1021微微微乙乙乙01-05班班班期期期末末末考考考解解解答答答 1. (15%) 令 In =
ˆ
(ln x)ndx, n ≥ 1 (a) (6%) 求 I1 =?
(b) (6%) 把 In+1 用 In 表出。
(c) (3%) 利用(b)的遞迴式求 I4。
Solution:
令 In= ˆ
(ln x)ndx, n ≥ 1.
(a) 使用分部積分法, 令u = ln x, dv = dx (2%), 我們得到 ˆ
ln x dx = x ln x − ˆ
x d(ln x) (1%)
= x ln x − ˆ
x · 1
xdx (1%)
= x ln x − x + C (2%)
註:未加上積分常數扣1分, 沒有計算過程根據規則得0分.
(b) 使用分部積分法, 令u = (ln x)n+1, dv = dx (2%), 我們得到 In+1=
ˆ
(ln x)n+1dx = x(ln x)n+1− ˆ
x d(ln x)n+1 (1%)
= x ln x − ˆ
x · (n + 1)(ln x)n· 1
xdx (2%)
= x(ln x)n+1− (n + 1)In (1%)
註1:如果僅計算前幾個, 如I2, I3,然後直接推論猜測規律出最後結果的人, 得到4分.
註2:使用別的方法計算出來的遞迴式也會有分數, 上述只是其中一個較好計算的結果.
(c) 根據遞迴式我們得到
I4 = x(ln x)4− 4I3
= x(ln x)4− 4x(ln x)3− 3I2
= x(ln x)4− 4x(ln x)3+ 12x(ln x)2− 2I1
= x(ln x)4− 4x(ln x)3+ 12x(ln x)2− 24(x ln x − x + C)
= x(ln x)4− 4x(ln x)3+ 12x(ln x)2− 24x ln x + 24x + C
評分方法:
(i) (b)若計算錯, 則(c)得0分.
(ii) 計算錯誤, 如正負號錯誤, 括號有誤等, 或是未加上積分常數總和至多扣1分.
(iii) 未使用(b)小題中所得到的遞迴式的人得1分.
2. (12%) (a) (6%) 求
ˆ x + 1
x2+ x + 1dx。 (b) (6%) 求
ˆ dx
ex(e2x− 1)。
Solution:
(a) ˆ
x + 1 x2+ x + 1 dx
=
ˆ x +12
x2+ x + 1 dx +1 2
ˆ dx
x2+ x + 1 (2%)
= 1
2ln (x2+ x + 1) + 1
√3tan−1( 2
√3(x + 1
2)) + C (4%) 在答案接近完整時,沒加 C 扣一分,但不重複扣!
(b) ˆ
dx
ex(e2x− 1), 令 u = ex (2%)
= ˆ
(1 2
1 u − 1− 1
2 1
u + 1 − 1
u2) du (2%)
= 1
2ln |u − 1| − 1
2ln |u + 1| + 1 u + C
= 1
2ln |ex− 1| −1
2ln |ex+ 1| + e−x+ C (2%)
3. (12%) (a) (6%) 求 ˆ 1
2
0
x2√
1 − x2dx。
(b) (6%) 求 d dx
ˆ x2 x
dt 1 + t5。
Solution:
(a) [6pts]
ˆ 1
2
0
x2√
1 − x2dx
dx=cos θ
=====⇒
x=sin θ
ˆ π
6
0
sin2θ cos2θdθ [2pts]
sin2θ=1−cos 2θ2
=========⇒
cos2θ=1+cos 2θ2
ˆ π
6
0
1 − cos22θ 4 dθ
= ˆ π
6
0
sin22θ 4 dθ
sin22θ=1−cos 4θ2
=========⇒
ˆ π
6
0
1 − cos 4θ
8 dθ [2pts]
=θ 8 − 1
32sin 4θ
π 6
0
=π 48 −
√3
64 [2pts]
(b) [6pts]
d dx
ˆ x2
x
dt 1 + t5
F (x)=´x 0 f (t)dt
=========⇒
f (x)= 1
1+t5
d
dx{F (x2) − F (x)} [2pts]
=dF (x2) dx2
dx2
dx − dF (x)
dx [2pts]
Fundamental Calculus
==================⇒f (x2) · 2x − f (x)
= 2x
1 + x10 − 1
1 + x5 [2pts]
4. (12%) (a) (6%) 求 ˆ
sec3θdθ。(若知道 ˆ
sec θdθ 可以使用) (b) (6%) 求曲線 y = x2
2 + 1 由 x = 0 到 x = 2 之弧長。
Solution:
(a)
ˆ
sec3θ dθ = ˆ
(1 + tan2θ)secθ dθ
= ˆ
secθ dθ + ˆ
tan2θsecθ dθ (1 points)
= ˆ
secθ dθ + ˆ
tanθ dsecθ
= ˆ
secθ dθ + tanθsecθ − ˆ
sec3dθ (2 points)
= ln|secθ + tanθ| + tanθsecθ − ˆ
sec3dθ (2 points)
=⇒
ˆ
sec3θ dθ = 1
2ln|secθ + tanθ| +1
2tanθsecθ + C (1 points)
or ˆ
sec3θ dθ = ˆ
sec3θcosθ cosθ dθ
= ˆ
sec4θ dsinθ
=
ˆ 1
(1 − sin2θ)2 dsinθ (Let u = sinθ) =
ˆ 1
(1 − u2)2 du
=
ˆ 1
(1 − u)2(1 + u)2 du (2 points)
= 1 4
ˆ 1
(1 − u) + 1
(1 + u) + 1
(1 − u)2 + 1
(1 + u)2 du (2 points)
= 1 4
−ln|1 − u| + ln|1 + u| + 1
(1 − u) − 1 (1 + u)
+ C
= 1 4
ln
1 + u 1 − u
+ 2u 1 − u2
+ C 1
1 + sinθ
2sinθ
(b)
ˆ 2
0
p1 + (y0)2dx = ˆ 2
0
√1 + x2dx (2 points)
(Let x = tanθ) =
ˆ tan−12
tan−10
√
1 + tan2θ dtanθ
=
ˆ tan−12 tan−10
sec3θ dθ (2 points)
= 1
2ln |tanθ + secθ| +1
2tanθsecθ
tan−12 tan−10
= 1 2ln|√
5 + 2| +√
5 (2 points)
5. (10%) 求由 x = 0, x = 1, y = 0, y =√
x2+ 1 所圍區域對 y- 軸旋轉的旋轉體體積.
Solution:
M ethod I ˆ 1
0
2πx√
x2+ 1 dx (5%)
= 2
3π(x2+ 1)32|10 (3%)
= (4√
2 − 2)π
3 (2%)
M ethod II ˆ √2
1
π(12− (y2− 1)) dy + ˆ 1
0
π dy
= (4√
2 − 2)π 3
6. (15%)
(a) (8%) 寫出 f(x) = sin x 在 x = 0 處之第四次泰勒多項式,並寫出泰勒定理中之餘項。
(b) (7%) 求 sin 20◦ 之近似值,誤差小於 10−4。答案可用 π 表示,不必帶入 π = 3.14 · · · ,餘項 需仔細估計,以證明近似值之準確度。
Solution:
(a)
f (x) = x − x3/3! + cos(c)x5/5! , form some c (b)
sin(π/9) = (π/9) − (π/9)3/3! + cos(c)(π/9)5/5! ( 3 points )
|R(π/9)| ≤ (π/9)5/120 , where R(π/9) means the remainder term ( 2 points ) compute (π/9)5/120 ≤ (2/5)5/120 ≤ 4/46875 ≤ 1/10000 ( 2 points )
7. (12%) (a) (6%) 求 lim
x→∞x1x。 (b) (6%) 求 lim
x→0
ln(1 + x2) 1 − cos x 。 Solution:
(1)limx−>∞x1x = limx−>∞exp(1
xln(x))(3pt) If has some calculation error 2pt
= limx−>∞exp(1
x)(LHrule) = 1(3pt) (2) limx−>0ln(1 + x2)
1 − cos(x)=limx−>0
2x 1+x2
sin(x)(LH rule) (3PT) =limx−>0
2 1+x2 sin(x)
x
=2 (3PT)
8. (12%) 一人欲扛著長竹竿,要維持水平通過下圖之窄巷(寬為 1, 1 公尺),請問竹竿之長度最長可 以幾公尺?
1
1
Solution:
Observe that the length must be not larger than f (θ) := 1/ sin θ + 1/ cos θ, so it suffices to find the min. of f. (3pt)
f0(θ) = sin θ/ cos2θ − cos θ/ sin2θ = sin3θ − cos3θ
sin2θ cos2cos2θ, so f’=0 at θ = π/4 (3pt); f’> 0 for π/4 < θ < π/2 and f’< 0 for 0 < θ < π/4 (3pt).
So length ≤ f (π/4) = 2√
2 ≤ f (θ) (3pt)