.
We know
x
e =
∞
X
xn
0
n!
Therefore,
2
e−x =
∞
X
(−x2 )n
0
n!
b.
Z
1
2
e
0
−x2
1
2
x4
dx ≈
(1 − x + )dx
2!
0
443
=
960
Z
Page 8 of 8
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