Sean
f
:
I
⊆
R
→
R
{\displaystyle \,f\colon I\subseteq \mathbb {R} \to \mathbb {R} }
x
0
∈
I
{\displaystyle x_{0}\in I}
0
≤
f
(
x
)
≤
A
+
B
∫
x
o
x
f
(
s
)
d
s
∀
x
∈
I
{\displaystyle 0\leq f{(x)}\leq A+B\int _{x_{o}}^{x}{f{(s)}ds}\,\forall x\in I}
con A y B ≥ 0 constantes.
Entonces:
f
(
x
)
≤
A
e
B
(
x
−
x
0
)
{\displaystyle f{(x)}\leq A\,e^{B(x-x_{0})}}
Sea
G
(
x
)
=
∫
x
o
x
f
(
s
)
d
s
{\displaystyle \,G{(x)}=\int _{x_{o}}^{x}{f{(s)}ds}}
G
′
(
x
)
−
B
G
(
x
)
≤
A
{\displaystyle G'{(x)}-B\,G{(x)}\leq A}
e
−
B
(
x
−
x
0
)
{\displaystyle e^{-B(x-x_{0})}}
e
−
B
(
x
−
x
0
)
G
′
(
x
)
−
B
e
−
B
(
x
−
x
0
)
G
(
x
)
≤
e
−
B
(
x
−
x
0
)
A
{\displaystyle e^{-B(x-x_{0})}G'{(x)}-B\,e^{-B(x-x_{0})}G{(x)}\leq e^{-B(x-x_{0})}A}
(
e
−
B
(
x
−
x
0
)
G
(
x
)
)
′
≤
A
e
−
B
(
x
−
x
0
)
{\displaystyle (e^{-B(x-x_{0})}G{(x)})'\leq A\,e^{-B(x-x_{0})}}
x
0
{\displaystyle x_{0}}
x
{\displaystyle x}
∫
x
o
x
(
e
−
B
(
s
−
x
0
)
G
(
s
)
)
′
d
s
≤
A
∫
x
o
x
e
−
B
(
s
−
x
0
)
d
s
{\displaystyle \int _{x_{o}}^{x}{(e^{-B(s-x_{0})}G{(s)})'ds}\leq A\int _{x_{o}}^{x}{e^{-B(s-x_{0})}ds}}
e
−
B
(
x
−
x
0
)
G
(
x
)
−
G
(
x
0
)
≤
−
A
B
(
e
−
B
(
x
−
x
0
)
−
1
)
{\displaystyle e^{-B(x-x_{0})}G{(x)}-G{(x_{0})}\leq -{\frac {A}{B}}(e^{-B(x-x_{0})}-1)}
G
(
x
0
)
=
0
{\displaystyle G{(x_{0})}=0}
e
B
(
x
−
x
0
)
{\displaystyle e^{B(x-x_{0})}}
G
(
x
)
≤
A
B
(
e
B
(
x
−
x
0
)
−
1
)
{\displaystyle G{(x)}\leq {\frac {A}{B}}(e^{B(x-x_{0})}-1)}
G en la ecuación original:
f
(
x
)
≤
A
+
B
A
B
(
e
B
(
x
−
x
0
)
−
1
)
{\displaystyle f{(x)}\leq A+B\,{\frac {A}{B}}(e^{B(x-x_{0})}-1)}
f
(
x
)
≤
A
e
B
(
x
−
x
0
)
{\displaystyle f{(x)}\leq A\,e^{B(x-x_{0})}}
Datos: Q510733