Lehenengo oinarrizko teorema
aldatu
Izan bitez
[
a
,
b
]
{\displaystyle \,[a,b]}
tartean integragarria den
f
{\displaystyle \,f}
funtzio bat eta
F
{\displaystyle \,F}
beste funtzio bat honela definitua:
F
(
x
)
=
∫
α
x
f
(
t
)
d
t
{\displaystyle F(x)={\int _{\alpha }^{x}f(t)dt}}
non
α
∈
[
a
,
b
]
{\displaystyle \alpha \in [a,b]}
den.
Teoremak hau esaten du:
f
{\displaystyle \,f}
funtzioa
c
∈
[
a
,
b
]
{\displaystyle c\in [a,b]}
puntuan jarraitua bada , orduan
F
{\displaystyle \,F}
funtzioa
c
{\displaystyle \,c}
puntuan deribagarria da eta
F
′
(
c
)
=
f
(
c
)
{\displaystyle \,F'(c)=f(c)}
betetzen da.
Lema garrantzitsua:
Demagun
f
{\displaystyle f}
funtzioa
[
a
,
b
]
{\displaystyle [a,b]}
tartean integragarria dela eta
m
≤
f
(
x
)
≤
M
∀
x
∈
[
a
,
b
]
{\displaystyle m\leq f(x)\leq M\forall x\in [a,b]}
, orduan:
m
(
b
−
a
)
≤
∫
a
b
f
(
t
)
d
t
≤
M
(
b
−
a
)
{\displaystyle m(b-a)\leq {\int _{a}^{b}f(t)dt}\leq M(b-a)}
Frogapenaren hasiera
Hipotesia:
Biz
c
∈
(
a
,
b
)
{\displaystyle c\in (a,b)}
.
Biz
[
a
,
b
]
{\displaystyle [a,b]}
tartean integragarria eta c puntuan jarraitua den
f
{\displaystyle f}
funtzioa.
Biz
[
a
,
b
]
{\displaystyle [a,b]}
tartean honela definitutako
F
{\displaystyle F}
funtzioa:
F
(
x
)
=
∫
α
x
f
(
t
)
d
t
{\displaystyle F(x)=\int _{\alpha }^{x}f(t)dt}
, non
α
∈
[
a
,
b
]
{\displaystyle \alpha \in [a,b]}
den.
Tesia:
F'(c)=f(c)
Definizioz hau daukagu:
F
′
(
c
)
=
lim
h
→
0
F
(
c
+
h
)
−
F
(
c
)
h
{\displaystyle F'(c)={\lim _{h\rightarrow 0}{\frac {F(c+h)-F(c)}{h}}}}
.
Demagun h>0 dela, orduan
F
(
c
+
h
)
−
F
(
c
)
=
∫
c
c
+
h
f
(
t
)
d
t
{\displaystyle F(c+h)-F(c)={\int _{c}^{c+h}f(t)dt}}
.
m
h
{\displaystyle m_{h}}
eta
M
h
{\displaystyle M_{h}}
honela definituta:
m
h
=
inf
{
f
(
x
)
|
c
≤
x
≤
c
+
h
}
{\displaystyle m_{h}=\inf\{f(x)|c\leq x\leq c+h\}}
,
M
h
=
sup
{
f
(
x
)
|
c
≤
x
≤
c
+
h
}
{\displaystyle M_{h}=\sup\{f(x)|c\leq x\leq c+h\}}
'Lema' aplikatuta, hau daukagu:
m
h
⋅
h
≤
∫
c
c
+
h
f
(
t
)
d
t
≤
M
h
⋅
h
{\displaystyle m_{h}\cdot h\leq {\int _{c}^{c+h}f(t)dt}\leq M_{h}\cdot h}
.
Beraz,
m
h
≤
F
(
c
+
h
)
−
F
(
c
)
h
≤
M
h
{\displaystyle m_{h}\leq {\frac {F(c+h)-F(c)}{h}}\leq M_{h}}
Orain demagun
h
<
0
{\displaystyle h<0}
dela, izan bitez:
m
∗
h
=
inf
{
f
(
x
)
|
c
+
h
≤
x
≤
c
}
{\displaystyle {m^{*}}_{h}=\inf\{f(x)|c+h\leq x\leq c\}}
,
M
∗
h
=
sup
{
f
(
x
)
|
c
+
h
≤
x
≤
c
}
{\displaystyle {M^{*}}_{h}=\sup\{f(x)|c+h\leq x\leq c\}}
.
'Lema' aplikatuta, hau daukagu:
m
∗
h
⋅
(
−
h
)
≤
∫
c
+
h
c
f
(
t
)
d
t
≤
M
∗
h
⋅
(
−
h
)
{\displaystyle {m^{*}}_{h}\cdot (-h)\leq {\int _{c+h}^{c}f(t)dt}\leq {M^{*}}_{h}\cdot (-h)}
.
Honako hau betetzen denez:
F
(
c
+
h
)
−
F
(
c
)
=
∫
c
c
+
h
f
(
t
)
d
t
=
−
∫
c
+
h
c
f
(
t
)
d
t
{\displaystyle F(c+h)-F(c)={\int _{c}^{c+h}f(t)dt}=-{\int _{c+h}^{c}f(t)dt}}
,
Orduan:
m
∗
h
⋅
h
≥
F
(
c
+
h
)
−
F
(
c
)
≥
M
∗
h
⋅
h
{\displaystyle {m^{*}}_{h}\cdot h\geq F(c+h)-F(c)\geq {M^{*}}_{h}\cdot h}
.
h
<
0
{\displaystyle h<0}
denez gero, orduan hau daukagu:
m
∗
h
≤
F
(
c
+
h
)
−
F
(
c
)
h
≤
M
∗
h
{\displaystyle {m^{*}}_{h}\leq {\frac {F(c+h)-F(c)}{h}}\leq {M^{*}}_{h}}
.
Eta
f
{\displaystyle f}
funtzioa c puntuan jarraitua denez, orduan:
lim
h
→
0
m
h
=
lim
h
→
0
M
h
=
lim
h
→
0
m
∗
h
=
lim
h
→
0
M
∗
h
=
f
(
c
)
{\displaystyle \lim _{h\rightarrow 0}m_{h}=\lim _{h\rightarrow 0}M_{h}=\lim _{h\rightarrow 0}{m^{*}}_{h}=\lim _{h\rightarrow 0}{M^{*}}_{h}=f(c)}
,
Azkenean, guzti horrek teorema frogatzera garamatza:
F
′
(
c
)
=
lim
h
→
0
F
(
c
+
h
)
−
F
(
c
)
h
=
f
(
c
)
{\displaystyle F'(c)={\lim _{h\rightarrow 0}{\frac {F(c+h)-F(c)}{h}}}=f(c)}
.
F
(
x
)
=
∫
0
x
t
2
d
t
⇒
F
′
(
x
)
=
x
2
{\displaystyle F(x)=\int _{0}^{x}t^{2}dt\Rightarrow F'(x)=x^{2}}
H
(
x
)
=
∫
10
exp
3
x
sin
(
t
)
d
t
⇒
H
′
(
x
)
=
sin
(
e
3
x
)
e
3
x
3
{\displaystyle H(x)=\int _{10}^{\exp {3x}}\sin(t)dt\Rightarrow H'(x)=\sin(e^{3x})e^{3x}3}
G
(
x
)
=
∫
0
x
2
arcsin
(
t
)
d
t
⇒
G
′
(
x
)
=
arcsin
(
x
2
)
2
x
{\displaystyle G(x)=\int _{0}^{x^{2}}\arcsin(t)dt\Rightarrow G'(x)=\arcsin(x^{2})2x}
Bigarren oinarrizko teorema
aldatu
Barrow ren erregela ere deiturikoa, Isaac Barrowren omenez.
Izan bitez
[
a
,
b
]
{\displaystyle \,[a,b]}
tartean jarraitua den
f
{\displaystyle \,f}
funtzio bat eta
g
{\displaystyle \,g}
haren edozein jatorrizko funtzio bat, hau da
g
′
(
x
)
=
f
(
x
)
{\displaystyle \,g'(x)=f(x)}
. Orduan:
∫
a
b
f
(
x
)
d
x
=
g
(
b
)
−
g
(
a
)
{\displaystyle \int _{a}^{b}f(x)dx=g(b)-g(a)}
Teorema hau askotan erabiltzen da integral mugatuak ebaluatzeko.
Hipotesia:
Biz
[
a
,
b
]
{\displaystyle [a,b]}
tartean jarraitua den
f
{\displaystyle f}
funtzioa
Biz
[
a
,
b
]
{\displaystyle [a,b]}
tartean diferentziagarria den
g
{\displaystyle g}
funtzioa, non
g
′
(
x
)
=
f
(
x
)
∀
x
∈
[
a
,
b
]
{\displaystyle g'(x)=f(x){\ }\forall x\in [a,b]}
den
Tesia:
∫
a
b
f
(
x
)
d
x
=
g
(
b
)
−
g
(
a
)
{\displaystyle \int _{a}^{b}f(x)dx=g(b)-g(a)}
Frogapena:
Izan bedi
F
(
x
)
=
∫
a
x
f
(
t
)
d
t
{\displaystyle F(x)=\int _{a}^{x}f(t)dt}
.
Kalkuluaren lehenengo oinarrizko teorema dela medio, hau daukagu:
F
′
(
x
)
=
f
(
x
)
=
g
′
(
x
)
∀
x
∈
[
a
,
b
]
{\displaystyle F'(x)=f(x)=g'(x){\ }\forall x\in [a,b]}
.
Beraz:
∃
K
∈
R
{\displaystyle \exists K\in \mathbb {R} {\ }}
non
∀
x
∈
[
a
,
b
]
,
F
(
x
)
=
g
(
x
)
+
K
{\displaystyle \forall x\in [a,b],F(x)=g(x)+K}
den.
Hau aintzat hartuta:
0
=
F
(
a
)
=
g
(
a
)
+
K
{\displaystyle 0=F(a)=g(a)+K}
Eta hortik segitzen denez
c
=
−
g
(
a
)
{\displaystyle c=-g(a)}
da; beraz:
F
(
x
)
=
g
(
x
)
−
g
(
a
)
{\displaystyle F(x)=g(x)-g(a)}
.
Bereziki, baldin
x
=
b
{\displaystyle x=b}
bada, hau dugu:
∫
a
b
f
(
t
)
d
t
=
F
(
b
)
=
g
(
b
)
−
g
(
a
)
{\displaystyle \int _{a}^{b}f(t)dt=F(b)=g(b)-g(a)}
∫
0
π
cos
(
x
)
d
x
=
sin
(
π
)
−
sin
(
0
)
=
0
{\displaystyle \int _{0}^{\pi }\cos(x)dx=\sin(\pi )-\sin(0)=0}
∫
1
e
d
x
x
=
ln
(
e
)
−
ln
(
1
)
=
1
{\displaystyle \int _{1}^{e}{\frac {dx}{x}}=\ln(e)-\ln(1)=1}