11.
Z
1
0
dx
=2
√
x(1 − x)
Z
Z
1/2
s
0
1/2
dx
1
1 2
− x−
4
2
dx
c→0+ c
1 2
1
− x−
4
2
1/2
= 2 lim sin−1 (2x − 1) = π.
= 2 lim
s
c→0+
The integral converges.
c
14.
Z
π/2
0
=
C
sec x d x =
lim ln | sec x + tan x|
C→(π/2)−
0
lim
C→(π/2)−
ln | sec C + tan C| = ∞.
This integral diverges to infinity.
√ √ 2
x
x
x
38. Since 0 ≤ 1 − cos x = 2 sin2
= , for x ≥ 0, therefore
≤ 2
2
2
2
Z π2
Z π2
dx
dx
, which diverges to infinity.
√ ≥2
x
1 − cos x
0
0
√
37. Since sin x ≥
2x
on [0, π/2], we have
π
Z
∞
0
Z
π/2
sin x
dx
x2
0
Z
2 π/2 d x
≥
= ∞.
π 0
x
| sin x|
dx ≥
x2
The given integral diverges to infinity.
y
y=sin x
2x
y= π
π
2
Fig. 5-37
x
40. Since ln x grows more slowly than any positive power of x, therefore we haveZln x ≤ kx 1/4
∞
1
dx
1
≥ 3/4 for x ≥ 2 and
for some constant k and every x ≥ 2. Thus, √
√
kx
x ln x
2
Z x ln x
1 ∞ dx
.
diverges to infinity by comparison with
k 2 x 3/4
4.
Z
∞
dx
√
+ x +1
1
t2
1
2 dt
dx = − 3
t Z 0
1
2 dt
=
− 3
2 r
t
1
1
1
+
1
+
t2
t2
Z 1
t dt
.
=2
4
3
0 t +t +1
x2
Let x =
5.
Z
π/2
dx
√
sin x
0
=2
Z
1
0
1
=2
Z
1
=4
Z
=4
Z
0
0
0
1
Let sin x = u 2
√
2u du = cos x d x = 1 − u 4 d x
u du
√
u 1 − u4
du
p
(1 − u)(1 + u)(1 + y 2
Let 1 − u = v 2
−du = 2v dv
v dv
p
v (1 + 1 − v 2 )(1 + (1 − v 2 )2 )
dv
p
.
(2 − v 2 )(2 − 2v 2 + v 4 )
3. One possibility: let x = sin θ and get
I =
Z
1
−1
ex d x
=
√
1 − x2
Z
π/2
esin θ dθ.
−π/2
Another possibility:
I =
Z
0
−1
ex d x
+
√
1 − x2
Z
1
0
ex d x
= I1 + I2 .
√
1 − x2
In I1 put 1 + x = u 2 ; in I2 put 1 − x = u 2 :
so I = 2
Z
0
1
eu
2 −1
I1 =
Z
I2 =
Z
+ e1−u
√
2 − u2
2
du.
1
0
1
0
Z 1
2
2eu −1 u du
=2
√
u 2 − u2
0
Z 1
2
1−u
2e
u du
=2
√
u 2 − u2
0
eu
√
2 −1
du
2 − u2
2
e1−u du
√
2 − u2
8. Let
∞
1
t
1
dt
dx = − 2
t
Z 0
Z 1 −1/t 2
1
e
−(1/t)2
e
− 2 dt =
=
dt.
t
t2
1
0
I =
Observe that
Z
2
e−x d x
Let x =
2
e−1/t
t −2 h ∞ i
lim
=
lim
2
t→0+
t→0+ e1/t
t2
∞
−3
−2t
= lim
t→0+ e1/t 2 (−2t −3 )
1
= lim
= 0.
2
t→0+ e1/t
Hence,
i
1 1 h
−4
−1
0 + 4(4e ) + e
S2 =
3 2
≈ 0.1101549
1 1
0 + 4(16e−16 ) + 2(4e−4 )
S4 =
3 4
16
−16/9
−1
e
+e
+4
9
≈ 0.1430237
64
64
1 1
0 + 4 64e−64 + e−64/9 + e−64/25 +
S8 =
3 8
9
25
64 −64/49
16 −16/9 −16
−4
−1
e
+ 2 16e
+ 4e + e
+e
49
9
≈ 0.1393877.
Hence, I ≈ 0.14, accurate to 2 decimal places. These approximations do not converge very
2
quickly, because the fourth derivative of e−1/t has very large values for some values of t
near 0. In fact, higher and higher derivatives behave more and more badly near 0, so higher
order methods cannot be expected to work well either.
14. Let y = f (x). We are given that m 1 is the midpoint of [x0 , x1 ] where x1 − x0 = h. By
tangent line approximate in the subinterval [x0 , x1 ],
f (x) ≈ f (m 1 ) + f ′ (m 1 )(x − m 1 ).
The error in this approximation is
E(x) = f (x) − f (m 1 ) − f ′ (m 1 )(x − m 1 ).
If f ′′ (t) exists for all t in [x0 , x1 ] and | f ′′ (t)| ≤ K for some constant K , then by Theorem
11 of Section 4.9,
K
|E(x)| ≤ (x − m 1 )2 .
2
Hence,
K
| f (x) − f (m 1 ) − f ′ (m 1 )(x − m 1 )| ≤ (x − m 1 )2 .
2
We integrate both sides of this inequlity. Noting that
x1 − m 1 = m 1 − x0 = 21 h, we obtain for the left side
Z
x1
Z
x1
f (x) d x −
f (m 1 ) d x
x0
Z x1
′
−
f (m 1 )(x − m 1 ) d x x0
x0
Z
x1
2 x1 (x
−
m
)
1 =
f (x) d x − f (m 1 )h − f ′ (m 1 )
x0
2
x0
Z x
1
= f (x) d x − f (m 1 )h .
x0
Integrating the right-hand side, we get
K (x − m 1 )3 x1
K
2
(x − m 1 ) d x =
2
3
x0 2
x0
3
K h
h3
K 3
=
+
=
h .
6
8
8
24
Z
Hence,
Z
x1
x0
Z
=
x1
f (x) d x − f (m 1 )h x1
x0
[ f (x) − f (m 1 ) − f ′ (m 1 )(x − m 1 )] d x K 3
h .
≤
24
2
1
1
1 1
1
1
x d x = . M1 =
13. I =
.
(1) = . The actual error is I − M1 = − =
3
2
4
3 4
12
0
Since the second derivative of x 2 is 2, the error estimate is
Z
1
2
|I − M1 | ≤
2
1
(1 − 0)2 (12 ) =
.
24
12
Thus the constant in the error estimate for the Midpoint Rule cannot be improved; no
smaller constant will work for f (x) = x 2 .
46. Ŵ(x) =
Z
∞
t x−1 e−t dt.
0
a) Since limt→∞ t x−1 e−t/2 = 0, there exists T > 0 such that t x−1 e−t/2 ≤ 1 if t ≥ T .
Thus
Z ∞
Z ∞
x−1 −t
e−t dt = 2e−T /2
t
e dt ≤
0≤
T
T
and
Z
∞
t x−1 e−t dt converges by the comparison theorem.
T
If x > 0, then
0≤
Z
0
T
t
x−1 −t
e
dt <
Z
T
t x−1 dt
0
converges by Theorem 2(b). Thus the integral defining Ŵ(x) converges.
Z ∞
t x e−t dt
b) Ŵ(x + 1) =
0
Z R
t x e−t dt
= lim
c→0+
R→∞
c
d V = e−t dt
U = tx
dU = xt x−1 d x
V = −e−t
!
R
Z R
t x−1 e−t dt
= lim −t x e−t + x
c→0+
c
c
R→∞
Z ∞
t x−1 e−t dt = xŴ(x).
=0+x
0
c) Ŵ(1) =
Z
∞
0
e−t dt = 1 = 0!.
By (b), Ŵ(2) = 1Ŵ(1) = 1 × 1 = 1 = 1!.
In general, if Ŵ(k + 1) = k! for some positive integer k, then
Ŵ(k + 2) = (k + 1)Ŵ(k + 1) = (k + 1)k! = (k + 1)!.
Hence Ŵ(n + 1) = n! for all integers n ≥ 0, by induction.
Z ∞
1
d) Ŵ
t −1/2 e−t dt Let t = x 2
=
2
0
dt =Z2x d x
Z ∞
∞
√
1 −x 2
2
e−x d x = π
e
2x d x = 2
=
0
0 x 3
1
1
1√
Ŵ
= Ŵ
=
π.
2
2
2
2