Here are the solutions to some of the problems in HW1 and HW2 from Spivak’s chapters
13 and 14:
Chapter 13.
31. Let f be integrable on [ a, b]. Recall that
Rb
a
f ≥ 0 if f ( x ) ≥ 0 for all x in [ a, b].
(a) Give an example where f ( x ) ≥ 0 for all x, and f ( x ) > 0 for some x in [ a, b], and
Rb
a f = 0.
Consider for example the function
(
0 x 6= 0
.
1 x=0
R1
Then clearly f ≥ 0 and f (0) > 0, yet −1 f = 0.
f (x) =
(b) Suppose f ( x ) ≥ 0 for all x in [ a, b] and f is continuous at x0 in [ a, b] and f ( x0 ) > 0.
Rb
Prove that a f > 0.
Since f is continuous at x0 , for every given ε > 0 we can find a δ > 0 such that
| f ( x ) − f ( x0 )| < ε whenever | x − x0 | < δ. Taking ε = f ( x0 )/2 > 0, say, it follows
that there is a δ > 0 such that if | x − x0 | < δ then
f ( x0 )
f ( x0 )
f ( x0 )
=⇒ −
< f ( x ) − f ( x0 ) <
2
2
2
3 f ( x0 )
f ( x0 )
< f (x) <
=⇒
2
2
Note that if x0 happens to be the endpoint a, the interval ( x0 − δ, x0 + δ) above
must be replaced by ( a, a + δ). Similarly, if x0 = b, replace ( x0 − δ, x0 + δ) by
(b − δ, b). Whatever the case, it follows that there is an interval I containing x0
such that f ( x ) > f ( x0 )/2 for all x in I.
| f ( x ) − f ( x0 )| <
3 f(x0 )/2
Now consider the function
(
f ( x0 )/2 if x is in I
g( x ) =
0
otherwise
f
f(x0 )
(compare the figure). Then g is
integrable and f ≥ g everywhere on
[ a, b]. Hence
Z b
a
f ≥
Z b
a
g=
f(x0 )/2
g
1
f ( x0 ) · length( I ) > 0.
2
a
x0
I
b
Remark. Here is another proof based on FTC (note however that since
R x the problem is in
chapter 13, we are not supposed to use FTC for it!): Set F ( x ) = a f . The assumption
f ≥ 0 implies F is non-decreasing, for if a ≤ x1 < x2 ≤ b, then
F ( x2 ) − F ( x1 ) =
Z x2
a
f−
Z x1
a
f =
Z x2
x1
f ≥ 0,
Rb
or F ( x1 ) ≤ F ( x2 ). We know that F ( a) = 0. Suppose F (b) = a f were zero also. Since F is
non-decreasing, it would then follow that F ( x ) = 0 for all x in [ a, b]. Hence F 0 ( x ) = 0 for
all x in [ a, b]. But by FTC1, F 0 ( x0 ) = f ( x0 ) > 0, which is a contradiction. Thus, we must
Rb
have F (b) = a f > 0.
37. Prove that if f is integrable on [ a, b], then
Z b
Z b
≤
f
(
t
)
dt
| f (t)| dt.
a
a
Integrability of | f | follows from the previous problem. The trivial inequalities
−| f (t)| ≤ f (t) ≤ | f (t)|
together with the property that integration respects the order imply
−
Z b
| f (t)| dt ≤
a
Z b
a
f (t) dt ≤
Z b
a
| f (t)| dt.
(∗)
Recall that a double inequality −r ≤ A ≤ r is equivalent to the single inequality | A| ≤ r.
Thus the double inequality (*) can be written as
Z b
Z b
f (t) dt ≤
| f (t)| dt
a
a
which is what we wanted to show.
Chapter 14.
In the following problems, we will use FTC1 according to which
Z x
d
f (t) dt = f ( x )
dx
a
provided that f is continuous. Combined with the chain rule, this implies the more
general formula
Z u( x )
d
f (t) dt = f (u( x )) · u0 ( x ) − f (v( x )) · v0 ( x )
dx
v( x )
whenever f is continuous and u, v are differentiable.
1. Find the derivatives of each of the following functions.
(i)
F(x) =
Z x3
a
sin3 t dt =⇒ F 0 ( x ) = sin3 ( x3 ) · 3x2
(iii)
F(x) =
Z x Z y
15
8
Z x
1
1
0
dt dy =⇒ F ( x ) =
dt.
2
2
2
2
8 1 + t + sin t
1 + t + sin t
(iv)
F(x) =
Z b
x 1 + t2
1
1
dt =⇒ F 0 ( x ) = −
2
2
+ sin t
1 + x + sin2 x
(v)
F(x) =
Z b
a
x
dt
1 + t2 + sin2 t
Pulling x out of the integral, it follows that F is the linear function
F(x) = x
Hence,
0
F (x) =
Note that F 0 ( x ) is a constant.
Z b
a
1
dt.
1 + t2 + sin2 t
Z b
a 1 + t2
1
dt.
+ sin2 t
3. Show that the values of following expressions do not depend on x:
(ii)
Z sin x
√
1
dt, x in (0, π/2).
1 − t2
Let us call this expression F ( x ). Then,
1
1
· (sin x )0 − √
· (− cos x )0
F0 (x) = p
2x
2
1
−
cos
1 − sin x
1
1
=√
· (cos x ) − √
· (sin x )
cos2 x
sin2 x
cos x
sin x
=
−
.
| cos x | | sin x |
Since 0 < x < π/2 by the assumption, sin x and cos x are both positive, so
| sin x | = sin x and | cos x | = cos x. It follows that F 0 ( x ) = 0, which means F is
a constant independent of x.
− cos x
Task: Show that this constant is π/2.
4. Find ( f −1 )0 (0) if
(i) f ( x ) =
Z x
0
[1 + sin(sin t)] dt.
The formula for the derivative of the inverse function tells us that
1
( f −1 ) 0 ( x ) = 0 −1
f ( f ( x ))
The definition of f shows f (0) = 0, so f −1 (0) = 0. Hence
1
( f −1 ) 0 (0 ) = 0 .
f (0)
By FTC1,
f 0 ( x ) = 1 + sin(sin x ) =⇒ f 0 (0) = 1 + sin(sin 0) = 1.
Hence,
( f −1 ) 0 (0 ) =
1
= 1.
1