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1. (10%) 令 f (x) =
∫ (^) tan− (^1) x
x^2
dt 1 + t^7
。 求 f ′(x) 。
Sol: Then by Fundamental Theorem of Calculus
f ′(x) =
1 + (tan−^1 x)^7 (tan−^1 x)′^ −
1 + (x^2 )^7 (x^2 )′
=
1 + (tan−^1 x)^7
1 + x^2
2 x 1 + x^14 (10 points)
1 − x^2 )^3 dx 。
Sol:
令 x = sin θ ∵ − 1 ≤ x ≤ 1 ∴ −
π 2
≤ θ ≤ π 2
則 dx = cos θdθ.
1 − x^2 )^3 dx =
| cos θ|^3 cos θdθ =
cos^4 θdθ
=
1 + cos 2θ 2 )^2 dθ
=
1 + 2 cos 2θ + cos^2 2 θ 4 )dθ
=
1 + 2 cos 2θ 4
1 + cos 4θ 8
)dθ
=
2 cos 2θ 4
cos 4θ 8 )dθ
=
θ + sin 2θ 4
sin 4θ 32
+ C
θ + 2 sin θ cos θ 4
2 sin 2θ cos 2θ 32
+ C
θ + sin θ cos θ 2
4 sin θ cos θ(1 − 2 sin^2 θ) 32
+ C
sin−^1 x +
x
1 − x^2 +
x
1 − x^2 (1 − 2 x^2 ) + C
=
sin−^1 x +
x
1 − x^2 (5 − 2 x^2 ) + C, where C is a constant.
cos^4 θdθ 或 −
sin^4 θdθ · · · · · · (4%)
把 θ 的函數換回成 x 的函數 · · · · · · · · · (4%)
3. (20%) (a) 求
0
dx x^2 − x + 1
。 (10%) (b) 求
x 1 + x^3
dx 。 (10%)
Sol:
(a) ∫ (^1)
0
dx x^2 − x + 1
0
dx (x − 12 )^2 + (^34)
0
√ 3 2 d^ √^2 ( 3 x √^2 3 (x^ −^
1 2 )
[
tan−^1
(x −
)]^1
0
√^2
tan−^1
√^1
− tan−^1 −
√^1
(π 6
π 6
2 π 3
(b) From (a), we have ∫ 1 1 − x + x^2 dx =
x − (^12)
3 2
) 2 dx^ =^
√^1
3 2
tan−^1
x − (^12) √ 3 2
tan−^1
2 x √ − 1 3
Letting x 1 + x^3
x (1 + x)(1 − x + x^2 )
A
1 + x
Bx + C 1 − x + x^2 (3 points)
A(1 − x + x^2 ) + (Bx + C)(1 + x) (1 + x)(1 − x + x^2 ) and setting x = 0, 1 , −1 into
x = A(1 − x + x^2 ) + (Bx + C)(1 + x),
we obtain
0 = A + C
1 = A + 2B + 2C
−1 = 3A
⇒ (A, B, C) =
. (3 points)
6. (15%) (a) 求 sin x 在 x = 0 的 7 次泰勒多項式及餘項 。 (10%)
(b) 求 lim x→ 0
sin x − x + x 3 3! −^
x^5 5! x^7
Sol:
(a) P 7 (x) = x − x^3 3!
x^5 5!
x^7 7!
R 7 (x) =
sin ξ · x^8 8! , for ξ lies between 0 and x (3%).
(b) lim x→ 0
sin x − x + x 3 3! −^
x^5 5! x^7
= lim x→ 0
−x 7 7! +^
sin ξ·x^8 8! x^7
= lim x→ 0
sin ξ · x 8!
×
5 的近似值 , 使誤差 < 10 −^4 。
Sol:
consider binomial expansion. √ 5 =
[
)((1 + ξ)
− 23 )
]
where (ξ between 0 and
then
[1 + (
)] =
Note that the error≤ | (
) ∗ 2 |< 10 −^4 (3%)
since (ξ between 0 and
8. (10%) 求 lim
x→ π 2 −
π 2 −^ x
)cos x
。 ( “左極限” x →
π 2
−
意指 x <
π 2
, x 趨近於
π 2
Sol:
lim x→(π/2)−
π/ 2 − x
)cos x
= lim x→(π/2)−^ exp [− cos x · ln(π/ 2 − x)] (+1pts) : Taking logarithm
= exp
[
lim x→(π/2)−
− ln(π/ 2 − x) sec x
]
= exp
[
lim x→(π/2)−
(π/ 2 − x) sec x tan x
]
(+3pts) : LHospital’s Rule
= exp
[
lim x→(π/2)−
cos^2 x π/ 2 − x
· lim x→(π/2)−
sin x
]
= exp
[
lim x→(π/2)−
2 cos x sin x 1
]
(+5pts) : LHosptial’s Rule
= exp
[
]
= exp(0) = 1 (+1pts) : The Answer
