98學年度第1學期 微積分甲二組期末考評分標準
1. (10%) Evaluate the integral
Z dx x3− 1. 評分標準:
Deduct 1% for each computational error.
Step 1. Apply the partial fraction decomposition of the integrand, we have (11) (1%)
1
x3− 1 = a
x − 1 + bx + c x2+ x + 1. (12) (2%)
Determine the coefficients a, b, c by the following identity
1 = a(x2+ x + 1) + (bx + c)(x − 1).
Note that a + b = 0.
Let x = 1, we have 1 = 3a. Hence,
a = 1/3 and b = −1/3.
Set x = 0, we have 1 = a − c or c = −2/3.
Step 2. (2%) Evaluate the integral aR 1
x−1dx.
Note that
1 3
Z 1
x − 1dx = 1
3ln |x − 1| + C1. Step 3. (5%) Evaluate the integral
Z bx + c x2+ x + 1dx by writing
1 3
Z x + 2
x2+ x + 1dx = 1 6
Z 2x + 1
x2+ x + 1dx + 1 2
Z 1
x2+ x + 1dx.
(31) (2%) Note that
1 6
Z 2x + 1
x2+ x + 1dx = 1
6ln(x2+ x + 1) + C2.
(32) (3%) Write
x2+ x + 1 =
x +1
2
2
+3 4 and denote x + 1/2 by (√
3/2) tan θ. Hence,
x2+ x + 1 = 3
4(tan2θ + 1) = 3 4sec2θ and dx =√
3/2 sec2θ dθ.
1 2
Z 1
x2+ x + 1dx = 1 2
√2 3
Z
dθ = 1
√3θ+C3 = 1
√3tan−1θ+C3 = 1
√3tan−1 x + 1/2
√3/2 +C3.
2. (10%) Evaluate the integral
Z x
√x2+ 2x + 2dx.
評分標準:
Z x
√x2+ 2x + 2dx =
Z x + 1
p(x + 1)2+ 1 − 1
p(x + 1)2+ 1dx
=
Z u
u2+ 1du −
Z du
√u2+ 1
=√
u2+ 1 − ln u +√
1 + u2 + c
=√
x2+ 2x + 2 − ln
(x + 1) +√
x2+ 2x + 2 + c Where the 3rd equality follows by:
u = tan θ; du = sec θ; tan2θ + 1 = sec2θ
Z sec2θ sec θdθ =
Z
sec θdθ
= ln |sec θ + tan θ| + c
= ln u +√
1 + u2 + c
eq1 = 3 credits , eq2 = 1 credits, eq3 = 5 credits, eq4 = 1 credits 3. (10%) Let In =
Z ∞ 0
xne−xdx.
(a) Find the recursive relation between In and In−1.
(b) Compute I3.
(c) Find the general formula of In. 評分標準:
(a) (5%)
In = Z ∞
0
xne−xdx = lim
a→∞
Z a 0
xne−xdx
= lim
a→∞−xne−x
a 0
+ n Z a
0
xn−1e−xdx
= lim
a→∞−xne−x
a 0
+ nIn−1
Since lim
a→∞−xne−x
a 0
= lim
a→∞
an
ea = 0 (1%) Hence In= nIn−1
Note : If you did not use ”Improper Intergal” to compute this probelm, then your socore will be deducted 1 point.
(b) (3%) Follow (a) we have I3 = 3I2 = 3 · 2I1 = 3 · 2 · 1I0 Moreover, we have I0 =
Z ∞ 0
e−xdx = lim
a→∞
Z a 0
e−xdx = lim
a→∞−e−x
a 0 = 1 Hence I3 = 3! = 6.
(c) (2%) Same discuss as (b) we can get the general formula of In In= nIn−1 = n · (n − 1) · In−1 = · · · = n!I0 = n!
4. (15%) Given 0 < a < b, find the arc length of y = lnex+ 1
ex− 1 for a ≤ x ≤ b.
評分標準:
dy
dx = (eexx+1−1)0
ex+1 ex−1
=
ex(ex−1)−ex(ex+1) (ex−1)2
ex+1 ex−1
= −2ex
(ex− 1)2 × ex− 1
ex+ 1 = −2ex
e2x− 1 (3%)
L = Z b
a
r
1 + (dy
dx)2dx = Z b
a
r
1 + ( −2ex
e2x− 1)2dx (2%)
= Z b
a
s
e4x+ 2e2x+ 1 (e2x− 1)2 dx =
Z b a
e2x+ 1
e2x− 1dx (3%)
Let t = ex, then dt
dx = ex → dx = dt t . x = a → t = ea; x = b → t = eb
L = Z eb
ea
t2+ 1
t(t + 1)(t − 1)dt (4%)
= Z eb
ea
(−1 t + 1
t + 1+ 1 t − 1)dt
= (− ln t + ln (t + 1) + ln (t − 1))
eb ea
= ln e2b− 1
e2a− 1 + a − b (3%)
5. (10%) The region in the first quadrant enclosed by x = y2, x-axis and x = 1 is rotated about the line y = 3. Find the volume of the solid.
評分標準:
(1) Disk Method:
Area = Z 1
0
π · 32− π(3 −√ x)2dx
= Z 1
0
6π√
x − πxdx
= 4πx3/2− π 2x2
1 0
= 7 2π (2) Shell Method:
Area = Z 1
0
2π(3 − y)(1 − y2)dy
= Z 1
0
2π(3 − y − 3y2+ y3)dy
= 2π
3y − y2
2 − y3+y4 4
1 0
= 7 2π (3) Assessment criteria:
The integral is worthy of 4 points, and the integration techniques are of 5 points. But
you will not get any credit for your integration process unless you use the right method to write down an integral. The remained 1 point is for the right answer.
6. (10%) Find the orthogonal trajectories of {y2 = kx3, k ∈ R} which passes through the point (1, 2).
評分標準:
dy
dx = −2x
3y (4%)
−x2+ c = 3
2y2 (4%) 3
2y2 = −x2+ 7 (2%)
7. (10%) (a) Find the particular solution of 3xy0− y = ln x + 1, x > 0, satisfying y(1) = −2.
(b) Find the value y(e).
評分標準:
(a) y0− 1
3xy = ln x + 1
3x (1%)
integrating factor V = eR−3x1dx = x−13 (2%) then
x−13y = 1 3
Z
(ln x + 1)x−43dx
= −x−13(ln x + 1) + Z
x−43dx (integration by part)
= −x−13(ln x + 1) − 3x−13 + C
So y = −(ln x + 4) + Cx13 (4%).
When x = 1, y = 2 then −2 = −4 + C so C = 2 i.e. y = 2x13 − ln x − 4 (2%)
(b) y(e) = 2e13 − 5 (1%)
8. (10%) Given a > 0, one arch of the cycloid x = a(θ − sin θ), y = a(1 − cos θ), 0 ≤ θ ≤ 2π, is rotated about the x-axis. Find the area of the resulting surface.
評分標準:
A = Z 2π
0
2πa(1 − cos θ) q
a2(1 − cos θ)2+ a2sin2θ (3%)
= Z 2π
0
2πa(1 − cos θ)√ 2a√
1 − cos θ
= 2πa2 Z 2π
0
4 sin3(θ 2)dθ
= 16πa2 Z π
0
sin3αdα (3%)
= 16πa2 Z π
0
(1 − cos2α) sin αdα
= 16πa2 Z 1
−1
(1 − u2)du
= 16πa24 3
= 64
3 πa2 (4%)
9. (15%) (a) Plot the region A which is inside the circle r = 6 cos θ and outside the cardioid r = 2(1 + cos θ).
(b) Find the area of A.
(c) Find the length of the boundary of the region A.
評分標準:
(a) (3%) The graph r = 6 cos θ is a circle with radius 3 and center (3, 0) The graph r = 2 + 2 cos θ is shaped like a heart symmetric at x-axis The graph outside the cardiod and inside the circle is shaped like moon (b) First we find their intersection points.
2(1 + cos θ) = 6 cos θ ⇒ cos θ = 0.5, θ = π
3 and − π 3 By symmetry
Area = 2 × 0.5hZ π3
0
(6 cos θ)2dθ − Z π3
0
4(1 + cos θ)2dθi (2%)
= 36 Z π3
0
(cos θ)2dθ − 4 Z π3
0
(1 + cos θ)2dθ
= Z π3
0
h
18 cos 2θ + 18 − 4 − 8 cos θ − 2 cos 2θ − 2i
dθ (2%)
= (9 sin 2θ + 18θ − 4θ − 8 sin θ − sin 2θ − 2θ)
π 3
0
= 4π (2%)
(c) The boundary of the region has two parts.
The first part is formed by the circle r = 6 cos θ −π
3 ≤ θ ≤ π 3 The length is 2 × 3 × 2π
3 = 4π (2%)
The second part is formed by the cardioid. Using the formula ds = r
r2+ (dr dθ)2dθ
ds =p
(2 + 2 cos θ)2 + (−2 sin θ)2dθ
=p
4 + 8 cos θ + 4(cos θ)2+ (4 sin θ)2dθ
=√
8 + 8 cos θdθ (2%)
The length is
2 Z π3
0
√
8 + 8 cos θdθ = 2 Z π3
0
4 cosθ
2dθ (2%)
= 16 sinθ 2
π 3
0 = 8 So the total length is 8 + 4π