Page 32 - RMGO 5

32 Rezolvarea problemelor din num˘arul anterior

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MGO 158. Fie n ∈ N , a ∈ R s , i b > 0. Calculat ,i integrala
Z 2n−1 2n−1
x (x + a)
dx, x ∈ (0, ∞).
(x + a) 4n + bx 4n
Daniel Jinga, Pites , ti

Solut ,ie. Notˆand cu I integrala dat˘a, avem
a 1 a
2n−1 2n−1
Z x 4n−2 1 + Z 2 · 1 +
I = x dx = x x dx
a 4n a 4n
x 4n 1 + + b 1 + + b
x x
0
0
a a
2n−1 a 2n
Z 1 + 1 + Z 1 +
1 x x 1 x
= − · dx = − · dx
a a 4n 2an a 2n 2
1 + + b 1 + + b
x x
a √
2n
1 + 2n
1 1 x b (x + a)
= − · √ · arctg √ + C = − · arctg √ + C.
2an b b 2abn x 2n · b
MGO 159. Fie f : [−1, 1] → R o funct ,ie de dou˘a ori derivabil˘a, cu derivata a
a
doua continu˘a, astfel ˆıncˆat f(−1) = f(1) = 0. Demonstrat ,i c˘ are loc inegalitatea
1 Z 1 2
00
2
· f (x) dx ≥ max f (x).
6 −1 x∈[−1,1]
Florin St˘anescu, G˘aes , ti

a
a
Solut ,ie. Concluzia fiind evident˘ dac˘ funct , ia f este identic nul˘a, presupunem ˆın
2
continuare c˘a f nu este identic nul˘a. Atunci exist˘a c ∈ (0, 1) a.ˆı. max f (x) =
x∈[−1,1]
0
0
2
f (c) > 0. Astfel 0 = f 2 0 (c) = 2f (c)f(c), deci f (c) = 0.
c c
Z Z
0
0
00
Integrˆand prin p˘art , i obt , inem (1 + x) f (x)dx = (1+c)f (c)− f (x)dx =
−1 −1
−f(c), deci aplicˆand Inegalitatea Cauchy-Buniakowski-Schwarz avem
c c c 2
Z 2 Z Z
00
00
2
2
f (c) = (1 + x) f (x) dx ≤ (1 + x) dx · f (x) dx,
−1 −1 −1
2
Z c 3f (c)
00
de unde rezult˘ c˘ f (x) 2 dx ≥ 3 .
a
a
−1 (1 + c)
1 1 3f (c)
2
Z Z
00
00
a
Analog obt , inem c˘ (1 − x) f (x)dx = −f(c) s , i f (x) 2 dx ≥ 3 .
c c (1 − c)