Math1024 Answer to Homework 3
Exercise 3.4.2 (2)
R x2
d
log(1 + t2 )dt = 2x log(1 + x4 )
dx 1
Exercise 3.4.2 (6)
d
dx
Z
cot x
d
(1 + t ) dt =
dx
2
tan x
3
2
Z
cot x
2
3
2
Z
(1 + t ) dt −
0
tan x
(1 + t ) dt
2
3
2
0
3
3
1
1
2
2
= − 2 (1 + cot2 x) 2 −
(1
+
tan
x)
cos2 x
sin x
−5
−5
= − sin x − cos x
Exercise 3.4.3 (6)
d
dx
Z
cos x
sin x
Z cos x
Z sin x
d
f (t)dt =
f (t)dt −
f (t)dt
dx 0
0
= − sin xf (cos x) − cos xf (sin x)
Exercise 3.4.3 (7)
R f (x)
d
f (t)dt = f 0 (x)f [f (x)]
dx 0
Exercise 3.4.3 (8)
R f −1 (x)
d
f (t)dt =
dx 0
1
f [f −1 (x)]
f 0 (f −1 (x))
=
x
f 0 (f −1 (x))
Exercise 3.4.4 (3)R
2
x
Let f (x) = √2π 0 e−t dt. Then
2
2
f 0 (x) = √ e−x > 0
π
4x
2
f 00 (x) = − √ e−x
π
f (x) is an increasing function, so there is no local extrema. x = 0 is the inflection point,
when x < 0, f (x) is convex and when x > 0,f (x) is concave.
Exercise 3.4.5 (1)
1
Differentiate both sides, we have f (x) = −f (x), the f (x) ≡ 0.
Exercise 3.4.5 (2)
Differentiate both sides, we have
Z
x
Axf (x) = xf (x) +
f (t)dt.
(∗)
0
If A = 1, then
Z
x
f (t)dt = 0.
0
Differentiate it, then f (x) ≡ 0.
If A 6= 1,
Rx
f (t)dt
,
(A − 1)x
so f (x) is differentiable on (0, ∞). Differentiate (∗),
f (x) =
0
(A − 1)xf 0 (x) = (2 − A)f (x).
If A = 2, then f 0 (x) ≡ 0 ⇒ f (x) ≡ C, here C is a constant. We need to plug the constant
function into the equation of this question, then we have AC = 2C, but A = 2, so f (x) can be
an arbitrary constant function.
,
If A 6= 2, then let k , 2−A
A−1
f 0 (x)
k
= .
f (x)
x
Remember that
f 0 (x)
0
((log f (x)) =
,
f (x)
therefore
(log f (x))0 = k(log x)0
and we get f (x) = Cxk , where C is a constant. Again plug this function into the equation of
the question, then C can be an arbitrary constant.
Exercise 3.4.6 (2)
By l’Hospital rule
1
lim 3
x→0 x
Z
x
sin x2
x→0 3x2
2x cos x2
= lim
x→0
6x
cos x2
= lim
x→0
3
1
=
3
sin t2 dt = lim
0
2
Exercise 3.4.8 (3)
1 0
1
1
1
lim (x sin ) = lim 2x sin − cos
= lim cos ,
x→0
x→0
x→0
x
x
x
x
2
so the limit does not exist. f (x) is not continuous at x = 0. And
x
Z
f (t)dt =
0
x2 sin
0
1
x 6= 0
x
x = 0,
which is differentiable, since
x2 sin x1 − 0
1
= lim x sin = 0
x→0
x→0
x−0
x
lim
Exercise 3.4.9 (1)
Z
√
4
5
4
1 − xdx = − (1 − x) 4 + C
5
Exercise 3.1.11 (1)
The function f (−x) would take opposite values of the independent variable x over [-a,-b],
thus the graph of f (−x) over [-a,-b] would be the same as graph of f (x) over [a,b]. In other
Z −a
Z b
words, the area
f (−x)dx just equals to the area
f (x)dx.
−b
a
Exercise 3.4.10 (4)(6)
You can check it by differentiating the right hand side. Here I show how to go from the left
hand to the right hand side, pretending that we don’t know the shape of the right hand side.
Z √
x:=a sin t
2
Z
a2 − x2 ======= a
cos2 tdt
Z
1 + cos 2t
2
=a
2
t sin 2t
2
=a
+
+C
2
4
t:=arcsin x a2
x 1 √
=======a=
arcsin + x a2 − x2 + C
2
a 2
3
Z
Z d
√
dx
x2 +a:=t−x
√
=========
x2 + a
Z
dt
=
t
= log |t| + C
t2 −a
2t
t2 +a
2t
√
t:=x+ x2 +a
========= log |x +
√
x2 + a| + C
Exercise
3.4.11 (4)
Z
If
f (x)dx = F (x) + C, then F 0 (x) = f (x), so that F (ax + b)0 = F 0 (ax + b) · (ax + b)0 =
af (ax + b). Therefore
Z
1
f (ax + b)dx = F (ax + b) + C.
a
By Example 3.4.8, we have
Z
√
dx
= arcsin x + C.
1 − x2
Then we get
Z
Exercise 3.4.11 (5)
As
Z
dx
p
= arcsin(x − 1) + C.
1 − (x − 1)2
dx
Z
p
=
x(1 − x)
dx
q
1
22
− x−
1 2
2
.
For a > 0, we have
Z
Z
dx
dx
1a
bx + c
1
bx + c
p
q
=
=
arcsin
+ C = arcsin
+ C.
2
2
2
ab
a
b
a
a − (bx + c)
a 1 − bx+c
a
Therefore
Z
Z
x−
dx
dx
p
q
=
=
arcsin
1
2
1
x(1 − x)
2
− x − 12
22
Exercise
Z 3.4.14 (1)
Z
1 3
2
By x dx = x + C and sin xdx = − cos x + C.
3
4
1
2
+ C = arcsin(2x − 1) + C.
Z
Besides the constant C here should be the same, we also require
x > 0 because 31 x3 |x=0 = 0. So we have

Z
 1 x3 ,
ifx ≤ 0
f (x)dx = 3
1 − cos x, if x > 0
f (x)dx = 1 − cos x when
+ C.
Exercise 3.5.1 (3)
When p 6= −1, −2, −3 :
Z
Z
1
2
p
x (ax + b) dx =
((ax + b)2 − 2abx − b2 )(ax + b)p dx
a2
Z
1
b2
2b
= ( 2 (ax + b)p+2 − 2 (ax + b)p − 2 (ax + b − b)(ax + b)p )dx
a
a
a
Z
Z
Z
2b
b2
1
p+2
p+1
= 2 (ax + b) dx − 2 (ax + b) dx + 2 (ax + b)p dx
a
a
a
2b
b2
1
(ax + b)p+3 − 2
(ax + b)p+2 + 2
(ax + b)p+1 + C.
= 2
a (p + 3)
a (p + 2)
a (p + 1)
Then for p = −1, −2, −3:
Z
Z
Z
Z
1
2b
b2
2
−1
x (ax + b) dx = 2 (ax + b)dx − 2 1dx + 2 (ax + b)−1 dx
a
a
a
2
1
2b
b
= 2 (ax + b)2 − 2 x + 3 log |ax + b| + C
2a
a
a
Z
Z
Z
Z
Z
1
2b
b2
−1
x (ax + b) dx = 2 1dx − 2 (ax + b) dx + 2 (ax + b)−2 dx
a
a
a
2
1
2b
b
= 2 x − 3 log |ax + b| − 3 (ax + b)−1 + C.
a
a
a
−2
2
Z
Z
Z
1
2b
b2
−1
−2
x (ax + b) dx = 2 (ax + b) dx − 2 (ax + b) dx + 2 (ax + b)−3 dx
a
a
a
2
1
2b
b
= 3 log |ax + b| + 3 (ax + b)−1 − 3 (ax + b)−2 + C.
a
a
2a
−3
2
Exercise 3.5.1 (7)
Z
x2 − x + 1
dx =
(x + 1)10
(x + 1)2 − 3(x + 1) + 3
dx
(x + 1)10
Z
Z
Z
1
1
1
=
dx − 3
dx + 3
dx
8
9
(x + 1)
(x + 1)
(x + 1)10
1
3
1
=−
+
−
+ C.
7
8
7(x + 1)
8(x + 1)
3(x + 1)9
Z
5
Exercise 3.5.1 (8)
Z
x−1
√ dx =
x
Z
√
1
x − √ dx
x
1
2 3
= x 2 − 2x 2 + C.
3
Exercise
3.5.3 (2) Z
Z
4x
6x
9x
(2x + 3x )2 dx = (4x + 2 · 6x + 9x )dx =
+2
+
+ C.
log 4
log 6 log 9
Exercise 3.5.4 (6)
Z
x2
dx =
x2 + 3x + 2
x2 − 1 + 1
dx
(x + 1)(x + 2)
Z
Z
x−1
1
=
dx +
dx
x+2
(x + 1)(x + 2)
Z
Z
3
1
1
= (1 −
)dx + (
−
)dx
x+2
x+1 x+2
= x − 4 log |x + 2| + log |x + 1| + C.
Z
Exercise 3.5.4 (8)
Z
Z
1
1
1
( 2
− 2
)dx
3 x +1 x +4
1
1
x
= arctanx − arctan + C.
3
6
2
dx
=
2
(x + 1)(x2 + 4)
Exercise 3.5.6 (8)
Z
π
Z
| sin x cos 2x|dx =
0
π
4
3π
4
Z
sin x cos 2xdx −
Z
π
4
0
π
sin x cos 2xdx +
sin x cos 2xdx
3π
4
Z π
Z 3π
Z π
4
4
1
= (
(sin 3x − sin x)dx −
(sin 3x − sin x)dx +
(sin 3x − sin x)dx)
π
3π
2 0
4
4
√
2 1 √
2
= + (4 2 − 2) = 2 2 −
6 2
3
Exercise 3.5.7 (3)
R0
Now the function is f (x) = −x2
et −1
dt.
t
6
According to the 3-th order Taylor expansion of et − 1 at 0 which is P (t) = t + 12 t2 + 16 t3 ,
then for any > 0, there is δ > 0 such that
√
|x| < δ → |t| < δ → |et − 1 − P (t)| ≤ |t|3
Therefore we have
Z
0
0
et − 1 P (t)
−
)dt|
t
t
−x2
Z 0
≤
|t|2 dt = x6
3
−x2
1
1
|f (x) −
(1 + t + t2 )dt| = |
2
6
−x2
Z
(
which means
1
1
f (x) = x2 − x4 + x6 + o(x6 ).
4
18
This just shows that the integration of 5-th order approximation is 6-th order approximation
and the above polynomial is the high order approximation of f (x) at 0.
Exercise 3.5.9
As usual, we start with the partition that evenly divides the interval
a = x0 < x1 = a + h < · · · < xk = a + kh < · · · < xn = b = a + nh.
.
where h = b−a
n
According to the estimation in Example 3.5.8, we use the linear approximation at the middle
point ck = xk +x2 k+1 with the Lagrange form of the remainder to estimate f (x) on [xk , xk+1 ] which
has interval length b−a
.
n
Then we would have the following error bound for f (x) on [xk , xk+1 ] if f 00 (x) is bounded by
K2 :
Z xk+1
b − a K2 b − a 3
f (x)dx − f (ck )(
) ≤
(
)
n
24 n
xk
Thus
Z
Z b
k=n−1
X f (x)dx − Mn = a
≤
xk+1
xk
k=0
k=n−1
X Z xk+1
k=0
xk
K2 b − a 3
(
)
24 n
K2 (b − a)3
=
24 n2
≤n
7
b − a f (x)dx − f (ck )(
) n
b − a )
f (x)dx − f (ck )(
n