70. Use the partial fraction decomposition
A
Bx + C
1
= + 2
2
+x +x
x
x +x +1
2
A(x + x + 1) + B x 2 + C x
=
x3 + x2 + x
(
A+B =0
⇒ A + C = 0 ⇒ A = 1, B = −1, C = −1.
A=1
x3
Therefore,
Z
dx
2
Z +x +
Zx
dx
x +1
=
−
dx
x
x2 + x + 1
Z
u + 21
= ln |x| −
du
u 2 + 34
1
= ln |x| − ln x 2 + x + 1 −
2
x3
Let u = x +
du = d x
1
√ tan−1
3
1
2
2x + 1
√
3
+ C.
8. Let I =
Z
b
f (x) d x, and the interval [a, b] be subdivided into 2n subintervals of equal
a
length h = (b − a)/2n. Let y j = f (x j ) and x j = a + j h for 0 ≤ j ≤ 2n, then
S2n
1
=
3
b−a
2n
h
y0 + 4y1 + 2y2 + · · ·
i
+ 2y2n−2 + 4y2n−1 + y2n
2n−1
n−1
i
X
X
1 b−a h
=
y0 + 4
yj − 2
y2 j + y2n
3
2n
j =1
and
T2n
1
=
2
1
2
Tn =
b−a
2n
y0 + 2
b−a
n
y0 + 2
j =1
2n−1
X
j =1
n−1
X
j =1
y j + y2n
y2 j + y2n .
Since T2n = 21 (Tn + Mn ) ⇒ Mn = 2T2n − Tn , then
Tn + 2(2T2n − Tn )
4T2n − Tn
Tn + 2Mn
=
=
3
3
3
2T2n + 2T2n − Tn
4T2n − Tn
2T2n + Mn
=
=
.
3
3
3
Hence,
2T2n + Mn
4T2n − Tn
Tn + 2Mn
=
=
.
3
3
3
Using the formulas of T2n and Tn obtained above,
4T2n − Tn
3
2n−1
X
1 4 b−a y0 + 2
y j + y2n
=
3 2
2n
j =1
1
−
2
1
=
3
= S2n .
b−a
n
b−a
2n
h
y0 + 2
y0 + 4
2n−1
X
j =1
n−1
X
j =1
y2 j + y2n
yj − 2
n−1
X
j =1
y2 j + y2n
Hence,
S2n =
Tn + 2Mn
2T2n + Mn
4T2n − Tn
=
=
.
3
3
3
i
44.
∞
X
(−1)n−1 x 2n−1
n=1
∞
X
(2n − 1)!
= sin x
(−1)n π 2n−1
= − sin π = 0
(2n − 1)!
n=1
∞
X
1
1
(−1)π
(−1)n π 2n−4
= 3 0−
= 2.
(2n − 1)!
π
1!
π
n=2
46.
(x − tan−1 x)(e2x − 1)
x→0 2x 2 − 1 + cos(2x)
lim
= lim
!
x3
x5
4x 2
x−x+
−
+···
2x +
+···
3
5
2!
4x 2
16x 4
2x 2 − 1 + 1 −
+
−···
2!
4!
2
4
x
+···
3
= 1.
= lim
2
x→0
4
+···
x
3
x→0
7. Let f (x) =
a) Since
∞
X
ak x
2k+1
k=0
22k k!
, where ak =
.
(2k + 1)!
ak+1 x 2k+3 lim
k→∞ ak x 2k+1 22k+2 (k + 1)! (2k + 1)!
·
·
= |x| lim
k→∞ 22k
k!
(2k + 3)!
4k
+
4
=0
= |x|2 lim
k→∞ (2k + 3)(2k + 2)
2
for all x, the series for f (x) converges for all x. Its radius of convergence is infinite.
∞ 2k
∞
X
X
2 k! 2k
22k k!
2k
(2k + 1)x = 1 +
x
b) f (x) =
(2k + 1)!
(2k)!
′
k=1
k=0
∞
X
22k+1 k! 2k+2
1 + 2x f (x) = 1 +
x
(2k + 1)!
k=0
(replace k with k − 1)
∞ 2k−1
X
2
(k − 1)! 2k
x
=1+
(2k − 1)!
=1+
k=1
∞
X
k=1
22k k! 2k
x = f ′ (x).
(2k)!
d −x 2
2
2
e
f (x) = e−x f ′ (x) − 2x f (x) = e−x .
c)
dx
d) Since f (0) = 0, we have
e
−x 2
Z
x
Z x
d −t 2
2
e−t dt
e
f (t) dt =
dt
0
f (x) − f (0) =
Z x 0
2
2
e−t dt.
f (x) = e x
0
5.
Z
6
Z
6
(16 + 6z − z 2 ) dz
(2 + z)(8 − z) dz =
0
0
6
3
z = 16z + 3z 2 −
= 132 ft3
3 0
V =
12.
s=
Z
π/4 p
π/6
π/4
1 + tan2 x d x
π/4
sec x d x = ln | sec x + tan x|
=
π/6
π/6
√
1
2
= ln( 2 + 1) − ln √ + √
3
3
√
2+1
units.
= ln √
3
Z
√
√
4. The height of each triangular face is 2 3 m and the height of the pyramid
is
2
2 m. Let
r
2
1
and cos θ = √ .
the angle between the triangular face and the base be θ , then sin θ =
3
3
√
2 2
√
2 3
2 θ
4
4
Fig. 6-4
y
front view of
dy
√
10−2 2
one face
√
dy sec θ = 3dy
10
θ
x
√
√
x= 2y+10−2 2
60◦
2
side view of one face
4
Fig. 6-4
A vertical slice of water with thickness dy at a distance y from the vertex√
of the pyramid
2
exerts a force on the
√ shaded strip√shown in the front view, which has area 2 3y dy m and
which is at depth 2y + 10 − 2 2 m. Hence, the force exerted on the triangular face is
2
√ √
√
( 2y + 10 − 2 2)2 3y dy
0
h √2
√ 2 i2
√
3
= 2 3(9800)
y + (5 − 2)y 3
F = ρg
Z
0
5
≈ 6.1495 × 10 N.
4.
Z
Z
1 π/3 2
1 π/3
Area =
sin 3θ dθ =
(1 − cos 6θ ) dθ
2 0
4 0
π/3
1
π
1
=
=
sq. units.
θ − sin 6θ 4
6
12
0
y
π
θ= 3
A
x
r=sin 3θ
Fig. 6-4
27.
(n!)2
(1 · 2 · 3 · · · n)(1 · 2 · 3 · · · n)
=
(2n)!
1 · 2 · 3 · · · n · (n + 1) · (n + 2) · · · 2n
n
1
2
3
n
1
=
.
·
·
···
≤
n+1 n+2 n+3
n+n
2
Thus lim an = 0.
an =
36.
a) “If lim an = ∞ and lim bn = L > 0, then lim an bn = ∞” is TRUE. Let R be an
arbitrary, large positive number. Since lim an = ∞, and L > 0, it must be true that
L
2R
for n sufficiently large. Since lim bn = L, it must also be that bn ≥
for
an ≥
L
2
2R L
= R for n sufficiently large. Since R is
n sufficiently large. Therefore an bn ≥
L 2
arbitrary, lim an bn = ∞.
b) “If lim an = ∞ and lim bn = −∞, then lim(an + bn ) = 0” is FALSE. Let an = 1 + n
and bn = −n; then lim an = ∞ and lim bn = −∞ but lim(an + bn ) = 1.
c) “If lim an = ∞ and lim bn = −∞, then lim an bn = −∞” is TRUE. Let R be an
arbitrary, large
Since lim an = ∞ and lim bn = −∞, we must
√ positive number.
√
have an ≥ R and bn ≤ − R, for all sufficiently large n. Thus an bn ≤ −R, and
lim an bn = −∞.
d) “If neither {an } nor {bn } converges, then {an bn } does not converge” is FALSE. Let
an = bn = (−1)n ; then lim an and lim bn both diverge. But an bn = (−1)2n = 1 and
{an bn } does converge (to 1).
e) “If {|an |} converges, then {an } converges” is FALSE. Let an = (−1)n . Then limn→∞ |an | = limn→∞ 1 = 1,
but limn→∞ an does not exist.
dy
2y
1
12. We have
+
= 2 . Let µ =
dx
x
x
Z
2
d x = 2 ln x = ln x 2 , then eµ = x 2 , and
x
d 2
dy
(x y) = x 2
+ 2x y
dx d x
1
2y
2
2 dy
+
=x
=1
=x
dx
x
x2
Z
⇒ x2y =
dx = x + C
⇒
y=
1
C
+ 2.
x
x
1. The expected winnings on a toss of the coin are
$1 × 0.49 + $2 × 0.49 + $50 × 0.02 = $2.47.
If you pay this much to play one game, in the long term you can expect to break even.
41.
Z
∞
0
dx
=
xe x
Z
0
1
+
Z
1
∞
dx
. But
xe x
Z
0
1
dx
1
≥
x
xe
e
Z
1
0
Thus the given integral must diverge to infinity.
dx
= ∞.
x