1101模模模組組組17-19班班班 微微微積積積分分分2 期期期考考考解解解答答答和和和評評評分分分標標標準準準 1. 求以下定積分。 Evaluate the following definite integrals.
(a) (8%)∫
√3
1
1 x
√ 4 − x2
dx (b) (8%)∫
1 0
cos−1(
√x) dx
Solution:
(a) Let x = 2 sin t. Then dx = 2 cos t dt (1%). Since x ∈ [1,
√
3], we have that t ∈ [π/6, π/3] (1%) and cos t > 0 (1%). Thus
∫
√ 3 1
dx x
√
4 − x2dx =∫
π/3 π/6
2 cos t dt
2 sin t ⋅ 2 cos t (1%)
= 1 2∫
π/3 π/6
csc t dt (1%)
= − 1
2ln ∣ csc t + cot t∣∣
π/3 π/6 (2%)
= − 1 2ln
√ 3 2 +
√
3 (1%).
(b) Let t =√
x. Then dt = 1 2√
xdx = 1
2tdx (1%).
∫
1 0
cos−1(
√x) dx = 2∫
1 0
cos−1(t)t dt (1%)
= ∫
1 0
cos−1(t) dt2 (1%)
=cos−1(t)t2∣
1 0+ ∫
1 0
t2
√
1 − t2dt (2%) t ∶= sin u =0 +∫
π/2 0
sin2u
cos u cos u du (1%)
= ∫
π/2 0
1 − cos 2u
2 du (1%)
= π
4 (1%).
2. 求以下不定積分。 Evaluate the following indefinite integrals.
(a) (4%)∫
x + 2
x2+4x + 5dx (b) (4%)∫ 1
x2+4x + 5dx (c) (6%)∫
x − 1
x2(x2+4x + 5)dx
Solution:
(a) Let u = x2+4x + 5
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M)
. Then du = (2x + 4) dx
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M)
. Therefore,
∫
x + 2
x2+4x + 5dx =1 2∫
1 udu
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M)
= 1
2ln(x2+4x + 5) + C
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M)
Remark. Do it by inspection is OK.
(b) ∫ 1
x2+4x + 5dx =∫
1
(x + 2)2+1dx
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(2M)
=tan−1(x + 2) + C
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(2M)
.
(c) Suppose x − 1
x2(x2+4x + 5)= A x + B
x2+ Cx + D
x2+4x + 5 (1M).
By clearing the denominator,
(x − 1) = Ax(x2+4x + 5) + B(x2+4x + 5) + Cx3+Dx2
We obtain
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎩
A + C = 0 4A + B + D = 0 5A + 4B = 1 5B = −1
(1M).
Solving gives A = 9
25, B = −1
5, C = −9
25, D = −31
25 (0.5 each)
Hence, the given integral equals
∫ 9 25⋅1
x−1 5⋅ 1
x2− 9x + 31
x2+4x + 5dx =∫ 9 25⋅ 1
x−1 5⋅ 1
x2− 9
25⋅ x + 2
x2+4x + 5−13 25⋅ 1
x2+4x + 5dx
= 9
25ln ∣x∣ + 1 5x−
9
50ln(x2+4x + 5) −13
25tan−1(x + 2) + C
(1M for attempt and 1M for final answer)
Page 2 of 7
3. 計算以下瑕積分或者說明為什麼它是發散的。
Evaluate each of the following improper integrals, or explain why it is divergent.
(a) (8%)∫
∞ 2
1
x(ln x)2dx (b) (4%)∫
2 1
1 x(ln x)2dx
Solution:
The antiderivative:
∫ 1
x(ln x)2 dx =∫ 1
(ln x)2 d(ln x) Substitution u = ln x
= ∫ u−2du = −u−1+C = −1 ln x+C (a) The improper integral is convergent and it converges to
lim
b→∞
1 ln 2−
1 ln b=
1 ln 2 (b) The improper integral is divergent because
lim
a→1+
− 1 ln 2+
1 ln a=
−1 ln 2+ lim
t→0+
1 t = ∞ The limit does not exist.
Grading scheme:
(4 pts) Finding the antiderivative. Or converting the improper integral to another one.
∫
∞ 2
1
x(ln x)2 dx =∫
∞ ln 2
1
u2 du and ∫
2 1
1
x(ln x)2 dx =∫
ln 2 0
1 u2 du (4 pts) Explaining why (a) is convergent and evaluating. (-2 pts) if student forgot to evaluate.
(4 pts) Explaining why (b) is divergent.
If the student found the wrong antiderivative or copy the problem wrong, a maximum of (8 pts) is possible. If the problem became too easy to determine convergence/divergence, then the maximum would be (4 pts). The explanation must involve a limit notation, otherwise (-2 pts).
4. 設 R 為曲線 y =1
3x32 和 y = x 在 x = 0 和 x = 9 之間圍成之區域。
Let R be the region enclosed by the curves y =1
3x32 and y = x between x = 0 and x = 9.
(a) (8%)求 R 繞 x-軸旋轉的旋轉體體積。
Find the volume of the solid of revolution obtained by rotating R about the x-axis.
(b) (8%)求 R 繞 y-軸旋轉的旋轉體體積。
Find the volume of the solid of revolution obtained by rotating R about the y-axis.
(c) (8%) 求 R 的總周長。（包括曲線和直線部份）
Find the perimeter of R (that is, the combined length of the two arcs).
x y
(0, 0)
(9, 9)
y = x
y = 1 3
√ x3
Solution:
(a) The integral that evaluates the volume can be set up in two ways.
∫
9 0
[π (x)2−π (1 3x3/2)
2
] dx =∫
9 0
2π (y) [(3y)2/3−y] dy We can evaluate either.
∫
9 0
[π (x)2−π (1 3x3/2)
2
] dx = π 9∫
9 0
(9x2−x3) dx = π 9[3x3−
x4 4 ]
9
0
= 243π
4 (b) The integral that evaluates the volume can be set up in two ways.
∫
9
0 2π (x) (x −1
3x3/2) dx =∫
9 0
[π ((3y)2/3)
2
−π (y)2] dy We can evaluate either.
∫
9 0
2π (x) (x −1
3x3/2) dx = 2π 3 ∫
9 0
(3x2−x5/2) =2π 3 [x3−
2 7x7/2]
9
0
= 486π
7 (c) The length of the line segment from (0, 0) to (9, 9) is 9
√ 2.
The length of the curve y =1
3x3/2 is given by
∫
9 0
¿ ÁÁ À1 + (dy
dx)
2
dx =∫
9 0
√ 1 +x
4 dx = 1
3[(x + 4)3/2]
9 0=
13√ 13 − 8 3 The combined length is 1
3(13
√ 13 + 27
√ 2 − 8).
Grading scheme:
(4 pts) for each correct integral setup. (-4 pts) if they confuse (a) and (b).
Students do not lose points if they notice a negative answer and add a negative sign to fix it. (-4 pts) for each negative volume or length.
(-2 pts) for each computational mistake. (-1 pt) if they over-simplified.
(-2 pts) if students forgot 9
√ 2.
Page 4 of 7
5. (a) (4%) 寫出 1
√
1 − x2 在 x = 0 的泰勒展開式。(你可以用 Cra 這個記號表示答案。）
Write down the Taylor series of 1
√
1 − x2 at x = 0. (You may express your answer in terms of Cra notation.) (b) (6%)假設 sin−1x在 x = 0 的泰勒展開式為 :
sin−1x = a1x + a3x3+a5x5+a7x7+ ⋯ 寫出 a1, a3 和 a5的值。 (答案請用分數表示)
Suppose the Taylor expansion of sin−1x at x = 0 is
sin−1x = a1x + a3x3+a5x5+a7x7+ ⋯ Write down the values of a1, a3 and a5 as a fraction.
Solution:
(a) By the binomial theorem, we have
(1 − x2)−1/2=C0−1/2+C1−1/2(−x2) +C2−1/2(−x2)2+ ⋯
=C0−1/2−C1−1/2x2+C2−1/2x4− ⋯ (4%).
(正負號打錯的話扣一分, x的次方打錯扣一分)
(b) We have that
sin−1x =∫
x 0
(1 − t2)−1/2dt (2%)
=C0−1/2x − C−1/2⋅x2
3 +C2−1/2⋅x5
5 − ⋯ (1%)
=x +x3 6 +
3
40x5+ ⋯.
Hence a1=1(1%), a3= 1
6(1%) and a5= 3 40(1%).
6. 計算以下極限。 Evaluate the following limits.
(a) (8%) lim
x→0+
1
√x∫
√x
−
√x
e−t2dt.
(b) (8%) lim
x→0
e−x3−1 + x3
x6 .
(提示 : 先寫出 e−x3在 x = 0 的泰勒展開式。) (Hint. Write down the Taylor series of e−x3 at x = 0) (c) (8%) lim
x→0
(cos x)1x
Solution:
(a)
lim
x→0+
∫
√x
−
√xe−t2dt
√x
(0/0)
= lim
x→0+
e−x⋅ 1
2√
x−e−x⋅ (− 1
2√ x)
1 2√ x
= lim
x→0+2e−x
=2 Marking scheme for (a)
1M in correctly identifying the indeterminate form
1M for using L’Hospital’s rule
4M for correct derivative of numerator (2M for applying FTC, 1M for the term from chain rule, 1M for the term arisen from the lower bound)
1M for correct derivative of denominator
1M for correct answer (b) Since ex=1 + x +x2
2! + ⋯, we have e−x3=1 − x3+ x6
2 + ⋯Hence,
x→0lim
e−x3−1 + x3 x6 =lim
x→0
(1 − x3+x
6
2 + ⋯) −1 + x3 x6
=lim
x→0
1 2+ ⋯
= 1 2
Marking scheme for (b)
2M for knowing the series expansion for ex
2M for the correct series expansion for e−x3
2M for simplifying the fraction by plugging in the series
2M for correct answer
We accept doing (b) by L’Hospital-ing six times. In that case, 1M for each correct application of L’H and the final 2M for the correct answer.
(c) Let y = (cos x)x1. Then ln y =ln(cos x)
x . Therefore, lim
x→0+ln y = lim
x→0+
ln(cos x) x
(0/0)
= lim
x→0+
−tan x 1
=0 Hence, lim
x→0(cos x)1x =e0=1.
Marking scheme for (c)
Page 6 of 7
2M for taking log
1M in correctly identifying the indeterminate form
1M for correct derivative of numerator
1M for correct derivative of denominator
2M for correct limit of ln y
1M for correct answer
