1021微微微甲甲甲01-04班班班期期期末末末考考考解解解答答答 1. (10%)
(a) Evaluate the limit
Im = lim
n→∞
nm
X
i=1
i2n3
n6+ i6 = lim
n→∞
n3
n6+ 16 + 22n3
n6+ 26 + · · · + (nm − 1)2n3
n6+ (nm − 1)6 + (nm)2n3 n6+ (nm)6
,
where m is a positive integer.
(b) Compute lim
m→∞Im. Solution:
(a)(6%)
The original equation = 1
nΣnmi=1 i2n4 n6+ i6
= 1
nΣnmi=1 (ni)2 1 + (ni)6
By the definition of Riemann sum,
= ˆ m
0
x2
1 + x6dx (3 points)
= 1 3
ˆ m
0
dx3
1 + (x3)2 (1 point)
= 1
3tan−1(x3) |m0 +C
= 1
3tan−1(m3) + C (2 points) Therefore, Im = 1
3tan−1(m3), where m is a positive integer . ps. Other methods for solving
ˆ m
0
x2
1 + x6dx are permitted.
(b)(4%)
m→∞lim Im = lim
m→∞
1
3tan−1(m3)
= 1 3 · π
2
= π 6.
2. (15%) Evaluate the integral.
(a) ˆ 1
0
x2√3
1 − xdx.
(b) ˆ
x√
3 − 2x − x2dx.
Solution:
(a)(8%) Let u = √3
1 − x, and we can find that u3 = 1 − x and dx = −3u2du . The original equation =
ˆ 0 1
(1 − u3)2 · u · (−3u2)du
= −3 ˆ 0
1
(u9− 2u6+ u3)du (4 points)
= −3( 1
10u10−2
7u7+ 1
4u4) |01 (2 points)
= 27
140. (2 points)
ps. Other methods for solving this integral are permitted.
(b)(7%)
The original equation = ˆ
xp
4 − (x + 1)2 dx ...Equation(1) Let x + 1 = 2 sin θ , and we can find that dx = 2 cos θ dθ ...Equation(2)
Substitute Equation(2) into Equation(1), (3 points for Trigonometric substitution) and we have
ˆ
(2 sin θ − 1) · 2 cos θ · 2 cos θ dθ. (3 points)
= −8
3 cos3θ − 2θ − sin 2θ + C
= −1
3 (3 − 2x − x2)32 − 2 sin−1(x + 1
2 ) − x + 1 2
√
3 − 2x − x2+ C. (1 point) ps. Other methods for solving this integral are permitted.
3. (20%) Evaluate the integral.
(a)
ˆ sin x cos x sin4x + cos4xdx.
(b) ˆ ∞
1
x2− 3
(x2− 2x + 3)(x2+ 2x + 3)dx.
Solution:
(a) (8%) Sol. (1)
ˆ sin x cos x
sin4x + cos4xdx =
ˆ 1
2sin 2x
(1−cos x2 )2+ (1+cos x2 )2dx with u = cos 2x, du = −2 sin 2xdx
= −1 2
ˆ 1
1 + u2du
= −1
2tan−1u + C
= −1
2tan−1(cos 2x) + C
Sol. (2)
ˆ sin x cos x
sin4x + cos4xdx =
ˆ tan x sec2x 1 + tan4x dx
with u = tan2x, du = 2 tan x sec2xdx
= ˆ 1
2du 1 + u2
= 1
2tan−1u + C
= 1
2tan−1(tan2x) + C
Sol. (3)
ˆ sin x cos x
sin4x + cos4xdx =
ˆ sin x cos x
sin4x + (1 − sin2x)2dx
with u = sin2x, du = 2 sin x cos xdx
=
ˆ 1
2du u2+ (1 − u)2
= 1 2
ˆ 1
2u2− 2u + 1du
= 1 2
ˆ 2
(2u − 1)2+ 1du
= 1
2tan−1(2u − 1) + C
= 1
2tan−1(2 sin2x − 1) + C
Grading policy:
(1) 4% for any proper change of variables (including correct du).
(2) 4% for correct integration. -1% if C is missing.
(b) (12%)
(1) Partial fraction (5%)
Define f (x) = x2− 3
(x2 − 2x + 3)(x2+ 2x + 3)
= Ax + B
x2− 2x + 3 + Cx + D
x2+ 2x + 3 (2% for the form)
⇒
x3: A + C = 0
x2: 2A + B − 2C + D = 1 x1: 3A + 2B + 3C − 2D = 0
x0: 3B + 3D = −3
⇒
A = 1 2 B = −1
2 C = −1 2 D = −1 2
∴ f (x) = 1
2( x − 1
x2− 2x + 3 − x + 1
x2 + 2x + 3) (3%)
(2) Integration (4 %)
ˆ 1
2( x − 1
x2− 2x + 3 − x + 1
x2 + 2x + 3)dx
= ˆ 1
4( 2x − 2
x2 − 2x + 3− 2x + 2 x2+ 2x + 3)dx
= 1 4(ln
x2− 2x + 3 − ln
x2+ 2x + 3 ) + C
= 1 4ln
x2− 2x + 3 x2 + 2x + 3
+ C
(3) Improper integral (3 %) ˆ ∞
1
f (x)dx
= lim
b→∞
ˆ b 1
1
4( 2x − 2
x2− 2x + 3 − 2x + 2 x2+ 2x + 3)dx
= lim
b→∞[1 4ln
x2− 2x + 3 x2+ 2x + 3 ]b1
= lim
b→∞
1 4ln
b2− 2b + 3 b2 + 2b + 3
−1 4ln
1 − 2 + 3 1 + 2 + 3
= 1
4ln 1 − 1 4ln1
3
= 1 4ln 3
(Upper limit: 2%; lower limit: 1%.) Grading policy:
The three parts are credited independently, e.g., you will still get full 4% in part (2) if the integration process is correct based on your result in (1), even though your result in (1) may be incorrect.
4. (10%) A function y = y(x) satisfies the equation
y(x) = x + ˆ x2
0
x − y(√ t) −√
t − 1 2√
t + 1
etdt, x ≥ 0.
(a) Find a differential equation with initial condition for y.
(b) Solve the differential equation.
Solution:
y = x + x ˆ x2
0
etdt − ( ˆ x2
0
(y(√ t) +√
t + 1 2√
t − 1)etdt) (1 point) Differentiate both side,
dy
dx = 1 + ˆ x2
0
etdt + x · 2xex2 − (y(x) + x + 1
2x − 1) · ex2 · 2x
dy
dx = 2xex2(1 − y(x)) or dy
dx+ 2xex2y(x) = 2xex2 (3 points)
and boundary condition y(0) = 0 (1 point) you may choose intergal factor or separable method,
• intergal fatcor: ˆ
2xex2dx = eex2(2 points)
⇒ (eex2y)0 = 2xex2eex2
⇒ eex2y(x) = ˆ
2xex2eex2dx = eex2 + C1 (1 point)
• separable mentod: ˆ 1
1 − ydy = ˆ
2xex2dx
⇒ − ln(1 − y) = ex2 + C2
⇒ y(x) = 1 − e−ex2−C2 apply boundary condition y(0) = 0,
⇒ y(x) = 1 − e1−ex2 (2 points) if you make a mistake at the first step:
y = 1 + (x − y(x) − x − 1
2x + 1) · 2xex2
we’ll give you 2 points , write correct boundary condition, giving 1 point and finally you calculate integral factor, giving 2 points; totally 5 points.
5. (10%)
(a) If the infinite curve y = e−x, x ≥ 0, is rotated about the x-axis, find the area of the resulting surface.
(b) Find the arc length of the infinite curve with polar equation r = θ−1, θ ≥ 1.
Solution:
(a)
The area of surface is
S = 2π ˆ ∞
0
e−xp
1 + (−e−x)2dx(2pt)
= 2π ˆ 0
1
−√
1 + u2du
= 2π ˆ π
4
0
p1 + tan2y sec2ydy
= 2π ˆ π
4
0
sec3ydy(1pt)
= π[log(√
2 + 1) +√
2](2pt) (b)
The arc length is
L = ˆ ∞
1
r
r2+ (dr dθ)2dθ ˆ ∞
1
√θ−2+ θ−4dθ(2pt) ˆ ∞
1
1 θ−2
√1 + θ2dθ
>
ˆ ∞ 1
1 θ−2
√ θ2dθ
= ˆ ∞
1
1
θdθ = ∞(3pt)
6. (20%) The figure shows a curve C with the property that, for every point P on the middle curve y = 2x2, the area of B is twice the area of A.
P
A B y
2x2
y yx2
x )
( :x gy C
(a) Find an equation x = g(y) for C.
(b) Let R be the region bounded by the curve C, y = x2, x = 2 and y = 8. Find the volume of the solid obtained by rotating R about the x-axis.
(c) Find the y-coordinate of the centroid of R.
Solution:
(a) Assume P (t, 2t2) is a point on the curve y = 2x2. The area of A =
ˆ t 0
(2x2− x2)dx = ˆ t
0
x2dx = 1
3x3|t0 = 1 3t3 The area of B =
ˆ 2t2
0
(r y
2 − g(y))dy = 1
√2 ·2
3y32|2t02 − ˆ 2t2
0
g(y)dy
=
√2 3 (2√
2t3) − ˆ 2t2
0
g(y)dy = 4 3t3−
ˆ 2t2 0
g(y)dy B = 2A ⇒
ˆ 2t2 0
g(y)dy = 2
3t3 3 pts d
dt ˆ 2t2
0
g(y)dy = d dt(2
3t3) 2 pts
By the Fundamental Theorem of Calculus, g(2t2) · 4t = 2t2 ⇒ g(2t2) = 2t Let y = 2t2 ⇒ t =r y
2 ⇒ g(y) = 1 2
r y
2. 3pts
That is, y = 8x2.
(b) Note that the point (2, 8) is on the curve y = 2x2. Sol 1.
V =
ˆ 8 0
2πy(r y 2 −1
2 r y
2)dy + ˆ 2
0
π((2x2)2− (x2)2)dx 2pts
= 2π ˆ 8
0
1 2√
2y32dy + π ˆ 2
0
3x4dx
= 1
√2π(2
5y52)|80+ π(3 · 1 5x5)|20
= 1
√2π ·2 5(128√
2) + π(3 5 · 32)
= 256
5 π + 96
5 π = 352
5 π 4pts
Sol 2. On the curve C, when y = 8 ⇒ x = 1 2
r8 2 = 1
V = ˆ 1
0
π((8x2)2 − (x2)2)dx + ˆ 2
1
π((8)2− (x2)2)dx 2pts
= π ˆ 1
0
63x4dx + π ˆ 2
1
(64 − x4)dx
= π(63
5 x5)|10+ π(64x −1 5x5)|21
= 63
5 π + (128 − 32
5 − 64 + 1
5)π = 352
5 π 4pts
(c)
The area of R = ˆ 1
0
(8x2− x2)dx + ˆ 2
1
(8 − x2)dx
= 7
3x3|10+ (8x − 1
3x3)|21 = 7
3 + 16 − 8
3− 8 + 1
3 = 8 3pts or The area of R = 3
ˆ 2 0
(2x2− x2)dx = 3(1
xx3)|20 = 3 · 8
3 = 8 3pts
Sol 1. By the Pappus Theorm, V = 2π ¯yArea. Hence ¯y = V
2πArea = (352
5 π)/(2π · 8) = 22
5 . 3pts
Sol 2.
moment = ˆ 1
0
1
2((8x2)2− (x2)2)dx + ˆ 2
1
1
2((8)2− (x2)2)dx
= 1 2 ·63
5 x5|10+1
2(64x − 1 5x5)|21
= 1 2 ·63
5 +1
2(128 − 32
5 − 64 + 1
5) = 176 5 Therefore ¯y = moment
Area = 176 5 ·1
8 = 22
5 . 3pts
7. (15%) Let the curve C defined by
x = t2 y = t3
3 − t, t ∈ R.
(a) Find the point P where the curve intersects itself.
(b) Find the equation of the tangent lines at the point P . (c) For which values of t is the curve increasing?
(d) For which values of t is the curve concave upward?
(e) Sketch the curve C.
Solution:
(a) (3分)
假設C上的兩點為(t21,t31
3 − t1) , (t22,t32
3 − t2)。 由P 為自交點我們可以得到兩條方程式:
t21 = t22 (方程式一)
和 t31
3 − t1 = t32
3 − t2 (方程式二) 由方程式一可以得到(t1− t2)(t1+ t2) = 0，即t1 = t2或t1 = −t2
(t1 = t2不合，因為自交點的定義是不同的參數t會得到同一點)。
因此t1 = −t2。接著將t1 = −t2代入方程式二，得到
−t32
3 + t2 = t32 3 − t2 解出來可得到t2 = 0或t2 = ±√
3(t2 = 0不合，因為t2 = 0會導致t1 = t2)。 令t2 =√
3(所以t1 = −√
3)，就可以得到x = 3 , y = 0，所以P = (3, 0) (註:令t2 = −√
3也可以 算出相同答案，它們是兩種不同C的參數化)。
(b) (3分)
曲線C上的每點切線斜率:
dy dx =
dy dt dx dx
= t2− 1 2t
由(a)知道切點為(3, 0)，我們可以假設切線方程式為y − 0 = t2− 1
2t (x − 3)，接著把從(a)算出 來的t1 , t2代入該方程式:
t = t2 =√
3 ⇒ y − 0 = (√
3)2− 1 2√
3 (x − 3) ⇒ y = 1
√3(x − 3)
t = t1 = −√
3 ⇒ y − 0 = (−√
3)2− 1 2(−√
3) (x − 3) ⇒ y = −1
√3(x − 3) 因此 y = 1
√3(x − 3) 和 y = −1
√3(x − 3) 即為該點切線方程式。
(c) (3分)
由一階檢定可知 dy
dx ≥ 0 ⇒ C 遞增，所以 dy
dx ≥ 0 ⇐⇒ t2− 1
2t ≥ 0 ⇐⇒ 2t(t2 − 1) ≥ 0但t 6= 0 ⇐⇒ t(t − 1)(t + 1) ≥ 0
但t 6= 0 ⇐⇒ t ≥ 0或 − 1 ≤ t < 0。
因此當 t ≥ 0或 − 1 ≤ t < 0 時，曲線 C 遞增。
(d) (3分)
由二階檢定可知 d2y
dx2 > 0 ⇒ C 開口向上，所以 d2y
dx2 > 0 ⇐⇒ d2y
dx2 = d(dydx) dx =
d dt(dxdy)
dx dt
> 0 ⇐⇒
d dt(dydx)
dx dt
=
(2t)2−2(t2−1) (2t)2
2t > 0
⇐⇒ t2+ 1
2t3 > 0 ⇐⇒ t3 > 0 ⇐⇒ t > 0 因此當 t > 0 時，曲線 C 開口向上。
(e) (3分)
利用(a)(b)(c)(d)來畫圖(註:可以試著把參數式寫成x和y的表示式:y2 = x3
9 − 2x2
3 + x，不難發 現它是對x軸對稱)
8. (10%)
(a) Find the points of intersection of the curves r = 1 + 2 cos θ and r2 = cos θ.
(b) Find the area of the region in the second quadrant that lies inside r2 = cos θ and outside r = 1 + 2 cos θ.
Solution:
(a)
Claim the intersection points for for polar coordinate are [0, 0], [1, 0], [1
2, ± cos−1−1 4].
Equally, for Cartesian coordinate, they are
(0, 0), (1, 0), (−1 8, ±
√15 8 ).
Part 1: the origin.
Claim [0, 0] is an intersection point.
For r = 1 + 2 cos θ, there is θ = 2π
3 such that r = 1 + 2 cos θ = 1 + 2 cos 2π 3 = 0.
For r2 = cos θ, there is θ = π
2 such that r2 = cos θ = cosπ 2 = 0.
We find that [0, 0] is on the both curves so is an intersection point.
Part 2: other points.
A point [r0, θ0] on r = 1 + 2 cos θ is also on r2 = cos θ if and only if Case 1: [r0, θ0] satisfies r2 = cos θ.
Case 2: [−r0, θ0+ π] satisfies r2 = cos θ.
Case 1:
We have
(1 + 2 cos θ0)2 = r20 = cos θ0 1 + 3 cos θ0+ 4 cos2θ0 = 0 But there is no solution for such θ0.
Case 2:
We have
(1 + 2 cos θ0)2 = r20 = (−r0)2 = cos (θ0+ π) = − cos θ0
1 + 5 cos θ0+ 4 cos2θ0 = 0 We find cos θ0 = −1, −1
4.
When cos θ0 = −1, θ0 = π and r0 = 1 + 2 cos θ0 = −1, so [r0, θ0] = [−1, π] = [1, 0].
When cos θ0 = −1
4, θ0 = ± cos−1−1
4 and r0 = 1 + 2 cos θ0 = 1 2, so [r0, θ0] = [1
2, ± cos−1−1 4].
In this, the arc cosine function is denoted as cos−1.
(b)
Observe that the boundary of the desired region are x-axis, r = −
√
cos θ for θ from π + cos−1−1
4 to 2π, and
r = 1 + 2 cos θ for θ from cos−1−1 4 to 2π
3 . Therefore, the area of the region is
ˆ 2π π+cos−1−1
4
1 2(−√
cos θ)2dθ − ˆ 2π
3
cos−1−1
4
1
2(1 + 2 cos θ)2dθ
= 1 2
ˆ 2π π+cos−1−14
cos θdθ − 1 2
ˆ 2π
3
cos−1−14
1 + 4 cos θ + 4 cos2θdθ
= 1
2 sin θ|2ππ+cos−1−14 − 1
2(θ + 4 sin θ + sin 2θ + 2θ)
2π 3
cos−1−14
= 1 2
√15 4 −1
2 2√ 3 −
√3
2 + 2π −√ 15 +
√15
8 − 3 cos−1−1 4
!
= 9√ 15 16 − 3√
3
4 − π + 3
2cos−1−1 4
Criteria:
(a)
There are four criteria:
1. Explain that the origin [0, 0] point is an intersection point. 1 point.
2. Use the technique of polar coordinate to find that [1, 0] is an intersection point. 1 point.
3. Solve and find 2 intersection points for case of arc cosine. 1 point for each.
There are other situations:
1. Only write down the correct answer. 1 point.
2. Write some meaning computation but no identify any intersection point. 1 point.
(b)
1. Write down the correct integral range. 1 point for each.
2. Write down the correct integrated functions.1 point . 3. Compute this integral for first part: cos θ/2. 1 point .
4. Compute this integral for second part: (1 + 2 cos θ)2/2. 1 point . 5. Find the final answer. 1 point.
