AM1
Mathematical Analysis 1
Oct. 2011 – Feb. 2012
Exercises Lecture 6
Date: November 11
Theorem 6.1 (Mean-value Theorem for Integrals). Let f be a continuous function on [a, b]. Then
there exists x0 ∈ [a, b] such that
Z b
f (x) dx = f (x0 )(b − a)
(6.1)
a
Theorem 6.2 (Weighted Mean-value Theorem for Integrals). Assume f and g are continuous
functions on [a, b]. If g never changes sign in [a, b], then there exists x0 ∈ [a, b] such that
Z b
Z b
f (x)g(x) dx = f (x0 )
g(x) dx
(6.2)
a
a
The proofs of Theorems 6.1 and 6.2 make use of the following theorem, which is an easy consequence
of Bolzano’s Theorem.
Theorem 6.3 (Intermediate-value Theorem for continuous functions). Let f be continuous at each
point of a closed interval [a, b]. Choose two arbitrary points x1 , x2 in [a, b] such that f (x1 ) 6= f (x2 ).
Then for any given number c between f (x1 ) and f (x2 ) there exists a point x0 ∈ [x1 , x2 ] such that
f (x0 ) = c.
Proof of Theorem 6.1. Let m = minx∈[a,b] f (x) and M = maxx∈[a,b] f (x) (f is continuous, thus m
and M are attained by f in certain points c and d in [a, b] — recall the Extreme-value Theorem!).
Then m ≤ f (x) ≤ M
∀ x ∈ [a, b]. Integrating these inequalities and dividing by (b − a) we obtain
Z b
Z b
Z b
1
1
1
m=
m dx ≤
f (x) dx ≤
M dx = M
(6.3)
b−a a
b−a a
b−a a
Rb
1
Let c = b−a
a f (x) dx, by (6.3) and Theorem 6.3 (take the points where m and M are achieved by
f as x1 and x2 ) we have that there exists x0 such that f (x0 ) = c, that is the thesis.
Proof of Theorem 6.2. Since g never changes sign on [a, b] let us assume without loss of generality
that g is non-negative on [a, b]. Then again since f is continuous there exists m = minx∈[a,b] f (x)
and M = maxx∈[a,b] f (x), and they are such that
mg(x) ≤ f (x)g(x) ≤ M g(x)
6-1
∀ x ∈ [a, b]
(6.4)
6-2
Integrating (6.4) we get
Z
b
Z
g(x) dx ≤
m
a
b
Z
f (x)g(x) dx ≤ M
a
b
g(x) dx
(6.5)
a
Rb
Rb
Now, if a g(x) dx = 0 (that is to say g ≡ 0) then (6.5) infers that a f (x)g(x) dx = 0 and
Rb
therefore (6.2) trivially holds for any c ∈ [a, b]. Otherwise, G = a g(x) dx is a positive quantity
and we can write
Z
Z
Z
m b
1 b
M b
m=
g(x) dx ≤
f (x)g(x) dx ≤
g(x) dx = M
(6.6)
G a
G a
G a
Rb
Let c = G1 a f (x)g(x) dx, by (6.6) and Theorem 6.3 (take the points where m and M are achieved
by f as x1 and x2 ) we have that there exists x0 such that f (x0 ) = c, that is the thesis.
Exercise 6.1. Establish the following inequalities
Z 1
1
1
x9
√
√ ≤
dx ≤
10
1+x
10 2
0
(6.7)
Solution:
1
Consider the functions f (x) = √1+x
and g(x) = x9 , they are continuous in the interval [0, 1] and g
is non-negative. Moreover, f is monotone strictly decreasing, so it has its maximum value, M = 1,
in 0 and its minimum, m = √12 , in 1.
Then (6.7) follows by (6.5), since
R1
0
g(x) dx =
1
10 .
Exercise 6.2. Establish the following inequalities
r
Z 1p
2
11
11 4
2
≤
1 − x dx ≤
24
24 3
0
(6.8)
Solution:
1
Consider the functions f (x) = √1−x
and g(x) = 1 − x2 , they are continuous in the interval 0, 21 ,
2
g is non-negative and the product f (x)g(x) equals the integrand in (6.8). Moreover, f q
is monotone
strictly decreasing, so it has its maximum value, M = 1, in 0 and its minimum, m =
Then (6.8) follows by (6.5), since
R1
0
g(x) dx =
1
2
−
11
38
=
4
3,
in 12 .
11
24 .
Exercise 6.3. Use the identity 1 + x6 = (1 + x2 )(1 − x2 + x4 ) to prove that for a > 0 we have
Z a
1
a3 a5
dx
a3 a5
+
+
(6.9)
a
−
≤
≤
a
−
2
1 + a6
3
5
3
5
0 1+x
6-3
Solution:
1
Consider the functions f (x) = √1+x
and g(x) = 1−x2 +x4 , they are continuous in the interval [0, a],
6
g is non-negative and the product f (x)g(x) equals the integrand in (6.9). Moreover, f is monotone
1
strictly decreasing, so it has its maximum value, M = 1, in 0 and its minimum, m = 1+a
6 , in a.
Ra
3
5
Then (6.9) follows by (6.5), since 0 g(x) dx = a − a3 + a5 .
Remark 6.1. To see why g(x) = 1 − x2 + x4 is a non-negative function, take the identity stated
1+x6
above, 1 + x6 = (1 + x2 )(1 − x2 + x4 ): then we have g(x) = 1+x
2 > 0.
Exercise 6.4. One of the following two statements is incorrect. Explain why it is wrong.
1.
R 4π
2.
R 4π
2π
2π
R 3π
sin t
t
R 4π
| sin t|
t
sin t
t
dt > 0 because
sin t
t
dt = 0 because, by Theorem 6.2 there exists c ∈ [2π, 4π] such that
2π
Z
4π
2π
dt >
sin t
1
dt =
t
c
3π
Z
dt
4π
sin t dt =
2π
cos 2π − cos 4π
=0
c
(6.10)
Solution:
Statement no.2 is false since Theorem 6.2 requires that g(x) = sinx x does not change sign in the
interval [2π, 4π] but g(x) > 0 for x ∈ (2π, 3π) and g(x) < 0 for x ∈ (3π, 4π).
Statement no.1 is true, see Figure 6.1 below to get convinced.
Exercise 6.5. Assume f is continuous in [a, b]. If
one c ∈ [a, b].
Rb
a
f (x) = 0, prove that f (c) = 0 for at least
Solution:
Rb
Using Theorem 6.1 we have that there exists c ∈ [a, b] such that (b − a) f (c) = a f (x) dx = 0.
Since b − a > 0 we must conclude that f (c) = 0.
Exercise 6.6. Assume that f is integrable and non-negative on [a, b]. If
f (x) = 0 at each point of continuity of f .
Rb
a
f (x)dx = 0, prove that
Solution:
Let us assume that for a given point of continuity x0 we have that f (x0 ) > 0. f being continuous
at x0 means that limx→x0 f (x) = f (x0 ), that is to say
∀ ε > 0 ∃ δ > 0 : |f (x) − f (x0 )| < ε ∀ 0 < |x − x0 | < δ
6-4
Figure 6.1: Plot of sint t between 2π and 4π. The role of 1t is dumping the oscillation of the sine.
This is why the positive area is bigger (in absolute value) than the negative one.
Then if we take ε = 21 f (x0 ) there must exist a number δ such that f (x) >
[x0 − δ, x0 + δ]. In that case
Z b
Z x0 +δ
f (x) dx ≥
f (x) dx
a
1
2 f (x0 )
for any x ∈
(6.11)
x0 −δ
1
> f (x0 )
2
Z
x0 +δ
dx
(6.12)
x0 −δ
= f (x0 ) δ > 0
(6.13)
which is a contradiction.
Remark 6.2. To obtain the inequality (6.11) it is crucial the hypothesis that f is non-negative!
Rb
Exercise 6.7. Assume f is continuous in [a, b]. Assume also that a f (x)g(x)dx = 0 for every
function g that is continuous in [a, b]. Prove that f (x) = 0 for all x ∈ [a, b].
Solution:
Rb
Take g(x) = f (x). Then we have that a f 2 (x)dx = 0. Next we realize that f 2 is a continuous
function as well as f , therefore from the previous exercise f 2 is identically zero on [a, b], that is to
say f (x) = 0 for all x ∈ [a, b].