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d dx
∫ (^) x 2
sin x
1 + t^4 dt 。
Sol:
Set F (x) =
∫ (^) x
a
1 + t^4 dt , where a is a constant.
∫ (^) x 2
sin(x)
1 + t^4 dt = F (x^2 ) − F (sin(x))
By using the Fundamental Calculus Theorem and the Chain Rule, we can get: d dx
∫ (^) x 2
sin(x)
1 + t^4 dt = F ′(x^2 )2x − F ′(sin(x)) cos(x)
= 2x
1 + x^8 − cos(x)
1 + sin^4 (x)
Basically, if the answer is correct, students can get 10 points. And substitution error will lose 1 ∼4 points.
2. (15%) (a) 將
x^3 + 1
(b) 求
dx x^3 + 1
Sol:
(a) 1 x^3 + 1
A
x + 1
Bx + C x^2 − x + 1
. (1 pts)
Then, A(x^2 − x + 1) + (Bx + C)(x + 1) = 1.
Take x = −1, we obtain 3A = 1, so A =
. (2 pts) Then,
(Bx + C)(x + 1) = 1 −
(x^2 − x + 1)
= −
(x^2 − x − 2)
= −
(x + 1)(x − 2).
We can get B = −
(2 pts) and C =
(2 pts). Hence,
1 x^3 + 1
3(x + 1)
−x + 2 3(x^2 − x + 1)
(b) ∫ 1 x^3 + 1 dx =
3(x + 1)
−x + 2 3(x^2 − x + 1) dx
=
[∫
x + 1
dx −
2 x + 1 − 3 x^2 − x + 1
dx
]
and we have ∫ 1 x + 1 dx = ln | x + 1 |. (2 pts) ∫ 2 x + 1 x^2 − x + 1 dx =
(x^2 − x + 1)^2 d(x^2 − x + 1) = ln | x^2 − x + 1 |. (2 pts) ∫ 3 (x^2 − x + 1) dx =
(x − 12 )^2 + (
√ 3 2 ) 2
= 2
(2(x−^
(^12) ) √ 3 )^2 + 1
d(
2(x − 12 ) √ 3
3 arctan( 2(x − 12 ) √ 3
). (2 pts)
Hence, ∫ 1 x^3 + 1 dx =
ln | x + 1 | −
ln | x^2 − x + 1 | +
√^2
arctan(
√^2
x −
√^1
) + C
Each coefficient of the first two terms worths 1 pts.
x^2
x^2 + 1
dx 。
Sol: Let x = tan(θ) (+4%) Since dx dθ
= sec^2 (θ) dθ,
we have (^) ∫ 1 x^2
x^2 + 1
dx =
sec(θ) tan^2 (θ) dθ
Now let u = sin θ, then the original integral now become :
=
u^2
du. (+2%)
Hence we have the answer:, − 1 u
= − sin−^1 (θ) + C =
x^2 + 1 x
+ C, (+1%)
where C is a constant in R
and x=
2.wrute down the integration function 5 % + integration 3 %
V =
0
2 xπ(6 −
x^5 + x^2 ) dx
0
12 πxdx − 2 π
0
x^2 (
x^3 + 1) dx
= 24π − 2 π 3
0
x^3 + 1 d(x^3 + 1)
= 112 π 9
6. (15%) (a) 寫出 ln (1 + x) 在 x = 0 的泰勒展式 。 (5%)
(b) 令 f (x) = ln
1 + x 1 − 2 x
, 求 f (6)(0) 。 (10%)
Sol:
(a) Let f (x) = ln(1 + x),hence f (0) = ln(1) = 0 ln(1 + x)(1)^ =
1 + x =⇒ f (1)(0) = 1 ln(1 + x)(2)^ = −(1 + x)(−2)^ =⇒ f (2)(0) = − 1 ln(1 + x)(3)^ = 2(1 + x)(−3)^ =⇒ f (3)(0) = 2 ln(1 + x)(4)^ = −6(1 + x)(−4)^ =⇒ f (4)(0) = − 6
continue this way,we can get general items
(−1)(n−1) n xn.
Hence ln(1+x) = 0+x+
x^2 +
x^3 +· · ·+ (−1)(n−1) n xn^ +· · · for the first some items,you get 3 points, for the general items,you get 2 points.
(b)
f (x) = ln
( (^) 1 + x 1 − 2 x
)^12
[ln(1 + x) − ln(1 − 2 x)] (1 points)
=
[0 + x +
x^2 +
x^3 −
x^4 +
x^5 −
x^6 + · · · ]
−
[− 2 x − 2 x^2 −
x^3 − 4 x^4 −
x^5 −
x^6 − · · · ]
(2 points and 4 points respectively)
By Taylor expression,we know that
f (6)(0) 6!
[−
] =
hence f (6)(0) = 3780 (3 points).
(a) 求以下函數在 x = 0 之泰勒展式非零的前三項 。
(i) sin x 。 (5%)
(ii) tan−^1 x 。 (5%)
(b) 求 lim
x→ 0
2 sin x − tan−^1 x − x x^5
Sol:
(a) (i) Let f (x) := sin x,
f (n)(x) =
cos x if n ∈ 4 N + 1 − sin x if n ∈ 4 N + 2 − cos x if n ∈ 4 N + 3 sin x if n ∈ 4 N + 4 Then at x = 0,
f (n)(0) =
1 if n ∈ 4 N + 1 0 if n ∈ 4 N + 2 − 1 if n ∈ 4 N + 3 0 if n ∈ 4 N + 4 The Taylor series of f (x) at x = 0 is ∑^ ∞
n=
f (n)(0) n! xn^ =
∑^ ∞
m=
(−1)m (2m + 1)! x^2 m+1^ = x − x^3 3!
x^5 5!
(ii) Let f (x) := tan−^1 x,
f ′(x) =
1 + x^2
1 − (−x^2 )
∑^ ∞
n=
(−x^2 )n
f (x) =
f ′(x) + C =
n=
(−x^2 )n^ + C
∑^ ∞
n=
(−1)n 2 n + 1
x^2 n+1^ + tan−^1 (0) = x − x^3 3
x^5 5
(tan−^1 (0) = 0)
(c) Tangent Plane at (2, 2 , 4):
z − 4 = ∂f ∂x
¯(x,y)=(2,2)^ (x^ −^ 2) +^
∂f ∂y
¯(x,y)=(2,2)^ (y^ −^ 2)^ (3 pts)
z − 4 = 4(x − 2) + 4 ln 2(y − 2)
or
4 x + 4(ln 2)y − z − 4(2 ln 2 + 1) = 0
(2 pts)
