Page 29 - MATINF Nr. 3

Inegalit˘at , i integrale 29

0
Aplicat , ia 2. Fie f : [0, 1] → R o funct , ie derivabil˘a cu f (1) = 0 s , i f continu˘a. Ar˘atat , i c˘a:

1 Ñ 1 é 2
Z Z
2
0
(f (x)) d x ≥ 3 · f (x) d x .
0 0
Solut ,ie. Utilizˆand formula de integrare prin p˘art , i, avem:
1 1 1 1
R R 0 R 0 R 0
f (x) d x = x ·f (x) d x = f (1)− xf (x) d x = − xf (x) d x, de unde, prin intermediul
0 0 0 0
inegalit˘at , ii C-B-S, avem:

Ñ é 2 Ñ é 2
1 1 1 1 1
Z Z Z Z 1 Z
2
2
0
2
0
0
f (x) d x = xf (x) d x ≤ x d x · (f (x)) d x = · (f (x)) d x.
3
0 0 0 0 0
Aplicat , ia 3. Fie f : [0, 1] → R o funct , ie continu˘a. Ar˘atat , i c˘a :
Ñ é 2
1 1
Z 7 Z
f 2 x 2 d x ≥ xf (x) d x .
4
0 0
1 1
R
2
3
2
·
Solut ,ie. Cu schimbarea de varibil˘a x = t, obt , inem x f (x ) d x = 1 R xf (x) d x, de unde
0 2 0
Ñ é 2
1 1 1 1
1 Z Z Z 1 Z
6
· xf (x) d x ≤ x d x · f 2 x 2 d x = · f 2 x 2 d x ⇒
2 7
0 0 0 0
Ñ é 2
1 1
Z 7 Z
f 2 x 2 d x ≥ xf (x) d x .
4
0 0
Aplicat , ia 4. Dac˘a f : [a, b] → R este o funct , ie derivabil˘a, cu derivata continu˘a, astfel ˆıncˆat
b
R a+b
f (x) d x = f = 0, demonstrat , i inegalitatea:
2
a
b b
Z 1 8 Z
2
2
0
2
(f (x)) d x − (f (a) + f (b)) ≥ (f (x)) d x.
b − a (b − a) 2
a a
Solut ,ie. Folosind f a+b = 0, inegalitatea C-B-S s , i formula integr˘arii prin p˘art , i, putem scrie:
2
!2 Ñ é 2
a+b a+b
b b x
R 2 2 R 2 R 0 R R 0
(f (x)) d x = f (t) d t d x + f (t) d t d x
a a x a+b a+b
2 2
a+b  a+b   
2 Z b Å ã Z x
 a + b 2 a + b 2
2 Z
ã Z
Å
0
0
≤ − x (f (t)) d t d x +  x − (f (t)) d t d x



2 2

a x a+b a+b
2 2