Page 34 - MATINF Nr. 1

34 F. St˘anescu

Aplicat , ii

1. Se consider˘a f : [a, b] → R o funct , ie derivabil˘a, cu derivata continu˘a, astfel ˆıncˆat f (a) =
f (b) = 0. Ar˘atat , i c˘a are loc inegalitatea:

1 Z b
0
max f (t) ≤ · |f (x)| dx.
a≤t≤b 2 a

R x 0 b 0 R x 0
R
Solut ,ie. Avem |f (x)| = | a f (t) dt| s , i |f (x)| = x f (t) dt , deci 2 |f (x)| = | a f (t) dt| +

b f (t) dt ≤ R x 0 R b 0 R b |f (x)| dx, (∀) x ∈ [a, b] . Astfel, max f (t) ≤
0
0
R

x a x a
|f (t)| dt + |f (t)| dt =
a≤t≤b
0
1 · R b |f (x)| dx.
2 a
0
2. Dac˘a f : [a, b] → R este o funct , ie derivabil˘a cu f continu˘a pe [a, b], demonstrat , i inegalita-
tea:

1 Z b 1 Z b b − a
0
0 ≤ · |f (x)| dx − · f (x) dx ≤ · max |f (x)| .

b − a a b − a a 3 a≤x≤b

b f (x) dx ≤ R b |f (x)| dx, deci 0 ≤ 1 · R b 1 · R b
R

Solut ,ie. Avem a a b−a a |f (x)| dx − b−a a f (x) dx .
Mai departe, vom calcula:
b x b
Z Z Z
0
0
f (x) dx + (t − a) f (t) dt − (b − t) f (x) dt
a a x
b x b
Z Z Z
= f (x) dx + (x − a) f (x) − f (x) dx + (b − x) f (x) − f (x) dx = (b − a) f (x) ,
a a x
de unde obt , inem c˘a

Z b Z x Z b

f (x) dx + (t − a) f (t) dt −
0 0
|(b − a) f (x)| = (b − t) f (x) dt
a a x

b x b
Z Z Z

f (x) dx + (t − a) |f (t)| dt + (b − t) |f (t)| dt.
0 0
≤
a a x

Rezult˘a

Z b Z b
(b − a) · |f (x)| dx − (b − a) f (x) dx
a a
Z b ÅZ x ã Z b Ç Z b å
0
0
≤ (t − a) |f (t)| dt dx + (b − t) |f (t)| dt dx
a a a x
3
(b − a)
0
≤ · max |f (x)| ,
3 a≤x≤b
ceea ce ˆıncheie demonstrat , ia.
3. Fie f : [a, b] → R derivabil˘a cu derivata continu˘a. S˘a se arate c˘a:
Z a+b
2
2
Z b (b − a)
f (x) dx − f (x) dx ≤ · max |f (x)| .
0

a
a+b 4 x∈[0,1]
2
R b R b 0 R b 0
Solut ,ie. Avem: f (x) dx = (x − a) f (x) dx = (b − a) f (b) − (x − a) f (x) dx,
a a a
R b 0 0
iar din teorema de medie exist˘a c 1 ∈ (a, b) astfel ˆıncˆat a (x − a) f (x) dx = f (c 1 ) ·
R b 0 (b−a) 2 R b 0 (b−a) 2 ˆ
a (x − a) dx = f (c 1 ) 2 , de unde a f (x) dx = (b − a) f (b) − f (c 1 ) 2 . In mod
R b 0 (b−a) 2
analog, exist˘a c 2 ∈ (a, b) astfel ˆıncˆat f (x) dx =(b − a) f (a) + f (c 2 ) .
a 2