微甲08-13班 統一教學期末考解答
1. (10%) Evaluate the integral Z
xαln x dx, α ∈ R.
Sol:
if α 6= −1, Z
xαlnxdx= Z
(xα+1
α+ 1)0lnxdx= (xα+1
α+ 1)lnx − Z
(xα+1
α+ 1)(lnx)0dx
= (xα+1
α+ 1)lnx −
Z xα α+ 1dx
= 1
α+ 1xα+1lnx− 1
(α + 1)2xα+1+ c if α = −1, let t = lnx ⇒ dt
dx = 1 x, Z
x−1lnxdx= Z
tdt= 1
2t2+ c = 1
2(lnx)2+ c
2. (12%) Evaluate the integral Z p1 + √x
x dx. (Hint. Make a substitution to express the inte- grand as a rational function.)
Sol:
u=p 1 +√
x⇒ u2− 1 =√
x⇒ x = (u2− 1)2 ⇒ dx = 2(u2− 1) · 2udu
∴Z p1 + √x x dx=
Z u
(u2− 1)2 · 2(u2− 1) · 2udu =
Z 4u2 (u2− 1)du
=Z 4(u2− 1)
(u2− 1) du+ 4
Z du u2− 1
= Z
4du + 2
Z 1
u− 1 − 1 u+ 1du
= 4u + 2
ln |u − 1| − ln |u + 1|
+ C
= 4 q
1 +√ x+ 2
ln(
q 1 +√
x− 1) − ln(
q 1 +√
x+ 1)
+ C or
= 4 q
1 +√
x+ 2 ln
p1 +√ x− 1 p1 +√x+ 1 + C.
3. (14%)
1
(a) Evaluate Z
sec3θ dθ.
(b) Find the area of the surface obtained by rotating y = sin x, 0 ≤ x ≤ π, about the x-axis.
Sol:
(a)
∵ Z
sec3xdx= Z
sec xd tan x
= sec x tan x − Z
tan xd sec x
= sec x tan x − Z
tan2xsec xdx
= sec x tan x − Z
sec2x− 1
sec xdx
= sec x tan x − Z
sec3xdx+ Z
sec xdx
∴R
sec3xdx= 12 sec x tan x +R
sec xdx
= 12(sec x tan x + ln |sec x + tan x| + C) .
(b) y = sin x, y0 = cos x.
Area= Z
2πyds
= Z π
0
2πy q
1 + (y0)2dx
= Z π
0
2π sin x√
1 + cos2xdx
= −2π Z π
0
√1 + cos2xdcos x
= −2π Z π
0
√1 + cos2xdcos x (Let cos x = tan θ)
= −2π Z −4π
π 4
p1 + tan2θdtan θ
= 2π Z π4
−π 4
sec θd tan θ
= 2π Z π4
−π 4
sec3θdθ
= 2π · 1
2[sec θ tan θ + ln |sec θ + tan θ|]
π 4
−π 4
= π ·
secπ
4tanπ 4 + ln
sec π
4 + tanπ 4
− sec−π
4 tan −π 4 − ln
sec −π
4 + tan−π 4
π4
−π 4
= π ·h√
2 · 1 + ln
√2 + 1 −
√2 · (−1) − ln
√2 − 1 i
= π ·
"
2√ 2 + ln
√2 + 1
√2 − 1
#
= π ·
2√
2 + ln√
2 + 12
= 2π ·h√
2 + ln√
2 + 1i
4. (14%)
(a) Find the values of a for which the improper integral Z ∞
1
dx xa(1 +√
x) converges.
(b) Evaluate the integral Z π
0
sec2x dx.
Sol:
(a) 1
xa(1 +√
x) = 1
xa+ xa+12
3
1
2xa+12 < 1 xa(1 +√
x) < 1
xa+12 when 1 ≤ x hence,
Z ∞ 1
1 2xa+12 <
Z ∞ 1
dx
xa(1 +√x) <
Z ∞ 1
1 xa+12 Z ∞
1
1
xa+12 converges if and only if a > 1 2 hence,
Z ∞ 1
dx xa(1 +√
x) converges if and only if a > 1 2 (b) lim
x→π2 sec2x= ∞ This is an improper integral.
Z π 0
sec2xdx= Z π2
0
sec2xdx+ Z π
π 2
sec2xdx
= lim
t→π2−
Z t 0
sec2xdx+ lim
t→π2+
Z π t
sec2xdx
= lim
t→π2−
tan t + lim
t→π2+
tan t
= ∞ Z π
0
sec2xdx diverges.
5. (12%) Find the volume of the solid obtained by rotating about the y-axis the region between x-axis and y = cos x, 0 ≤ x ≤ π
2.
Sol: Sol.(I) Cutting into disks:
Volume = Z 1
0
π(cos−1y)2dy let x = cos−1y
= Z 0
π/2
πx2(− sin x)dx
= Z π/2
0
πx2sin xdx integration by parts
= π(−x2cos x)
π/2
0
| {z }
=0
+ Z π/2
0
π· 2x cos xdx integration by parts
= 2π sin x
π/2
0
| {z }
=π2
− Z π/2
0
2π sin xdx
= π2+ 2π cos x
π/2
0
= π2− 2π
Sol.(II) Cutting into cylindrical shells:
Volume = Z π/2
0
2πx cos xdx integration by parts
= 2πx sin x
π/2
0
| {z }
=π2
− Z π/2
0
2π sin xdx
= π2+ 2π cos x
π/2
0
= π2− 2π
6. (12%) Solve the initial-value problem y0 = (y − 4)(2x + 1)
(x2+ 1) , y(0) = −1.
Sol:
dy
dx = (y − 4)2x + 1
x2+ 1 is seprable dy
y− 4 = 2x + 1 x2+ 1 Z dy
y− 4 =Z 2x + 1 x2+ 1 5
ln|y − 4| = ln(x2+ 1) + tan−1x+ c, c ∈ R y= 4 + A(x2+ 1)etan−1x, A∈ R y(0) = −1 = 4 + A ⇒ A = −5
y = 4 − 5(x2 + 1)etan−1x
7. (12%) Solve the differential equation x(x+1)y0+y +(x+1)2cos x = 0, x > 0, with y(π) = π +1.
Sol:
y0+ y
x(x + 1) = −x+ 1 x cos x I(x) = exp
R 1
x(x+1)dx
= expln|x+1x |= x x+ 1
⇒ I(x)y0 + 1
(x + 1)2y= − cos x
⇒ (I(x)y)0 = − cos x
⇒ xy
x+ 1 = − sin x + c ⇒ y = −x+ 1
x sin x + cx+ 1 x Since y(π) = π + 1 ⇒ c = π
⇒ y = x+ 1
x (π − sin x)
8. (14%) Consider the cardioid given by r = 1 − cos θ, 0 ≤ θ ≤ 2π.
(a) Find the area enclosed by this curve.
(b) Find the length of this curve.
Sol:
(a)
Area= 1 2
Z 2π 0
r2dθ
= 1 2
Z 2π 0
(1 − cos θ)2dθ
= 1 2
Z 2π
0 1 − 2 cos θ + cos2θdθ
= 1 2
Z 2π 0
3
2 − 2 cos θ + 1
2cos 2θdθ
= 1 2
h 3
2θ− 2 sin θ + 1
4sin 2θi
2π 0
= 3 2π
7
(b)
Length= Z 2π
0
r
r2+ (dr dθ)2dθ
= Z 2π
0
p1 − 2 cos θ + cos2θ+ sin2θdθ
= Z 2π
0
√2 − 2 cos θdθ
= Z 2π
0
p2(1 − cos θ)dθ
= Z 2π
0
r
2(2 sin2(θ 2))dθ
= Z 2π
0
2 sin(θ 2)dθ
=h
− 4 cos(θ 2)i
2π 0 = 8
