Rappresentazzjoni grafika tal-integral ta' Riemann Ejjew naqsmu l-intervall kompatt
[
a
,
b
]
{\displaystyle \ [a,b]}
partizzjoni
P
{\displaystyle \displaystyle {P}}
n
{\displaystyle \displaystyle {n}}
P
=
{
a
=
x
0
<
x
1
<
x
2
<
…
<
x
n
−
1
<
x
n
=
b
}
{\displaystyle P=\{a=x_{0}<x_{1}<x_{2}<\ldots <x_{n-1}<x_{n}=b\}}
Ħalli jkunu
m
k
=
inf
{
f
(
x
)
:
x
∈
[
x
k
−
1
,
x
k
]
}
{\displaystyle m_{k}=\inf\{f(x):x\in [x_{k-1},x_{k}]\}}
M
k
=
sup
{
f
(
x
)
:
x
∈
[
x
k
−
1
,
x
k
]
}
.
{\displaystyle M_{k}=\sup\{f(x):x\in [x_{k-1},x_{k}]\}.}
Niddefinixxu s-somma integrali inferjuri (relattiva għall-partizzjoni
P
{\displaystyle \displaystyle {P}}
s
(
f
,
P
)
=
∑
k
=
1
n
m
k
(
x
k
−
x
k
−
1
)
.
{\displaystyle s(f,P)=\sum _{k=1}^{n}m_{k}(x_{k}-x_{k-1}).}
Jekk nammettu li
f
{\displaystyle \displaystyle {f}}
s
(
f
,
P
)
{\displaystyle \displaystyle {s(f,P)}}
R
{\displaystyle \mathbb {R} }
f
{\displaystyle \displaystyle {f}}
Niddefinixxu s-somma integrali superjuri (relattiva għall-partizzjoni
P
{\displaystyle \displaystyle {P}}
S
(
f
,
P
)
=
∑
k
=
1
n
M
k
(
x
k
−
x
k
−
1
)
{\displaystyle S(f,P)=\sum _{k=1}^{n}M_{k}(x_{k}-x_{k-1})}
Analogament,
S
(
f
,
P
)
{\displaystyle \displaystyle {S(f,P)}}
R
{\displaystyle \mathbb {R} }
Jidher ċar li jekk
m
≤
f
(
x
)
≤
M
,
∀
x
∈
[
a
,
b
]
{\displaystyle m\leq f(x)\leq M,\ \forall x\in [a,b]}
P
{\displaystyle \displaystyle {P}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}\ }
m
(
b
−
a
)
≤
s
(
f
,
P
)
≤
S
(
f
,
P
)
≤
M
(
b
−
a
)
{\displaystyle m(b-a)\leq s(f,P)\leq S(f,P)\leq M(b-a)}
Dawn iż-żewġ lemmata wieħed jista' jipprovahom faċilment:
Lemma 1. :
Jekk
P
{\displaystyle \displaystyle {P}}
Q
{\displaystyle \displaystyle {Q}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}}
Q
{\displaystyle \displaystyle {Q}}
P
{\displaystyle \displaystyle {P}\ }
s
(
f
,
P
)
≤
s
(
f
,
Q
)
≤
S
(
f
,
Q
)
≤
S
(
f
,
P
)
{\displaystyle s(f,P)\leq s(f,Q)\leq S(f,Q)\leq S(f,P)}
Lemma 2. :
Għal kull żewġ partizzjonijiet
P
,
Q
{\displaystyle \displaystyle {P,Q}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}\ }
s
(
f
,
P
)
≤
S
(
f
,
Q
)
{\displaystyle s(f,P)\leq S(f,Q)}
Ħalli jkunu
s
(
f
)
=
sup
{
s
(
f
,
P
)
:
P
{\displaystyle s(f)\displaystyle {=}\sup\{s(f,P):P}
[
a
,
b
]
}
{\displaystyle \displaystyle {[a,b]}\}}
S
(
f
)
=
inf
{
S
(
f
,
P
)
:
P
{\displaystyle S(f)\displaystyle {=}\inf\{S(f,P):P}
[
a
,
b
]
}
{\displaystyle \displaystyle {[a,b]}\}}
s
(
f
)
{\displaystyle \displaystyle {s(f)}}
integral inferjuri u
S
(
f
)
{\displaystyle \displaystyle {S(f)}}
integral superjuri . Mill-lemma preċidenti nistgħu niddeduċu li dawn jissodisfaw
s
(
f
)
≤
S
(
f
)
.
{\displaystyle s(f)\leq S(f).}
Definizzjoni :Integral skont Riemann
In-numri
a
{\displaystyle \displaystyle {a}}
b
{\displaystyle \displaystyle {b}}
f
{\displaystyle \displaystyle {f}\ }
a
{\displaystyle \displaystyle {a}}
b
{\displaystyle \displaystyle {b}}
varjabbli muta jiġifieri
∫
f
(
x
)
d
x
{\displaystyle \int \!\!\!f(x){\rm {d}}x}
∫
f
(
t
)
d
t
{\displaystyle \int \!\!\!f(t){\rm {d}}t}
d
x
{\displaystyle \displaystyle {{\rm {d}}x}}
Jekk il-funzjoni integrabbli
f
{\displaystyle \displaystyle {f}}
{
(
x
,
y
)
|
0
≤
y
≤
f
(
x
)
,
x
∈
[
a
,
b
]
}
.
{\displaystyle \{(x,y)\ |\ 0\leq y\leq f(x),\ x\in [a,b]\}.}
Jekk il-funzjoni
f
{\displaystyle \displaystyle {f}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}}
[
a
,
b
]
{\displaystyle \displaystyle {\ [a,b]}}
(
b
−
a
)
/
n
{\displaystyle \displaystyle {(b-a)/n}}
s
(
f
,
P
n
)
{\displaystyle \displaystyle {s(f,P_{n})}}
S
(
f
,
P
n
)
{\displaystyle \displaystyle {S(f,P_{n})}}
n
{\displaystyle \displaystyle {n}}
s
(
f
)
≥
lim
n
→
∞
s
(
f
,
P
n
)
=
lim
n
→
∞
S
(
f
,
P
n
)
≥
S
(
f
)
{\displaystyle s(f)\geq \lim _{n\to \infty }s(f,P_{n})=\lim _{n\to \infty }S(f,P_{n})\geq S(f)}
u allura, la
s
(
f
)
≤
S
(
f
)
{\displaystyle \displaystyle {s(f)\leq S(f)}}
s
(
f
)
=
S
(
f
)
.
{\displaystyle \displaystyle {s(f)=S(f).}}
Eżempju 1.
Ħalli
f
(
x
)
=
x
2
{\displaystyle \displaystyle {f(x)=x^{2}}}
[
0
,
1
]
{\displaystyle \displaystyle {\ [0,1]}}
M
k
=
sup
{
x
2
|
x
∈
[
(
k
−
1
)
/
n
,
k
/
n
]
}
=
(
k
/
n
)
2
{\displaystyle \displaystyle {M_{k}=\sup\{x^{2}\,|\,x\in [(k-1)/n,k/n]\}=(k/n)^{2}}}
u
m
k
=
inf
{
x
2
|
x
∈
[
(
k
−
1
)
/
n
,
k
/
n
]
}
=
(
(
k
−
1
)
/
n
)
2
.
{\displaystyle \displaystyle {m_{k}=\inf\{x^{2}\,|\,x\in [(k-1)/n,k/n]\}=((k-1)/n)^{2}}.}
Mela
S
(
f
,
P
n
)
=
1
n
∑
k
=
1
n
(
k
n
)
2
=
1
n
3
∑
k
=
1
n
k
2
=
1
3
(
n
+
1
n
)
(
2
n
+
1
2
n
)
,
{\displaystyle \displaystyle {S(f,P_{n})={\frac {1}{n}}\sum _{k=1}^{n}\left({\frac {k}{n}}\right)^{2}={\frac {1}{n^{3}}}\sum _{k=1}^{n}k^{2}={\frac {1}{3}}\left({\frac {n+1}{n}}\right)\left({\frac {2n+1}{2n}}\right),}}
fejn użajna l-formula
1
2
+
2
2
+
…
+
n
2
=
n
(
n
+
1
)
(
2
n
+
1
)
/
6
{\displaystyle \displaystyle {1^{2}+2^{2}+\ldots +n^{2}=n(n+1)(2n+1)/6}}
Bl-istess mod
s
(
f
,
P
n
)
=
1
n
∑
k
=
1
n
(
k
−
1
n
)
2
=
1
n
3
∑
k
=
1
n
−
1
k
2
=
1
3
(
n
−
1
n
)
(
2
n
−
1
2
n
)
.
{\displaystyle \displaystyle {s(f,P_{n})={\frac {1}{n}}\sum _{k=1}^{n}\left({\frac {k-1}{n}}\right)^{2}={\frac {1}{n^{3}}}\sum _{k=1}^{n-1}k^{2}={\frac {1}{3}}\left({\frac {n-1}{n}}\right)\left({\frac {2n-1}{2n}}\right).}}
Allura
∫
0
1
x
2
d
x
=
lim
n
→
∞
s
(
f
,
P
n
)
=
lim
n
→
∞
S
(
f
,
P
n
)
=
1
3
.
{\displaystyle \int _{0}^{1}x^{2}{\rm {d}}x=\lim _{n\to \infty }s(f,P_{n})=\lim _{n\to \infty }S(f,P_{n})={\frac {1}{3}}.}
Eżempju 2.
B'kuntrast mal-eżempju ta' qabel, ejjew nikkunsidraw il-funzjoni
g
:
[
0
,
1
]
↦
R
{\displaystyle \displaystyle {g:[0,1]\mapsto \mathbb {R} }}
g
(
x
)
=
{
1
,
j
e
k
k
x
h
i
j
a
r
a
z
z
j
o
n
a
l
i
0
,
j
e
k
k
x
h
i
j
a
r
r
a
z
z
j
o
n
a
l
i
.
{\displaystyle {g(x)={\begin{cases}1,\ \mathrm {jekk} \ x\ \mathrm {hija\ razzjonali} \\0,\ \mathrm {jekk} \ x\ \mathrm {hija\ rrazzjonali} .\end{cases}}}}
Għal kull partizzjoni tal-intervall
[
0
,
1
]
{\displaystyle \displaystyle {[0,1]}}
[
x
k
−
1
,
x
k
]
{\displaystyle \displaystyle {[x_{k-1},x_{k}]}}
M
k
=
1
{\displaystyle \displaystyle {M_{k}=1}}
m
k
=
0
{\displaystyle \displaystyle {m_{k}=0}}
s
(
g
,
P
)
=
∑
k
=
1
n
m
k
(
x
k
−
x
k
−
1
)
=
0
{\displaystyle s(g,P)=\sum _{k=1}^{n}m_{k}(x_{k}-x_{k-1})=0}
s
(
g
,
P
)
=
∑
k
=
1
n
m
k
(
x
k
−
x
k
−
1
)
=
∑
k
=
1
n
(
x
k
−
x
k
−
1
)
=
1.
{\displaystyle s(g,P)=\sum _{k=1}^{n}m_{k}(x_{k}-x_{k-1})=\sum _{k=1}^{n}(x_{k}-x_{k-1})=1.}
Mela
s
(
g
)
=
0
{\displaystyle \displaystyle {s(g)=0}}
S
(
g
)
=
1
{\displaystyle \displaystyle {S(g)=1}}
g
{\displaystyle \displaystyle {g}}
La mhux kull funzjoni hi integrabbli, hemm bżonn li niddeċiedu meta l-integral ta' funzjoni
jeżisti jew le. Il-quddiem nagħtu żewġ klassijiet wiesa' ta' funzjonijiet li huma integrabbli. It-teorema li ġejja hi utli ħafna għal din id-deċizzjoni.
Teorema 1 :Kundizzjoni meħtieġa u biżżejjed għall-integrabbiltà
Halli
f
{\displaystyle \displaystyle {f}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}}
f
{\displaystyle \displaystyle {f}}
ϵ
>
0
{\displaystyle \displaystyle {\epsilon \,>\,0}}
P
{\displaystyle \displaystyle {P}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}}
S
(
f
,
P
)
−
s
(
f
,
p
)
<
ϵ
.
{\displaystyle \displaystyle {S(f,P)-s(f,p)<\epsilon .}}
Prova :
Nissoponu li
f
{\displaystyle \displaystyle {f}}
S
(
f
)
=
s
(
f
)
{\displaystyle \displaystyle {S(f)=s(f)}}
ϵ
>
0
{\displaystyle \displaystyle {\epsilon >0}}
P
1
{\displaystyle \displaystyle {P_{1}}}
[
a
.
b
]
{\displaystyle \displaystyle {[a.b]}}
s
(
f
,
P
1
)
>
s
(
f
)
−
ϵ
2
.
{\displaystyle \displaystyle {s(f,P_{1})>s(f)-{\frac {\epsilon }{2}}.}}
(Din issegwi mid-definizzjoni tas-supremum). Bl-istess mod teżisti partizzjoni
P
2
{\displaystyle \displaystyle {P_{2}}}
[
a
.
b
]
{\displaystyle \displaystyle {[a.b]}}
S
(
f
,
P
2
)
<
S
(
f
)
+
ϵ
2
.
{\displaystyle \displaystyle {S(f,P_{2})<S(f)+{\frac {\epsilon }{2}}.}}
Ħalli
P
=
P
1
∪
P
2
{\displaystyle \displaystyle {P=P_{1}\cup P_{2}}}
S
(
f
,
P
)
−
s
(
f
,
P
)
≤
S
(
f
,
P
2
)
−
s
(
f
,
P
1
)
<
(
S
(
f
)
+
ϵ
2
)
−
(
s
(
f
)
−
ϵ
2
)
=
S
(
f
)
−
s
(
f
)
+
ϵ
=
ϵ
.
{\displaystyle \displaystyle {S(f,P)-s(f,P)\leq S(f,P_{2})-s(f,P_{1})<\left(S(f)+{\frac {\epsilon }{2}}\right)-\left(s(f)-{\frac {\epsilon }{2}}\right)=S(f)-s(f)+\epsilon =\epsilon .}}
ϵ
>
0
{\displaystyle \displaystyle {\epsilon >0}}
P
{\displaystyle \displaystyle {P}}
[
a
.
b
]
{\displaystyle \displaystyle {[a.b]}}
S
(
f
,
P
)
<
s
(
f
,
P
)
+
ϵ
{\displaystyle \displaystyle {S(f,P)<s(f,P)+\epsilon }}
S
(
f
)
≤
S
(
f
,
P
)
<
s
(
f
,
P
)
+
ϵ
≤
s
(
f
)
+
ϵ
.
{\displaystyle \displaystyle {S(f)\leq S(f,P)<s(f,P)+\epsilon \leq s(f)+\epsilon .}}
La
ϵ
{\displaystyle \displaystyle {\epsilon }}
S
(
f
)
≤
s
(
f
)
{\displaystyle \displaystyle {S(f)\leq s(f)}}
S
(
f
)
=
s
(
f
)
{\displaystyle \displaystyle {S(f)=s(f)}}
f
{\displaystyle \displaystyle {f}}
Proprjetajiet tal-integral skont Riemann
immodifika
Proprijetà 1 :Il-monotonija hi biżżejjed għall-integrabbiltà
Jekk il-funzjoni
f
:
[
a
,
b
]
→
R
{\displaystyle \ f:[a,b]\to \mathbb {R} }
Prova:
Nissoponu li l-funzjoni
f
{\displaystyle \ f}
[
a
,
b
]
{\displaystyle \ [a,b]}
ϵ
>
0
{\displaystyle \epsilon >0}
δ
>
0
{\displaystyle \delta >0}
δ
<
ϵ
f
(
b
)
−
f
(
a
)
.
{\displaystyle \delta <{\frac {\epsilon }{f(b)-f(a)}}.}
Ħalli
P
{\displaystyle P}
[
a
,
b
]
{\displaystyle \ [a,b]}
[
x
k
−
1
,
x
k
]
{\displaystyle \ [x_{k-1},x_{k}]}
δ
{\displaystyle \delta }
M
k
=
f
(
x
k
)
{\displaystyle \ M_{k}=f(x_{k})}
m
k
=
f
(
x
k
−
1
)
{\displaystyle \ m_{k}=f(x_{k-1})}
0
<
S
(
f
,
P
)
−
s
(
f
,
P
)
=
∑
k
=
1
n
(
f
(
x
k
)
−
f
(
x
k
−
1
)
)
(
x
k
−
x
k
−
1
)
<
ϵ
f
(
b
)
−
f
(
a
)
∑
k
=
1
n
(
f
(
x
k
)
−
f
(
x
k
−
1
)
)
=
ϵ
.
{\displaystyle 0<S(f,P)-s(f,P)=\sum _{k=1}^{n}(f(x_{k})-f(x_{k-1}))(x_{k}-x_{k-1})<{\frac {\epsilon }{f(b)-f(a)}}\sum _{k=1}^{n}(f(x_{k})-f(x_{k-1}))=\epsilon .}
Allura mit-Teorema 1 nistgħu nikkonkludu li l-funzjoni hi integrabbli.
Proprijetà 2 :Il-kontinwità hi suffiċjenti għall-integrabbiltà
Jekk il-funzjoni
f
:
[
a
,
b
]
→
R
{\displaystyle \ f:[a,b]\to \mathbb {R} }
Prova:
La l-funzjoni
f
:
[
a
,
b
]
→
R
{\displaystyle \ f:[a,b]\to \mathbb {R} }
ϵ
>
0
{\displaystyle \epsilon >0}
δ
>
0
{\displaystyle \delta >0}
|
f
(
x
)
−
f
(
y
)
|
<
ϵ
b
−
a
{\displaystyle |f(x)-f(y)|<{\frac {\epsilon }{b-a}}}
kull meta
|
x
−
y
|
<
δ
{\displaystyle |x-y|<\delta }
P
{\displaystyle P}
[
a
,
b
]
{\displaystyle \ [a,b]}
[
x
k
−
1
,
x
k
]
{\displaystyle \ [x_{k-1},x_{k}]}
δ
{\displaystyle \delta }
0
<
M
k
−
m
k
<
ϵ
b
−
a
{\displaystyle 0<M_{k}-m_{k}<{\frac {\epsilon }{b-a}}}
u mela
0
<
S
(
f
,
P
)
−
s
(
f
,
P
)
=
∑
k
=
1
n
(
M
k
−
m
k
)
(
x
k
−
x
k
−
1
)
<
ϵ
b
−
a
∑
k
=
1
n
(
x
k
−
x
k
−
1
)
=
ϵ
.
{\displaystyle 0<S(f,P)-s(f,P)=\sum _{k=1}^{n}(M_{k}-m_{k})(x_{k}-x_{k-1})<{\frac {\epsilon }{b-a}}\sum _{k=1}^{n}(x_{k}-x_{k-1})=\epsilon .}
Allura mit-Teorema 1 nistgħu nikkonkludu li l-funzjoni hi integrabbli.
Proprijetà 3 :Il-proprijetà tal-linjarità
Ħalli
f
{\displaystyle f}
u
g
{\displaystyle g}
jkunu żewġ funzjonijiet kontinwi definiti f'intervall
[
a
,
b
]
{\displaystyle [a,b]}
u ħalli jkunu
α
,
β
∈
R
{\displaystyle \alpha ,\beta \in \mathbb {R} }
. Imbagħad:
∫
a
b
(
α
f
(
x
)
+
β
g
(
x
)
)
d
x
=
α
∫
a
b
f
(
x
)
d
x
+
β
∫
a
b
g
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}(\alpha f(x)+\beta g(x))\,dx=\alpha \int _{a}^{b}f(x)dx+\beta \int _{a}^{b}g(x){\rm {d}}x.}
Prova:
Jekk
α
≥
0
{\displaystyle \displaystyle {\alpha \geq 0}}
S
(
α
f
)
=
α
S
(
f
)
{\displaystyle \displaystyle {S(\alpha f)=\alpha S(f)}}
s
(
α
f
)
=
α
s
(
f
)
{\displaystyle \displaystyle {s(\alpha f)=\alpha s(f)}}
S
(
f
)
=
s
(
f
)
=
∫
a
b
f
(
x
)
d
x
,
{\displaystyle \displaystyle {S(f)=s(f)=\int _{a}^{b}f(x){\rm {d}}x,}}
għandna
∫
a
b
α
f
(
x
)
d
x
=
S
(
α
f
)
=
s
(
α
f
)
=
α
∫
a
b
f
(
x
)
d
x
.
{\displaystyle \displaystyle {\int _{a}^{b}\alpha f(x){\rm {d}}x=S(\alpha f)=s(\alpha f)=\alpha \int _{a}^{b}f(x){\rm {d}}x.}}
B'mod simili jekk
α
<
0
{\displaystyle \displaystyle {\alpha <0}}
S
(
α
f
)
=
α
s
(
f
)
{\displaystyle \displaystyle {S(\alpha f)=\alpha s(f)}}
s
(
α
f
)
=
α
S
(
f
)
{\displaystyle \displaystyle {s(\alpha f)=\alpha S(f)}}
∫
a
b
α
f
(
x
)
d
x
=
α
∫
a
b
f
(
x
)
d
x
.
{\displaystyle \displaystyle {\int _{a}^{b}\alpha f(x){\rm {d}}x=\alpha \int _{a}^{b}f(x){\rm {d}}x.}}
Mela issa biżżejjed li nipprovaw li
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
d
x
=
∫
a
b
f
(
x
)
d
x
+
∫
a
b
g
(
x
)
d
x
.
{\displaystyle \ \int _{a}^{b}(f(x)+g(x))\,{\rm {d}}x=\int _{a}^{b}f(x){\rm {d}}x+\int _{a}^{b}g(x){\rm {d}}x.}
Niftakru li
sup
x
∈
D
[
f
(
x
)
+
g
(
x
)
]
≤
sup
x
∈
D
f
(
x
)
+
sup
x
∈
D
g
(
x
)
u
inf
x
∈
D
(
[
f
(
x
)
+
g
(
x
)
]
)
≥
inf
x
∈
D
f
(
x
)
+
inf
x
∈
D
g
(
x
)
{\displaystyle \ \sup _{x\in D}[f(x)+g(x)]\leq \sup _{x\in D}f(x)+\sup _{x\in D}g(x)\ \ \ {\rm {u}}\ \ \ \inf _{x\in D}([f(x)+g(x)])\geq \inf _{x\in D}f(x)+\inf _{x\in D}g(x)}
u għalhekk għal kull partizzjoni
P
{\displaystyle \displaystyle {P}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}}
S
(
f
+
g
,
P
)
≤
S
(
f
,
P
)
+
S
(
g
,
P
)
u
s
(
f
+
g
,
P
)
≥
s
(
f
,
P
)
+
s
(
g
,
P
)
.
{\displaystyle \ S(f+g,P)\leq S(f,P)+S(g,P)\ \ \ {\rm {u}}\ \ \ s(f+g,P)\geq s(f,P)+s(g,P).}
Mit-Teorema 1 nafu li għall kull
ϵ
>
0
{\displaystyle \displaystyle {\epsilon >0}}
P
1
{\displaystyle \displaystyle {P_{1}}}
P
2
{\displaystyle \displaystyle {P_{2}}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}}
S
(
f
,
P
1
)
<
s
(
f
,
P
1
)
+
ϵ
2
u
S
(
g
,
P
2
)
<
s
(
g
,
P
2
)
+
ϵ
2
.
{\displaystyle \displaystyle {S(f,P_{1})<s(f,P_{1})+{\frac {\epsilon }{2}}\ \ \ {\rm {u}}\ \ \ S(g,P_{2})<s(g,P_{2})+{\frac {\epsilon }{2}}.}}
Ħalli
P
=
P
1
∪
P
2
{\displaystyle \displaystyle {P=P_{1}\cup P_{2}}}
S
(
f
,
P
)
<
s
(
f
,
P
)
+
ϵ
2
u
S
(
g
,
P
)
<
s
(
g
,
P
)
+
ϵ
2
.
{\displaystyle \displaystyle {S(f,P)<s(f,P)+{\frac {\epsilon }{2}}\ \ \ {\rm {u}}\ \ \ S(g,P)<s(g,P)+{\frac {\epsilon }{2}}.}}
Jekk nikkumbinaw id-diżugwaljanzi niksbu
S
(
f
+
g
,
P
)
<
S
(
f
,
P
)
+
S
(
g
,
P
)
<
s
(
f
,
P
)
+
s
(
g
,
P
)
+
ϵ
<
s
(
f
+
g
,
P
)
+
ϵ
{\displaystyle \displaystyle {S(f+g,P)<S(f,P)+S(g,P)<s(f,P)+s(g,P)+\epsilon <s(f+g,P)+\epsilon }}
u allura l-funzjoni
f
+
g
{\displaystyle \displaystyle {f+g}}
Nin-naħa l-oħra, la
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
d
x
=
S
(
f
+
g
)
≤
S
(
f
+
g
,
P
)
<
s
(
f
,
P
)
+
s
(
g
,
P
)
+
ϵ
≤
s
(
f
)
+
s
(
g
)
+
ϵ
=
∫
a
b
f
(
x
)
d
x
+
∫
a
b
g
(
x
)
d
x
+
ϵ
{\displaystyle {\begin{aligned}\int _{a}^{b}(f(x)+g(x))\,{\rm {d}}x&{}=S(f+g)\leq S(f+g,P)\\&{}<s(f,P)+s(g,P)+\epsilon \\&{}\leq s(f)+s(g)+\epsilon =\int _{a}^{b}f(x)\,{\rm {d}}x+\int _{a}^{b}g(x)\,{\rm {d}}x+\epsilon \end{aligned}}}
u
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
d
x
=
s
(
f
+
g
)
≥
s
(
f
+
g
,
P
)
>
S
(
f
,
P
)
+
S
(
g
,
P
)
−
ϵ
≥
S
(
f
)
+
S
(
g
)
−
ϵ
=
∫
a
b
f
(
x
)
d
x
+
∫
a
b
g
(
x
)
d
x
−
ϵ
{\displaystyle {\begin{aligned}\int _{a}^{b}(f(x)+g(x))\,{\rm {d}}x&{}=s(f+g)\geq s(f+g,P)\\&{}>S(f,P)+S(g,P)-\epsilon \\&{}\geq S(f)+S(g)-\epsilon =\int _{a}^{b}f(x)\,{\rm {d}}x+\int _{a}^{b}g(x)\,{\rm {d}}x-\epsilon \end{aligned}}}
nikkonkludu li
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
d
x
=
∫
a
b
f
(
x
)
d
x
+
∫
a
b
g
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}(f(x)+g(x))\,{\rm {d}}x=\int _{a}^{b}f(x)\,{\rm {d}}x+\int _{a}^{b}g(x)\,{\rm {d}}x.}
Proprijetà 4 :Il-proprijetà tal-additività
Jekk
f
{\displaystyle f}
tkun integrabbli fuq l-intervalli
[
a
,
c
]
{\displaystyle \displaystyle {[a,c]}}
u
[
c
,
b
]
{\displaystyle \displaystyle {[c,b]}}
, imbagħad tkun integrabbli fuq l-intervall
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}}
u
∫
a
b
f
(
x
)
d
x
=
∫
a
c
f
(
x
)
d
x
+
∫
c
b
f
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}f(x){\rm {d}}x=\int _{a}^{c}f(x){\rm {d}}x+\int _{c}^{b}f(x){\rm {d}}x.}
Prova:
Mit-Teorema 1 nafu li għall kull
ϵ
>
0
{\displaystyle \displaystyle {\epsilon >0}}
P
1
{\displaystyle \displaystyle {P_{1}}}
[
a
,
c
]
{\displaystyle \displaystyle {[a,c]}}
P
2
{\displaystyle \displaystyle {P_{2}}}
[
c
,
b
]
{\displaystyle \displaystyle {[c,b]}}
S
(
f
,
P
1
)
−
s
(
f
,
P
1
)
<
ϵ
2
u
S
(
f
,
P
2
)
−
s
(
f
,
P
2
)
<
ϵ
2
.
{\displaystyle \displaystyle {S(f,P_{1})-s(f,P_{1})<{\frac {\epsilon }{2}}\ \ \ {\rm {u}}\ \ \ S(f,P_{2})-s(f,P_{2})<{\frac {\epsilon }{2}}.}}
Ħalli
P
=
P
1
∪
P
2
{\displaystyle \displaystyle {P=P_{1}\cup P_{2}}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}}
S
(
f
,
P
)
−
s
(
f
,
P
)
=
S
(
f
,
P
1
)
+
S
(
f
,
P
2
)
−
s
(
f
,
P
1
)
−
s
(
f
,
P
2
)
=
(
S
(
f
,
P
1
)
−
s
(
f
,
P
1
)
)
+
(
S
(
f
,
P
2
)
−
s
(
f
,
P
2
)
)
<
ϵ
2
+
ϵ
2
=
ϵ
.
{\displaystyle \displaystyle {S(f,P)-s(f,P)=S(f,P_{1})+S(f,P_{2})-s(f,P_{1})-s(f,P_{2})=(S(f,P_{1})-s(f,P_{1}))+(S(f,P_{2})-s(f,P_{2}))<{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon .}}
Mela
f
{\displaystyle \displaystyle {f}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}}
Nin-naħa l-oħra, la
∫
a
b
f
(
x
)
d
x
≤
S
(
f
,
P
)
=
S
(
f
,
P
1
)
+
S
(
f
,
P
2
)
<
s
(
f
,
P
1
)
+
s
(
f
,
P
2
)
+
ϵ
≤
∫
a
c
f
(
x
)
d
x
+
∫
c
b
f
(
x
)
d
x
+
ϵ
{\displaystyle {\begin{aligned}\int _{a}^{b}f(x)\,{\rm {d}}x&{}\leq S(f,P)=S(f,P_{1})+S(f,P_{2})\\&{}<s(f,P_{1})+s(f,P_{2})+\epsilon \\&{}\leq \int _{a}^{c}f(x)\,{\rm {d}}x+\int _{c}^{b}f(x)\,{\rm {d}}x+\epsilon \end{aligned}}}
u
∫
a
b
f
(
x
)
d
x
≥
s
(
f
,
P
)
=
s
(
f
,
P
1
)
+
s
(
f
,
P
2
)
>
S
(
f
,
P
1
)
+
S
(
f
,
P
2
)
−
ϵ
≥
∫
a
c
f
(
x
)
d
x
+
∫
c
b
f
(
x
)
d
x
−
ϵ
{\displaystyle {\begin{aligned}\int _{a}^{b}f(x)\,{\rm {d}}x&{}\geq s(f,P)=s(f,P_{1})+s(f,P_{2})\\&{}>S(f,P_{1})+S(f,P_{2})-\epsilon \\&{}\geq \int _{a}^{c}f(x)\,{\rm {d}}x+\int _{c}^{b}f(x)\,{\rm {d}}x-\epsilon \end{aligned}}}
nikkonkludu li
∫
a
b
f
(
x
)
d
x
=
∫
a
c
f
(
x
)
d
x
+
∫
c
b
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x){\rm {d}}x=\int _{a}^{c}f(x){\rm {d}}x+\int _{c}^{b}f(x){\rm {d}}x}
kif nixiequ.
Proprijetà 5 :Il-propijetà tal-monotonija
Jekk
f
{\displaystyle {\displaystyle f}}
u
g
{\displaystyle \ {\displaystyle g}}
ikunu żewġ funzjonijiet integrabbli fuq l-intervall
[
a
,
b
]
{\displaystyle \ [a,b]}
u
f
(
x
)
≤
g
(
x
)
{\displaystyle f(x)\leq g(x)}
għal kull
x
∈
[
a
,
b
]
{\displaystyle \ x\in [a,b]}
, imbagħad
∫
a
b
f
(
x
)
d
x
≤
∫
a
b
g
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}f(x){\rm {d}}x\leq \int _{a}^{b}g(x){\rm {d}}x.}
Prova :
Jekk
f
(
x
)
≤
g
(
x
)
{\displaystyle f(x)\leq g(x)}
x
∈
[
a
,
b
]
{\displaystyle \ x\in [a,b]}
P
{\displaystyle \displaystyle {P}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}}
S
(
f
,
P
)
≤
S
(
g
,
P
)
u
s
(
f
,
P
)
≤
s
(
g
,
P
)
{\displaystyle \ S(f,P)\leq S(g,P)\ \ \ {\rm {u}}\ \ \ s(f,P)\leq s(g,P)}
Minn dawn id-diżugwaljanzi nikkonkludu l-monotonija tal-integral.
Teorema :Teorema tal-valur assolut
Jekk
f
{\displaystyle f}
tkun integrabbli fl-intervall
[
a
,
b
]
{\displaystyle \ [a,b]}
, imbagħad
|
f
|
{\displaystyle |f|}
hi wkoll integrabbli u
|
∫
a
b
f
(
x
)
d
x
|
≤
∫
a
b
|
f
(
x
)
|
d
x
.
{\displaystyle \left|\int _{a}^{b}f(x){\rm {d}}x\right|\leq \int _{a}^{b}\left|f(x)\right|{\rm {d}}x.}
Prova :
Ħalli
P
{\displaystyle \displaystyle {P}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}}
n
{\displaystyle \displaystyle {n}}
P
=
{
a
=
x
0
<
x
1
<
x
2
<
…
<
x
n
−
1
<
x
n
=
b
}
{\displaystyle P=\{a=x_{0}<x_{1}<x_{2}<\ldots <x_{n-1}<x_{n}=b\}}
u
m
~
k
=
inf
{
|
f
(
x
)
|
:
x
∈
[
x
k
−
1
,
x
k
]
}
,
M
~
k
=
sup
{
|
f
(
x
)
|
:
x
∈
[
x
k
−
1
,
x
k
]
}
.
{\displaystyle {\tilde {m}}_{k}=\inf\{|f(x)|:x\in [x_{k-1},x_{k}]\},\ \ {\tilde {M}}_{k}=\sup\{|f(x)|:x\in [x_{k-1},x_{k}]\}.}
Mid-diżugwaljanza
|
f
(
x
)
|
−
|
f
(
y
)
|
≤
|
f
(
x
)
−
f
(
y
)
|
≤
M
k
−
m
k
{\displaystyle |f(x)|-|f(y)|\leq |f(x)-f(y)|\leq M_{k}-m_{k}}
għal kull
x
,
y
∈
[
x
k
−
1
,
x
k
]
{\displaystyle \displaystyle {x,y\in [x_{k-1},x_{k}]}}
M
~
k
−
m
~
k
≤
M
k
−
m
k
{\displaystyle {\tilde {M}}_{k}-{\tilde {m}}_{k}\leq M_{k}-m_{k}}
u allura
S
(
|
f
|
,
P
)
−
s
(
|
f
|
,
P
)
≤
S
(
f
,
P
)
−
s
(
f
,
P
)
.
{\displaystyle S(|f|,P)-s(|f|,P)\leq S(f,P)-s(f,P).}
Mela la
f
{\displaystyle \displaystyle {f}}
|
f
|
{\displaystyle \displaystyle {|f|}}
Id-diżugwaljanza bejn l-integrali, niksbuha mir-relazzjoni
±
f
(
x
)
≤
|
f
(
x
)
|
{\displaystyle \pm f(x)\leq |f(x)|}
x
∈
[
a
,
b
]
{\displaystyle x\in [a,b]}
Teorema :Teorema integrali tal-medja
Jekk
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\to \mathbb {R} \!}
kontinwa imbagħad teżisti
c
∈
[
a
,
b
]
{\displaystyle c\in [a,b]\!}
1
b
−
a
∫
a
b
f
(
x
)
d
x
=
f
(
c
)
.
{\displaystyle {{1} \over {b-a}}\int _{a}^{b}f(x)\,{\rm {d}}x=f(c).\!}
Prova:
La
f
{\displaystyle f\!}
[
a
,
b
]
{\displaystyle [a,b]\!}
teorema ta' Weierstrass għandha massimu
M
{\displaystyle M\!}
minimu
m
{\displaystyle m\!}
[
a
,
b
]
{\displaystyle [a,b]\!}
sup
x
∈
[
a
,
b
]
f
(
x
)
=
M
u
inf
x
∈
[
a
,
b
]
f
(
x
)
=
m
.
{\displaystyle \sup _{x\in [a,b]}f(x)=M{\mbox{ u }}\inf _{x\in [a,b]}f(x)=m.\!}
Mela
m
≤
f
(
x
)
≤
M
.
{\displaystyle m\leq f(x)\leq M\!.}
Mill-proprijetà tal-monotonija tal-integral jirriżulta li
m
(
b
−
a
)
=
∫
a
b
m
d
x
≤
∫
a
b
f
(
x
)
d
x
≤
∫
a
b
M
d
x
=
M
(
b
−
a
)
{\displaystyle m(b-a)=\int _{a}^{b}m\,{\rm {d}}x\leq \int _{a}^{b}f(x)\,{\rm {d}}x\leq \int _{a}^{b}M\,{\rm {d}}x=M(b-a)\!}
u allura
m
≤
1
b
−
a
∫
a
b
f
(
x
)
d
x
≤
M
.
{\displaystyle m\leq {{1} \over {b-a}}\int _{a}^{b}f(x)\,{\rm {d}}x\leq M.\!}
Issa mill-proprijetajiet tal-funzjonijiet kontinwi nafu li
f
{\displaystyle f\!}
[
a
,
b
]
{\displaystyle [a,b]\!}
kollha f'
[
m
,
M
]
{\displaystyle [m,M]\!}
c
∈
[
a
,
b
]
{\displaystyle c\in [a,b]\!}
f
(
c
)
=
1
b
−
a
∫
a
b
f
(
x
)
d
x
{\displaystyle f(c)={{1} \over {b-a}}\int _{a}^{b}f(x){\rm {d}}x}
Kalkulu differenzjali u kalkulu integrali
immodifika
F'din is-sezzjoni nagħtu ż-żewġ teoremi fundamentali tal-kalkulu integrali li jistabillixxu il-konnessjoni intima li teżisti bejn il-kalkulu differenzjali u l-kalkulu integrali. Dawn it-teoremi huma s-sissien tal-analisi integrali fis-sens li huma l-ħolqa li tgħaqqad il-kalkulu differenzjali mal-kalkulu integrali.
Teorema fundamentali tal-kalkulu integrali I
immodifika
Teorema :Teorema fundamentali tal-kalkulu integrali I
Prova: Nieħdu
c
∈
[
a
,
b
]
{\displaystyle c\in [a,b]}
ϵ
>
0
{\displaystyle \displaystyle {\epsilon >0}}
δ
>
0
{\displaystyle \displaystyle {\delta >0}}
|
f
(
t
)
−
f
(
c
)
|
<
ϵ
,
{\displaystyle \displaystyle {|f(t)-f(c)|<\epsilon ,}}
jekk
|
t
−
c
|
≤
δ
.
{\displaystyle |t-c|\leq \delta .}
f
(
c
)
=
1
δ
∫
c
c
+
δ
f
(
c
)
d
t
{\displaystyle f(c)={\frac {1}{\delta }}\int _{c}^{c+\delta }f(c){\rm {d}}t}
u li
F
(
c
+
δ
)
−
F
(
c
)
δ
=
1
δ
∫
c
c
+
δ
f
(
t
)
d
t
.
{\displaystyle {\frac {F(c+\delta )-F(c)}{\delta }}={\frac {1}{\delta }}\int _{c}^{c+\delta }f(t){\rm {d}}t.}
Allura għandna
F
(
c
+
δ
)
−
F
(
c
)
δ
−
f
(
c
)
=
1
δ
∫
c
c
+
δ
(
f
(
t
)
−
f
(
c
)
)
d
t
{\displaystyle {\frac {F(c+\delta )-F(c)}{\delta }}-f(c)={\frac {1}{\delta }}\int _{c}^{c+\delta }(f(t)-f(c)){\rm {d}}t}
u għalhekk
|
F
(
c
+
δ
)
−
F
(
c
)
δ
−
f
(
c
)
|
≤
1
δ
∫
c
c
+
δ
|
f
(
t
)
−
f
(
c
)
|
d
t
.
{\displaystyle \left|{\frac {F(c+\delta )-F(c)}{\delta }}-f(c)\right|\leq {\frac {1}{\delta }}\int _{c}^{c+\delta }|f(t)-f(c)|{\rm {d}}t.}
Mela
F
(
c
+
δ
)
−
F
(
c
)
δ
{\displaystyle {\frac {F(c+\delta )-F(c)}{\delta }}}
f
(
c
)
{\displaystyle \displaystyle {f(c)}}
δ
{\displaystyle \displaystyle {\delta }}
F
′
(
c
)
:=
lim
δ
→
0
F
(
c
+
δ
)
−
F
(
c
)
δ
=
f
(
c
)
.
{\displaystyle F'(c):=\lim _{\delta \to 0}{\frac {F(c+\delta )-F(c)}{\delta }}=f(c).}
Nota:
Fil-kalkulu differenzjali hemm dan il-kunċett tal-primittiva :
Funzjoni
F
{\displaystyle \ F}
[
a
,
b
]
{\displaystyle \ [a,b]}
primittiva ta'
f
{\displaystyle \ f}
[
a
,
b
]
{\displaystyle \ [a,b]}
F
′
(
x
)
=
f
(
x
)
{\displaystyle \ F'(x)=f(x)}
għal kull
x
∈
[
a
,
b
]
{\displaystyle x\in [a,b]}
Mela dan it-teorema jiggarantixxi l-eżistenza ta' primittiva.
Teorema fundamentali tal-kalkulu integrali II
immodifika
Teorema :Teorema fundamentali tal-kalkulu integrali II
Jekk
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\to \mathbb {R} }
derivabbli , u d-derivata
f
′
{\displaystyle f'}
integrabbli , imbagħad
∫
a
b
f
′
(
t
)
d
t
=
f
(
b
)
−
f
(
a
)
.
{\displaystyle \int _{a}^{b}f'(t){\rm {d}}t=f(b)-f(a).}
Prova: Ħalli
P
{\displaystyle \displaystyle {P}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}}
n
{\displaystyle \displaystyle {n}}
P
=
{
a
=
x
0
<
x
1
<
x
2
<
…
<
x
n
−
1
<
x
n
=
b
}
,
{\displaystyle P=\{a=x_{0}<x_{1}<x_{2}<\ldots <x_{n-1}<x_{n}=b\},}
u
m
~
k
=
inf
{
f
′
(
x
)
:
x
∈
[
x
k
−
1
,
x
k
]
}
,
M
~
k
=
sup
{
f
′
(
x
)
:
x
∈
[
x
k
−
1
,
x
k
]
}
.
{\displaystyle {\tilde {m}}_{k}=\inf\{f'(x):x\in [x_{k-1},x_{k}]\},\ \ {\tilde {M}}_{k}=\sup\{f'(x):x\in [x_{k-1},x_{k}]\}.}
Billi napplikaw it-teorema tal-valur medju għall kull intervall
[
x
k
−
1
,
x
k
]
{\displaystyle \displaystyle {[x_{k-1},x_{k}]}}
t
k
∈
(
x
k
−
1
,
x
k
)
{\displaystyle \displaystyle {t_{k}\in (x_{k-1},x_{k})}}
f
(
x
k
)
−
f
(
x
k
−
1
=
f
′
(
t
k
)
(
x
k
−
x
k
−
1
)
.
{\displaystyle \displaystyle {f(x_{k})-f(x_{k-1}=f'(t_{k})(x_{k}-x_{k-1}).}}
Mela għandna
f
(
b
)
−
f
(
a
)
=
∑
k
=
1
n
[
f
(
x
k
)
−
f
(
x
k
−
1
)
]
=
∑
k
=
1
n
f
′
(
t
k
)
(
x
k
−
x
k
−
1
)
.
{\displaystyle f(b)-f(a)=\sum _{k=1}^{n}[f(x_{k})-f(x_{k-1})]=\sum _{k=1}^{n}f'(t_{k})(x_{k}-x_{k-1}).}
La
m
~
k
≤
f
(
t
k
)
≤
M
~
k
{\displaystyle \displaystyle {{\tilde {m}}_{k}}\leq f(t_{k})\leq {\tilde {M}}_{k}}
k
{\displaystyle \displaystyle {k}}
s
(
f
′
,
P
)
≤
f
(
b
)
−
f
(
a
)
≤
S
(
f
′
,
P
)
.
{\displaystyle s(f',P)\leq f(b)-f(a)\leq S(f',P).}
La din hi valida għal kull partizzjoni
P
{\displaystyle \displaystyle {P}}
s
(
f
′
)
≤
f
(
b
)
−
f
(
a
)
≤
S
(
f
′
)
.
{\displaystyle s(f')\leq f(b)-f(a)\leq S(f').}
Imma qegħdin nassumu li
f
′
{\displaystyle \displaystyle {f'}}
[
a
,
b
]
{\displaystyle \displaystyle {[a,b]}}
s
(
f
′
)
=
S
(
f
′
)
=
∫
a
b
f
′
(
t
)
d
t
.
{\displaystyle s(f')=S(f')=\int _{a}^{b}f'(t){\rm {d}}t.}
Allura
∫
a
b
f
′
(
t
)
d
t
=
f
(
b
)
−
f
(
a
)
.
{\displaystyle \int _{a}^{b}f'(t){\rm {d}}t=f(b)-f(a).}
Ngħidu li l-funzjoni
f
{\displaystyle f}
[
a
,
∞
)
{\displaystyle [a,\infty )}
f
{\displaystyle f}
Hemm ukoll teorema li tiggarantixxi li funzjoni li hi assolutament integrabbli hi integrabbli, fuq l-intervall tat-tip
[
a
,
∞
]
{\displaystyle [a,\infty ]}
Teorema :Teorema tal-integrabbiltà assoluta
Jekk
f
{\displaystyle f}
Prova:
Bit-teorema fuq l-eżistenza ta'l-integrali nafu li l-kondizzjoni neċessarja u suffiċjenti biex
∫
a
∞
f
(
x
)
d
x
{\displaystyle \int _{a}^{\infty }f(x){\rm {d}}x}
∀
ϵ
>
0
∃
γ
>
0
:
∀
x
1
,
x
2
<
γ
|
∫
x
1
x
2
f
(
x
)
d
x
|
<
ϵ
.
{\displaystyle \forall \epsilon >\ 0\quad \exists \gamma >\ 0:\quad \forall x_{1},x_{2}<\ \gamma \quad \left|\int _{x_{1}}^{x_{2}}f(x){\rm {d}}x\right|<\ \epsilon .}
|
f
|
{\displaystyle \ |f|}
|
f
(
x
)
|
{\displaystyle \ |f(x)|}
f
(
x
)
{\displaystyle \ f(x)}
∀
ϵ
>
0
∃
γ
>
0
:
∀
x
1
,
x
2
<
γ
|
∫
x
1
x
2
|
f
(
x
)
|
d
x
|
<
ϵ
.
{\displaystyle \forall \epsilon >\ 0\quad \exists \gamma >\ 0:\quad \forall x_{1},x_{2}<\ \gamma \quad \left|\int _{x_{1}}^{x_{2}}\left|f(x)\right|{\rm {d}}x\right|<\ \epsilon .}
Imma mill-proprietà tal-valur assolut għall-integrali għandna
|
∫
a
b
f
(
x
)
d
x
|
≤
∫
a
b
|
f
(
x
)
|
d
x
.
{\displaystyle \left|\int _{a}^{b}f(x){\rm {d}}x\right|\leq \int _{a}^{b}\left|f(x)\right|{\rm {d}}x.}
U mela nistgħu niktbu
∀
ϵ
>
0
∃
γ
>
0
:
∀
x
1
,
x
2
<
γ
∫
x
1
x
2
|
f
(
x
)
d
x
|
<
ϵ
.
{\displaystyle \forall \epsilon >\ 0\quad \exists \gamma >\ 0:\quad \forall x_{1},x_{2}<\ \gamma \quad \int _{x_{1}}^{x_{2}}\left|f(x){\rm {d}}x\right|<\ \epsilon .}
Mill-liema niksbu li
f
{\displaystyle \ f}
Hemm bżonn noqgħodu attenti li ma nħalltux dan it-teorema mal-kuntrarju tiegħu, li hu falz għax mhux il-funzjonijiet integrabbli kollha huma assolutament integrabbli. Eżempju ta' dan hi funzjoni ta' dan it-tip
sin
x
x
.
{\displaystyle {{\sin x} \over {x}}.}
![{\displaystyle \ [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f4c40bbeaa2f59e60b6259cebe2479bc24396f0)
![{\displaystyle \ [x_{s-1},x_{s}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5648acbd542dcfe8c80fbd0336db6474b0cb27b7)
![{\displaystyle m_{k}=\inf\{f(x):x\in [x_{k-1},x_{k}]\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/742fad88161c03d77ad57fd3864680a2e15b3278)
![{\displaystyle M_{k}=\sup\{f(x):x\in [x_{k-1},x_{k}]\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c89e7526eaa202832960a4acb1117c6a3f12be12)
![{\displaystyle m\leq f(x)\leq M,\ \forall x\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48c298893f4420bdd4210355dbe9cc851404c6ad)
![{\displaystyle \displaystyle {[a,b]}\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6efa92b98f335ec1a4fc15c1aed43c8cb6a6da7c)
![{\displaystyle \displaystyle {[a,b]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/174b2f0a217f666d656eaf32e3c612c773bac788)
![{\displaystyle \displaystyle {[a,b]}\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b39c1c43f1c7e7c1bb70b19d115fab4c3241fea)
![{\displaystyle \{(x,y)\ |\ 0\leq y\leq f(x),\ x\in [a,b]\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/377926527eeb7ad7e21aa2e13e61a4b012baf0fd)
![{\displaystyle \displaystyle {\ [a,b]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab38286c329e3ef7a5aa887d45c7160b7bb15dc9)
![{\displaystyle \displaystyle {\ [0,1]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/faf5da857d060bfea97ea8efaf11dfccce8fed08)
![{\displaystyle \displaystyle {M_{k}=\sup\{x^{2}\,|\,x\in [(k-1)/n,k/n]\}=(k/n)^{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3cfc3034d3f86c467390966c474d2ad404f1764)
![{\displaystyle \displaystyle {m_{k}=\inf\{x^{2}\,|\,x\in [(k-1)/n,k/n]\}=((k-1)/n)^{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3cf5a9c2324a617395daf628b5643fb03704cf97)
![{\displaystyle \displaystyle {g:[0,1]\mapsto \mathbb {R} }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f35010227515726b9261c89a76cd11efb47626ef)
![{\displaystyle \displaystyle {[0,1]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4275819f2051e4887b5ebe6318e68ed28ce05343)
![{\displaystyle \displaystyle {[x_{k-1},x_{k}]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9bc76575a41fb4529083c2ff2bece260cd9811cd)
![{\displaystyle \displaystyle {[a.b]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b3048c1bb6b432e949ceccb3fe63b0148ae0fd3)
![{\displaystyle \ f:[a,b]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/37ed514b298741a6018df22ac55bd1ba97657f24)
![{\displaystyle \ [x_{k-1},x_{k}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0315b27c062dbe87b7badba9114861162a540896)
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
![{\displaystyle \ \sup _{x\in D}[f(x)+g(x)]\leq \sup _{x\in D}f(x)+\sup _{x\in D}g(x)\ \ \ {\rm {u}}\ \ \ \inf _{x\in D}([f(x)+g(x)])\geq \inf _{x\in D}f(x)+\inf _{x\in D}g(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93167dd45a19472c690a432572cd9fa8fa43cdaa)
![{\displaystyle \displaystyle {[a,c]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72775f56956c2471585ce0489dc81c10cb223410)
![{\displaystyle \displaystyle {[c,b]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22bbc681078d068cb5e7be60cb1878ae1e918513)
![{\displaystyle \ x\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81c2ee6252d24b5f0a958ad066e393f803d7b427)
![{\displaystyle {\tilde {m}}_{k}=\inf\{|f(x)|:x\in [x_{k-1},x_{k}]\},\ \ {\tilde {M}}_{k}=\sup\{|f(x)|:x\in [x_{k-1},x_{k}]\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1cda537375545c1156d16805a0c17716f140136e)
![{\displaystyle \displaystyle {x,y\in [x_{k-1},x_{k}]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e96cf1b9464826404e05bc185627380b6c186f1)
![{\displaystyle x\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/026357b404ee584c475579fb2302a4e9881b8cce)
![{\displaystyle f:[a,b]\to \mathbb {R} \!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc6748327df1f5ecfad143a450b0868aae8f6ec0)
![{\displaystyle c\in [a,b]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c739b1276f3bba74847b56aa59c8b80e507e858)
![{\displaystyle [a,b]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6db8e01502a6bed98f89b7bb2127c620f288f605)
![{\displaystyle \sup _{x\in [a,b]}f(x)=M{\mbox{ u }}\inf _{x\in [a,b]}f(x)=m.\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d06e30cd1dafde1fa2badcdde553ab09869a6512)
![{\displaystyle [m,M]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbce868c92020fdefccee6b2fea25fbcdfffecca)
![{\displaystyle f:[a,b]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b592d102ccd1ba134d401c5b3ea177baaba3ffac)
![{\displaystyle F:[a,b]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3cacd74b0bb95ed9ddf45676b1ad0f3d618c4cdc)
![{\displaystyle c\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/997256364b06acf0710e5d24da39e8c42991a249)
![{\displaystyle {\tilde {m}}_{k}=\inf\{f'(x):x\in [x_{k-1},x_{k}]\},\ \ {\tilde {M}}_{k}=\sup\{f'(x):x\in [x_{k-1},x_{k}]\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f99ddbbcad1a7c58b54eb81b4f6c46259531efdd)
![{\displaystyle f(b)-f(a)=\sum _{k=1}^{n}[f(x_{k})-f(x_{k-1})]=\sum _{k=1}^{n}f'(t_{k})(x_{k}-x_{k-1}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20c0a63b75235dc5533d20dbe31caec72b6f38df)
![{\displaystyle [a,\infty ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6eb9b9f42399eda1ab1854611f5cc51cb2f2ea9b)
