Page 33 - MATINF Nr. 3

Inegalit˘at , i integrale 33

Solut ,ie. Din Teorema de medie, exist˘a c ∈ (a, b) astfel ˆıncˆat f (c) = 0. Putem scrie:

x x
Z Z
2
2
0
0
f (x) = f (t) d t ⇒ f (x) ≤ |x − c| (f (t)) d t
C−B−S
c c
b b  x  b b
Z Z Z Z Z
2
2
0
0
2
⇒ f (x) d x ≤  |x − c| (f (t)) d t d x ≤ (f (x)) d x · |x − c| d x

a a c a a
b b
2
Z Å a + b 2 ã (b − a) 2 Z
2
2
0
0
2
= (f (x)) d x · c − c (a + b) + ≤ (f (x)) d x.
2 2
a a
1
Aplicat , ia 10. Fie f : [a, b] → R de clas˘a C astfel ˆıncˆat f (a) = f (b) = 0. Ar˘atat , i c˘a:
b b
(b − a)
Z 2 Z
2
0
2
f (x) d x ≤ (f (x)) d x.
8
a a
x x
2
0
0
Solut ,ie. Pentru x ≤ b+a , avem f (x) = R f (t) d t ⇒ f (x) ≤ (x − a) R (f (t)) d t ⇒
2
2
a a
b+a b+a b+a b+a
2 2 Z
2 Z
2 Z
(b − a) 2
2 Z
0
2
0
2
f (x) d x ≤ (f (t)) d t · (x − a) d x = (f (t)) d t.
8
a a a a
b b b b
2
2
R (b−a) 2 R 2 R (b−a) 2 R 2
0
0
Analog, f (x) d x ≤ (f (t)) d t, de unde f (x) d x ≤ · (f (t)) d t.
8 8
b+a b+a a a
2 2
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