Keli svarbūs sumų ir integralų sąryšiai
Tarkime, kad A ir B yra bet kokie neneigiami skaičiai, o
p
{\displaystyle p}
p
′
{\displaystyle p'}
1
p
+
1
p
′
=
1
{\displaystyle {\frac {1}{p}}+{\frac {1}{p'}}=1\;}
A
B
≤
A
p
p
+
B
p
′
p
′
.
(
10.26
)
{\displaystyle AB\leq {\frac {A^{p}}{p}}+{\frac {B^{p'}}{p'}}.\quad (10.26)}
Rasime funkcijos
f
(
x
)
=
x
1
p
−
x
p
{\displaystyle f(x)=x^{1 \over p}-{\frac {x}{p}}\;}
x
≥
0.
{\displaystyle x\geq 0.}
f
′
(
x
)
=
1
p
(
x
1
p
−
1
−
1
)
=
1
p
(
x
−
1
p
′
−
1
)
,
{\displaystyle f'(x)={\frac {1}{p}}\left(x^{{1 \over p}-1}-1\right)={\frac {1}{p}}\left(x^{-{1 \over p'}}-1\right),}
f
′
(
x
)
>
0
,
{\displaystyle f'(x)>0,}
0
<
x
<
1
,
{\displaystyle 0<x<1,}
f
′
(
x
)
<
0
,
{\displaystyle f'(x)<0,}
x >1. Todėl funkcija turi maksimumą taške
x
=
1
,
{\displaystyle x=1,}
f
(
1
)
=
1
1
p
−
1
p
=
1
−
1
p
=
1
p
′
.
{\displaystyle f(1)=1^{1 \over p}-{\frac {1}{p}}=1-{\frac {1}{p}}={\frac {1}{p'}}.}
Taigi su visais
x
≥
0
{\displaystyle x\geq 0}
x
1
p
−
x
p
≤
1
p
′
.
{\displaystyle x^{1 \over p}-{\frac {x}{p}}\leq {\frac {1}{p'}}.}
Paėmę paskutinėje nelygybėje
x
=
A
p
B
p
′
{\displaystyle x={\frac {A^{p}}{B^{p'}}}}
B
p
′
,
{\displaystyle B^{p'},}
Tai atliekama šitaip:
(
A
p
B
p
′
)
1
p
−
A
p
B
p
′
⋅
1
p
≤
1
p
′
,
{\displaystyle \left({\frac {A^{p}}{B^{p'}}}\right)^{1 \over p}-{\frac {A^{p}}{B^{p'}}}\cdot {\frac {1}{p}}\leq {\frac {1}{p'}},}
A
B
p
′
p
−
A
p
B
p
′
⋅
1
p
≤
1
p
′
;
{\displaystyle {\frac {A}{B^{p' \over p}}}-{\frac {A^{p}}{B^{p'}}}\cdot {\frac {1}{p}}\leq {\frac {1}{p'}};}
1
−
1
p
=
1
p
′
,
{\displaystyle 1-{\frac {1}{p}}={\frac {1}{p'}},}
p
−
1
p
=
1
p
′
,
{\displaystyle {\frac {p-1}{p}}={\frac {1}{p'}},}
p
′
=
p
p
−
1
;
{\displaystyle p'={\frac {p}{p-1}};}
A
B
p
′
p
⋅
B
p
′
−
A
p
B
p
′
⋅
B
p
′
⋅
1
p
≤
1
p
′
⋅
B
p
′
,
{\displaystyle {\frac {A}{B^{p' \over p}}}\cdot B^{p'}-{\frac {A^{p}}{B^{p'}}}\cdot B^{p'}\cdot {\frac {1}{p}}\leq {\frac {1}{p'}}\cdot B^{p'},}
A
⋅
B
−
p
′
p
⋅
B
p
′
−
A
p
⋅
1
p
≤
1
p
′
⋅
B
p
′
,
{\displaystyle A\cdot B^{-{\frac {p'}{p}}}\cdot B^{p'}-A^{p}\cdot {\frac {1}{p}}\leq {\frac {1}{p'}}\cdot B^{p'},}
A
⋅
B
−
1
p
⋅
p
p
−
1
⋅
B
p
p
−
1
−
A
p
⋅
1
p
≤
1
p
′
⋅
B
p
′
,
{\displaystyle A\cdot B^{-{\frac {1}{p}}\cdot {\frac {p}{p-1}}}\cdot B^{\frac {p}{p-1}}-A^{p}\cdot {\frac {1}{p}}\leq {\frac {1}{p'}}\cdot B^{p'},}
A
⋅
B
−
1
p
−
1
⋅
B
p
p
−
1
−
A
p
⋅
1
p
≤
1
p
′
⋅
B
p
′
,
{\displaystyle A\cdot B^{-{\frac {1}{p-1}}}\cdot B^{\frac {p}{p-1}}-A^{p}\cdot {\frac {1}{p}}\leq {\frac {1}{p'}}\cdot B^{p'},}
A
⋅
B
p
p
−
1
−
1
p
−
1
−
A
p
⋅
1
p
≤
1
p
′
⋅
B
p
′
,
{\displaystyle A\cdot B^{{\frac {p}{p-1}}-{\frac {1}{p-1}}}-A^{p}\cdot {\frac {1}{p}}\leq {\frac {1}{p'}}\cdot B^{p'},}
A
⋅
B
p
−
1
p
−
1
−
A
p
⋅
1
p
≤
1
p
′
⋅
B
p
′
,
{\displaystyle A\cdot B^{\frac {p-1}{p-1}}-A^{p}\cdot {\frac {1}{p}}\leq {\frac {1}{p'}}\cdot B^{p'},}
A
⋅
B
−
A
p
⋅
1
p
≤
1
p
′
⋅
B
p
′
,
{\displaystyle A\cdot B-A^{p}\cdot {\frac {1}{p}}\leq {\frac {1}{p'}}\cdot B^{p'},}
A
⋅
B
≤
A
p
⋅
1
p
+
1
p
′
⋅
B
p
′
,
{\displaystyle A\cdot B\leq A^{p}\cdot {\frac {1}{p}}+{\frac {1}{p'}}\cdot B^{p'},}
A
⋅
B
≤
A
p
p
+
B
p
′
p
′
.
{\displaystyle A\cdot B\leq {\frac {A^{p}}{p}}+{\frac {B^{p'}}{p'}}.}
_______________________
* Čia laikome
B
>
0.
{\displaystyle B>0.}
B
=
0
,
{\displaystyle B=0,\;}
keisti
* Helderis (1859-1937) – vokiečių matematikas.https://en.wikipedia.org/wiki/Hölder%27s_inequality#Notable_special_cases Tarkime, kad
a
1
,
a
2
,
.
.
.
,
a
n
{\displaystyle a_{1},\;a_{2},\;...,\;a_{n}}
b
1
,
b
2
,
.
.
.
,
b
n
{\displaystyle b_{1},\;b_{2},\;...,\;b_{n}}
p ir
p
′
{\displaystyle p'}
∑
i
=
1
n
a
i
b
i
≤
[
∑
i
=
1
n
a
i
p
]
1
p
[
∑
i
=
1
n
b
i
p
′
]
1
p
′
,
(
10.27
)
{\displaystyle \sum _{i=1}^{n}a_{i}b_{i}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}\left[\sum _{i=1}^{n}b_{i}^{p'}\right]^{1 \over p'},\quad (10.27)}
kuri vadinama Helderio nelygybe sumoms . Iš pradžių įrodysime štai ką: jeigu
A
1
,
A
2
,
.
.
.
,
A
n
;
{\displaystyle A_{1},\;A_{2},\;...,\;A_{n};\;}
B
1
,
B
2
,
.
.
.
,
B
n
{\displaystyle B_{1},\;B_{2},\;...,\;B_{n}}
∑
i
=
1
n
A
i
p
≤
1
ir
∑
i
=
1
n
B
i
p
′
≤
1
,
(
10.28
)
{\displaystyle \sum _{i=1}^{n}A_{i}^{p}\leq 1\;\;{\text{ir}}\;\;\sum _{i=1}^{n}B_{i}^{p'}\leq 1,\quad (10.28)}
tai su tais skaičiais teisinga nelygybė
∑
i
=
1
n
A
i
B
i
≤
1.
(
10.29
)
{\displaystyle \sum _{i=1}^{n}A_{i}B_{i}\leq 1.\quad (10.29)}
Iš tikrųjų, parašę visoms skaičių
A
i
{\displaystyle A_{i}}
B
i
{\displaystyle B_{i}}
i nuo 1 iki n , gausime
∑
i
=
1
n
A
i
B
i
≤
1
p
∑
i
=
1
n
A
i
p
+
1
p
′
∑
i
=
1
n
B
i
p
′
≤
1
p
+
1
p
′
=
1.
{\displaystyle \sum _{i=1}^{n}A_{i}B_{i}\leq {\frac {1}{p}}\sum _{i=1}^{n}A_{i}^{p}+{\frac {1}{p'}}\sum _{i=1}^{n}B_{i}^{p'}\leq {\frac {1}{p}}+{\frac {1}{p'}}=1.}
Taigi (10.29) nelygybė įrodyta.
Imkime dabar
A
i
=
a
i
[
∑
i
=
1
n
a
i
p
]
1
p
,
B
i
=
b
i
[
∑
i
=
1
n
b
i
p
′
]
1
p
′
.
{\displaystyle A_{i}={\frac {a_{i}}{[\sum _{i=1}^{n}a_{i}^{p}]^{1 \over p}}}\;,\quad B_{i}={\frac {b_{i}}{[\sum _{i=1}^{n}b_{i}^{p'}]^{1 \over p'}}}.}
∑
i
=
1
n
A
i
p
=
∑
i
=
1
n
(
a
i
[
∑
i
=
1
n
a
i
p
]
1
p
)
p
=
∑
i
=
1
n
a
i
p
[
∑
i
=
1
n
a
i
p
]
p
p
=
∑
i
=
1
n
a
i
p
[
∑
i
=
1
n
a
i
p
]
=
∑
i
=
1
n
a
i
p
[
∑
i
=
1
n
a
i
p
]
=
1.
(
Paraboloido
)
{\displaystyle \sum _{i=1}^{n}A_{i}^{p}=\sum _{i=1}^{n}\left({\frac {a_{i}}{[\sum _{i=1}^{n}a_{i}^{p}]^{1 \over p}}}\right)^{p}=\sum _{i=1}^{n}{\frac {a_{i}^{p}}{[\sum _{i=1}^{n}a_{i}^{p}]^{p \over p}}}=\sum _{i=1}^{n}{\frac {a_{i}^{p}}{[\sum _{i=1}^{n}a_{i}^{p}]}}={\frac {\sum _{i=1}^{n}a_{i}^{p}}{[\sum _{i=1}^{n}a_{i}^{p}]}}=1.\quad ({\text{Paraboloido}})}
Nesunku įsitikinti, kad skaičiai
A
i
{\displaystyle A_{i}}
B
i
{\displaystyle B_{i}}
∑
i
=
1
n
A
i
B
i
=
∑
i
=
1
n
a
i
b
i
[
∑
i
=
1
n
a
i
p
]
1
p
[
∑
i
=
1
n
b
i
p
′
]
1
p
′
≤
1.
{\displaystyle \sum _{i=1}^{n}A_{i}B_{i}={\frac {\sum _{i=1}^{n}a_{i}b_{i}}{[\sum _{i=1}^{n}a_{i}^{p}]^{1 \over p}[\sum _{i=1}^{n}b_{i}^{p'}]^{1 \over p'}}}\leq 1.}
Iš paskutinės nelygybės išplaukia Helderio (10.27) nelygybė.
O iš (Paraboloido) lygybės išplaukia
∑
i
=
1
n
A
i
B
i
≤
1
p
∑
i
=
1
n
A
i
p
+
1
p
′
∑
i
=
1
n
B
i
p
′
≤
1
p
+
1
p
′
=
1.
{\displaystyle \sum _{i=1}^{n}A_{i}B_{i}\leq {\frac {1}{p}}\sum _{i=1}^{n}A_{i}^{p}+{\frac {1}{p'}}\sum _{i=1}^{n}B_{i}^{p'}\leq {\frac {1}{p}}+{\frac {1}{p'}}=1.}
Bet atsižvelgus, kad
∑
i
=
1
n
A
i
p
=
1
{\displaystyle \sum _{i=1}^{n}A_{i}^{p}=1\;}
∑
i
=
1
n
B
i
p
′
=
1
,
{\displaystyle \;\sum _{i=1}^{n}B_{i}^{p'}=1,}
∑
i
=
1
n
A
i
B
i
≤
1
p
⋅
1
+
1
p
′
⋅
1
=
1
p
+
1
p
′
=
1.
{\displaystyle \sum _{i=1}^{n}A_{i}B_{i}\leq {\frac {1}{p}}\cdot 1+{\frac {1}{p'}}\cdot 1={\frac {1}{p}}+{\frac {1}{p'}}=1.}
Taigi, vis tiek gauname, kad (10.27) nelygybė teisinga, nes
∑
i
=
1
n
A
i
B
i
=
∑
i
=
1
n
a
i
b
i
[
∑
i
=
1
n
a
i
p
]
1
p
[
∑
i
=
1
n
b
i
p
′
]
1
p
′
≤
1.
{\displaystyle \sum _{i=1}^{n}A_{i}B_{i}={\frac {\sum _{i=1}^{n}a_{i}b_{i}}{[\sum _{i=1}^{n}a_{i}^{p}]^{1 \over p}[\sum _{i=1}^{n}b_{i}^{p'}]^{1 \over p'}}}\leq 1.}
Pastaba. Atskiru atveju, kai p=p'=2, Helderio nelygybė virsta šitokia:
∑
i
=
1
n
a
i
b
i
≤
∑
i
=
1
n
a
i
2
∑
i
=
1
n
b
i
2
.
(
10.30
)
{\displaystyle \sum _{i=1}^{n}a_{i}b_{i}\leq {\sqrt {\sum _{i=1}^{n}a_{i}^{2}}}{\sqrt {\sum _{i=1}^{n}b_{i}^{2}}}.\quad (10.30)}
(10.30) nelygybė vadinama Buniakovskio ** nelygybe sumoms . _______________________
* Laikome, kad bent vienas iš skaičių
a
i
{\displaystyle a_{i}}
b
i
{\displaystyle b_{i}}
** V. Buniakovskis (1804-1889) – rusų matematikas.
keisti
* H. Minkovskis (1864-1909) – vokiečių matematikas ir fizikas.Tarkime, kad
a
1
,
a
2
,
.
.
.
,
a
n
;
{\displaystyle a_{1},\;a_{2},\;...,\;a_{n};\;}
b
1
,
b
2
,
.
.
.
,
b
n
{\displaystyle b_{1},\;b_{2},\;...,\;b_{n}}
p >1. Tada teisinga nelygybė
[
∑
i
=
1
n
(
a
i
+
b
i
)
p
]
1
p
≤
[
∑
i
=
1
n
a
i
p
]
1
p
+
[
∑
i
=
1
n
b
i
p
]
1
p
,
(
10.31
)
{\displaystyle \left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{1 \over p}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p},\quad (10.31)}
vadinama Minkovskio nelygybe sumoms . Visų pirma pertvarkykime sumą, esančią (10.31) nelygybės kairėje pusėje. Galima užrašyti
∑
i
=
1
n
(
a
i
+
b
i
)
p
=
∑
i
=
1
n
a
i
(
a
i
+
b
i
)
p
−
1
+
∑
i
=
1
n
b
i
(
a
i
+
b
i
)
p
−
1
.
{\displaystyle \sum _{i=1}^{n}(a_{i}+b_{i})^{p}=\sum _{i=1}^{n}a_{i}(a_{i}+b_{i})^{p-1}+\sum _{i=1}^{n}b_{i}(a_{i}+b_{i})^{p-1}.}
[
(
a
i
+
b
i
)
3
=
a
i
3
+
3
a
i
2
b
i
+
3
a
i
b
i
2
+
b
i
3
.
{\displaystyle (a_{i}+b_{i})^{3}=a_{i}^{3}+3a_{i}^{2}b_{i}+3a_{i}b_{i}^{2}+b_{i}^{3}.}
a
i
(
a
i
+
b
i
)
3
−
1
+
b
i
(
a
i
+
b
i
)
3
−
1
=
a
i
(
a
i
2
+
2
a
i
b
i
+
b
i
2
)
+
b
i
(
a
i
2
+
2
a
i
b
i
+
b
i
2
)
=
[
a
i
3
+
2
a
i
2
b
i
+
a
i
b
i
2
]
+
[
a
i
2
b
i
+
2
a
i
b
i
2
+
b
i
3
]
=
{\displaystyle a_{i}(a_{i}+b_{i})^{3-1}+b_{i}(a_{i}+b_{i})^{3-1}=a_{i}(a_{i}^{2}+2a_{i}b_{i}+b_{i}^{2})+b_{i}(a_{i}^{2}+2a_{i}b_{i}+b_{i}^{2})=[a_{i}^{3}+2a_{i}^{2}b_{i}+a_{i}b_{i}^{2}]+[a_{i}^{2}b_{i}+2a_{i}b_{i}^{2}+b_{i}^{3}]=}
=
a
i
3
+
3
a
i
2
b
i
+
3
a
i
b
i
2
+
b
i
3
.
{\displaystyle =a_{i}^{3}+3a_{i}^{2}b_{i}+3a_{i}b_{i}^{2}+b_{i}^{3}.}
Kiekvienai iš dešinės pusės sumų taikysime Helderio nelygybę. Kadangi
(
p
−
1
)
p
′
=
p
{\displaystyle (p-1)p'=p}
1
p
′
=
p
−
1
p
,
{\displaystyle {\frac {1}{p'}}={\frac {p-1}{p}},}
∑
i
=
1
n
(
a
i
+
b
i
)
p
≤
[
∑
i
=
1
n
a
i
p
]
1
p
[
∑
i
=
1
n
(
a
i
+
b
i
)
(
p
−
1
)
p
′
]
1
p
′
+
[
∑
i
=
1
n
b
i
p
]
1
p
[
∑
i
=
1
n
(
a
i
+
b
i
)
(
p
−
1
)
p
′
]
1
p
′
=
{\displaystyle \sum _{i=1}^{n}(a_{i}+b_{i})^{p}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}\left[\sum _{i=1}^{n}(a_{i}+b_{i})^{(p-1)p'}\right]^{1 \over p'}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p}\left[\sum _{i=1}^{n}(a_{i}+b_{i})^{(p-1)p'}\right]^{1 \over p'}=}
=
(
[
∑
i
=
1
n
a
i
p
]
1
p
+
[
∑
i
=
1
n
b
i
p
]
1
p
)
[
∑
i
=
1
n
(
a
i
+
b
i
)
p
]
p
−
1
p
.
{\displaystyle =\left(\left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p}\right)\left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{p-1 \over p}.}
Padaliję paskutinės nelygybės abi puses iš
[
∑
i
=
1
n
(
a
i
+
b
i
)
p
]
p
−
1
p
,
{\displaystyle \left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{p-1 \over p},}
∑
i
=
1
n
(
a
i
+
b
i
)
p
≤
(
[
∑
i
=
1
n
a
i
p
]
1
p
+
[
∑
i
=
1
n
b
i
p
]
1
p
)
[
∑
i
=
1
n
(
a
i
+
b
i
)
p
]
p
−
1
p
,
{\displaystyle \sum _{i=1}^{n}(a_{i}+b_{i})^{p}\leq \left(\left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p}\right)\left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{p-1 \over p},}
[
∑
i
=
1
n
(
a
i
+
b
i
)
p
]
[
∑
i
=
1
n
(
a
i
+
b
i
)
p
]
−
p
−
1
p
≤
[
∑
i
=
1
n
a
i
p
]
1
p
+
[
∑
i
=
1
n
b
i
p
]
1
p
,
{\displaystyle \left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]\left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{-{\frac {p-1}{p}}}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p},}
[
∑
i
=
1
n
(
a
i
+
b
i
)
p
]
1
−
p
−
1
p
≤
[
∑
i
=
1
n
a
i
p
]
1
p
+
[
∑
i
=
1
n
b
i
p
]
1
p
,
{\displaystyle \left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{1-{\frac {p-1}{p}}}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p},}
[
∑
i
=
1
n
(
a
i
+
b
i
)
p
]
p
−
(
p
−
1
)
p
≤
[
∑
i
=
1
n
a
i
p
]
1
p
+
[
∑
i
=
1
n
b
i
p
]
1
p
,
{\displaystyle \left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{\frac {p-(p-1)}{p}}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p},}
[
∑
i
=
1
n
(
a
i
+
b
i
)
p
]
1
p
≤
[
∑
i
=
1
n
a
i
p
]
1
p
+
[
∑
i
=
1
n
b
i
p
]
1
p
.
{\displaystyle \left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{\frac {1}{p}}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p}.}
4. Integruojamos funkcijos modulio bet kurio teigiamo laipsnio integruojamumas.
keisti
Įrodysime šitokią teoremą.
10.7 teorema. Jei funkcija
f
(
x
)
{\displaystyle f(x)}
[a; b], tai funkcija
|
f
(
x
)
|
r
,
{\displaystyle |f(x)|^{r},}
[a; b].Įrodymas. Teoremą užtenka įrodyti, kai
r
<
1.
{\displaystyle r<1.}
r
>
1
,
{\displaystyle r>1,}
|
f
(
x
)
|
r
,
{\displaystyle |f(x)|^{r},}
|
f
(
x
)
|
r
|
f
(
x
)
|
r
−
[
r
]
;
{\displaystyle |f(x)|^{r}|f(x)|^{r-[r]};}
|
f
(
x
)
|
{\displaystyle |f(x)|}
3
∘
{\displaystyle 3^{\circ }}
|
f
(
x
)
|
[
r
]
{\displaystyle |f(x)|^{[r]}}
|
f
(
x
)
|
r
−
[
r
]
{\displaystyle |f(x)|^{r-[r]}}
|
f
(
x
)
|
r
{\displaystyle |f(x)|^{r}}
r
=
1
p
{\displaystyle r={\frac {1}{p}}}
ϵ
>
0
{\displaystyle \epsilon >0}
T , kad
∑
i
=
1
n
(
M
i
−
m
i
)
Δ
x
i
<
ε
p
(
b
−
a
)
1
−
p
;
(
10.32
)
{\displaystyle \sum _{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}<\varepsilon ^{p}(b-a)^{1-p};\quad (10.32)}
∑
i
=
1
n
(
M
i
−
m
i
)
Δ
x
i
<
ε
p
(
b
−
a
)
p
−
1
<
ε
p
;
(
10.32
Paraboloido
)
{\displaystyle \sum _{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}<\varepsilon ^{p}(b-a)^{p-1}<\varepsilon ^{p};\quad (10.32\;\;{\text{Paraboloido}})\;}
čia
M
i
{\displaystyle M_{i}}
m
i
{\displaystyle m_{i}}
x
i
−
1
;
x
i
{\displaystyle x_{i-1};\;x_{i}}
S
−
s
=
∑
i
=
1
n
(
M
i
1
/
p
−
m
i
1
/
p
)
Δ
x
i
(
10.33
)
{\displaystyle S-s=\sum _{i=1}^{n}(M_{i}^{1/p}-m_{i}^{1/p})\Delta x_{i}\quad (10.33)}
yra mažesnė už
ε
.
{\displaystyle \varepsilon .}
Įvertinsime šią sumą, remdamiesi Helderio (10.27) nelygybe. Paėmę joje
a
i
=
(
M
i
1
/
p
−
m
i
1
/
p
)
(
Δ
x
i
)
1
/
p
,
{\displaystyle a_{i}=(M_{i}^{1/p}-m_{i}^{1/p})(\Delta x_{i})^{1/p},\;\;}
b
i
=
(
Δ
x
i
)
1
/
p
′
,
{\displaystyle b_{i}=(\Delta x_{i})^{1/p'},}
S
−
s
≤
[
∑
i
=
1
n
(
M
i
1
/
p
−
m
i
1
/
p
)
p
Δ
x
i
]
1
/
p
[
∑
i
=
1
n
Δ
x
i
]
1
/
p
′
.
(
10.34
)
{\displaystyle S-s\leq \left[\sum _{i=1}^{n}(M_{i}^{1/p}-m_{i}^{1/p})^{p}\Delta x_{i}\right]^{1/p}\left[\sum _{i=1}^{n}\Delta x_{i}\right]^{1/p'}.\quad (10.34)}
[
a
i
⋅
b
i
=
(
M
i
1
/
p
−
m
i
1
/
p
)
(
Δ
x
i
)
1
/
p
⋅
(
Δ
x
i
)
1
/
p
′
=
(
M
i
1
/
p
−
m
i
1
/
p
)
(
Δ
x
i
)
1
/
p
+
1
/
p
′
=
(
M
i
1
/
p
−
m
i
1
/
p
)
Δ
x
i
,
{\displaystyle a_{i}\cdot b_{i}=(M_{i}^{1/p}-m_{i}^{1/p})(\Delta x_{i})^{1/p}\cdot (\Delta x_{i})^{1/p'}=(M_{i}^{1/p}-m_{i}^{1/p})(\Delta x_{i})^{1/p+1/p'}=(M_{i}^{1/p}-m_{i}^{1/p})\Delta x_{i},}
Dabar įrodysime, kad
(
M
i
1
/
p
−
m
i
1
/
p
)
p
≤
(
M
i
−
m
i
)
.
(
10.35
)
{\displaystyle (M_{i}^{1/p}-m_{i}^{1/p})^{p}\leq (M_{i}-m_{i}).\quad (10.35)}
Paskutinę nelygybę padaliję iš
M
i
{\displaystyle M_{i}}
[
1
−
(
m
i
M
i
)
1
/
p
]
p
≤
1
−
m
i
M
i
.
{\displaystyle \left[1-\left({\frac {m_{i}}{M_{i}}}\right)^{1/p}\right]^{p}\leq 1-{\frac {m_{i}}{M_{i}}}.}
Jos teisingumu lengva įsitikinti, atsižvelgus į tai, kad
0
≤
m
i
M
i
≤
1
,
{\displaystyle 0\leq {\frac {m_{i}}{M_{i}}}\leq 1,}
∑
i
=
1
n
Δ
x
i
=
b
−
a
,
{\displaystyle \sum _{i=1}^{n}\Delta x_{i}=b-a,}
iš (10.34) nelygybės gauname
S
−
s
≤
[
∑
i
=
1
n
(
M
i
−
m
i
)
Δ
x
i
]
1
/
p
(
b
−
a
)
1
/
p
′
.
{\displaystyle S-s\leq \left[\sum _{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}\right]^{1/p}(b-a)^{1/p'}.}
Iš čia, pasirėmę (10.32) nelygybe ir prisiminę, kad
1
p
+
1
p
′
=
1
,
{\displaystyle {\frac {1}{p}}+{\frac {1}{p'}}=1,}
S
−
s
<
ε
.
(
10.35.1
)
{\displaystyle S-s<\varepsilon .\quad (10.35.1)}
Teorema įrodyta. (10.35.1) formulė gaunama sekančiu budu.
1
p
′
=
1
−
1
p
=
p
−
1
p
.
{\displaystyle {\frac {1}{p'}}=1-{\frac {1}{p}}={\frac {p-1}{p}}.}
S
−
s
≤
[
∑
i
=
1
n
(
M
i
−
m
i
)
Δ
x
i
]
1
/
p
(
b
−
a
)
1
/
p
′
.
{\displaystyle S-s\leq \left[\sum _{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}\right]^{1/p}(b-a)^{1/p'}.}
S
−
s
≤
[
∑
i
=
1
n
(
M
i
−
m
i
)
Δ
x
i
]
1
/
p
(
b
−
a
)
p
−
1
p
.
{\displaystyle S-s\leq \left[\sum _{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}\right]^{1/p}(b-a)^{\frac {p-1}{p}}.}
Pakeliame abi nelygybės puses p laipsniu.
(
S
−
s
)
p
≤
[
∑
i
=
1
n
(
M
i
−
m
i
)
Δ
x
i
]
(
b
−
a
)
p
−
1
.
{\displaystyle (S-s)^{p}\leq \left[\sum _{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}\right](b-a)^{p-1}.}
Prisimename formulę
∑
i
=
1
n
(
M
i
−
m
i
)
Δ
x
i
<
ε
p
(
b
−
a
)
p
−
1
<
ε
p
;
(
10.32
Paraboloido
)
{\displaystyle \sum _{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}<\varepsilon ^{p}(b-a)^{p-1}<\varepsilon ^{p};\quad (10.32\;\;{\text{Paraboloido}})\;}
Vadinasi,
(
S
−
s
)
p
≤
[
∑
i
=
1
n
(
M
i
−
m
i
)
Δ
x
i
]
(
b
−
a
)
p
−
1
<
ε
p
(
b
−
a
)
p
−
1
<
ε
p
,
{\displaystyle (S-s)^{p}\leq \left[\sum _{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}\right](b-a)^{p-1}<\varepsilon ^{p}(b-a)^{p-1}<\varepsilon ^{p},}
nes
(
b
−
a
)
p
−
1
<
1
{\displaystyle (b-a)^{p-1}<1\;}
Taigi,
(
S
−
s
)
p
<
ε
p
,
{\displaystyle (S-s)^{p}<\varepsilon ^{p},}
S
−
s
<
ε
.
{\displaystyle S-s<\varepsilon .}
____________________
* Galime laikyti
M
i
>
0.
{\displaystyle M_{i}>0.}
M
i
=
0
,
{\displaystyle M_{i}=0,}
m
i
=
0
,
{\displaystyle m_{i}=0,}
keisti
Tarkime, kad f(x) ir g(x) yra bet kokios dvi segmente [a; b] integruojamos funkcijos, o p ir p' – bet kokie du skaičiai, didesni už vienetą ir susiję sąryšiu
1
p
+
1
p
′
=
1.
{\displaystyle {\frac {1}{p}}+{\frac {1}{p'}}=1.}
|
∫
a
b
f
(
x
)
g
(
x
)
d
x
|
≤
[
∫
a
b
|
f
(
x
)
|
p
d
x
]
1
/
p
[
∫
a
b
|
g
(
x
)
|
p
′
d
x
]
1
/
p
′
,
(
10.36
)
{\displaystyle \left|\int _{a}^{b}f(x)g(x)\;dx\right|\leq \left[\int _{a}^{b}|f(x)|^{p}\;dx\right]^{1/p}\left[\int _{a}^{b}|g(x)|^{p'}\;dx\right]^{1/p'},\quad (10.36)}
vadinama Helderio nelygybe integralams . Pažymėsime, kad (10.36) nelygybės dešinės pusės integralai egzistuoja pagal 10.7 teoremą, o kairės pusės integralas – pagal 5 paragrafo
3
∘
{\displaystyle 3^{\circ }}
Iš pradžių įrodykime šitokį teiginį: jeigu A (x) ir B (x) – dvi neneigiamos ir segmente [a; b] integruojamos funkcijos, tenkinančios nelygybes
∫
a
b
A
p
(
x
)
d
x
≤
1
,
∫
a
b
B
p
′
(
x
)
d
x
≤
1
,
(
10.37
)
{\displaystyle \int _{a}^{b}A^{p}(x)\;dx\leq 1,\quad \int _{a}^{b}B^{p'}(x)\;dx\leq 1,\quad (10.37)}
tai
∫
a
b
A
(
x
)
B
(
x
)
d
x
≤
1.
(
10.38
)
{\displaystyle \int _{a}^{b}A(x)B(x)\;dx\leq 1.\quad (10.38)}
Iš tikrųjų bet kuriame segmento [a; b] taške x teisinga (10.26) nelygybė
A
(
x
)
B
(
x
)
≤
A
p
(
x
)
p
+
B
p
′
(
x
)
p
′
.
{\displaystyle A(x)B(x)\leq {\frac {A^{p}(x)}{p}}+{\frac {B^{p'}(x)}{p'}}.}
Iš čia, remiantis 6 paragrafo
3
∘
{\displaystyle 3^{\circ }}
∫
a
b
A
(
x
)
B
(
x
)
d
x
≤
1
p
∫
a
b
A
p
(
x
)
d
x
+
1
p
′
∫
a
b
B
p
′
(
x
)
d
x
≤
1
p
+
1
p
′
=
1.
{\displaystyle \int _{a}^{b}A(x)B(x)\;dx\leq {\frac {1}{p}}\int _{a}^{b}A^{p}(x)\;dx+{\frac {1}{p'}}\int _{a}^{b}B^{p'}(x)\;dx\leq {\frac {1}{p}}+{\frac {1}{p'}}=1.}
(10.38) nelygybė įrodyta.
Paėmę
A
(
x
)
=
|
f
(
x
)
|
[
∫
a
b
|
f
(
x
)
|
p
d
x
]
1
/
p
,
B
(
x
)
=
|
g
(
x
)
|
[
∫
a
b
|
g
(
x
)
|
p
′
d
x
]
1
/
p
′
,
{\displaystyle A(x)={\frac {|f(x)|}{[\int _{a}^{b}|f(x)|^{p}\;dx]^{1/p}}},\quad B(x)={\frac {|g(x)|}{[\int _{a}^{b}|g(x)|^{p'}\;dx]^{1/p'}}},}
[
[
∫
a
b
|
f
(
x
)
|
p
d
x
]
1
/
p
{\displaystyle \left[\int _{a}^{b}|f(x)|^{p}\;dx\right]^{1/p}\;}
[
∫
a
b
|
g
(
x
)
|
p
′
d
x
]
1
/
p
′
{\displaystyle \left[\int _{a}^{b}|g(x)|^{p'}\;dx\right]^{1/p'}\;}
gauname šitokią nelygybę:
∫
a
b
|
f
(
x
)
|
|
g
(
x
)
|
d
x
≤
[
∫
a
b
|
f
(
x
)
|
p
d
x
]
1
/
p
[
∫
a
b
|
g
(
x
)
|
p
′
d
x
]
1
/
p
′
.
{\displaystyle \int _{a}^{b}|f(x)||g(x)|\;dx\leq \left[\int _{a}^{b}|f(x)|^{p}\;dx\right]^{1/p}\left[\int _{a}^{b}|g(x)|^{p'}\;dx\right]^{1/p'}.}
Kadangi pagal 6 paragrafo 1 skirsnio 2 pastabą
|
∫
a
b
f
(
x
)
g
(
x
)
d
x
|
≤
∫
a
b
|
f
(
x
)
g
(
x
)
|
d
x
,
{\displaystyle |\int _{a}^{b}f(x)g(x)\;dx|\leq \int _{a}^{b}|f(x)g(x)|\;dx,}
tai (10.36) Helderio nelygybė integralams įrodyta.
∫
a
b
A
(
x
)
B
(
x
)
d
x
≤
1
p
∫
a
b
A
p
(
x
)
d
x
+
1
p
′
∫
a
b
B
p
′
(
x
)
d
x
≤
1
p
+
1
p
′
=
1
,
{\displaystyle \int _{a}^{b}A(x)B(x)\;dx\leq {\frac {1}{p}}\int _{a}^{b}A^{p}(x)\;dx+{\frac {1}{p'}}\int _{a}^{b}B^{p'}(x)\;dx\leq {\frac {1}{p}}+{\frac {1}{p'}}=1,}
∫
a
b
|
f
(
x
)
|
[
∫
a
b
|
f
(
x
)
|
p
d
x
]
1
/
p
|
g
(
x
)
|
[
∫
a
b
|
g
(
x
)
|
p
′
d
x
]
1
/
p
′
d
x
≤
1
p
∫
a
b
|
f
(
x
)
|
p
[
∫
a
b
|
f
(
x
)
|
p
d
x
]
p
/
p
d
x
+
1
p
′
∫
a
b
|
g
(
x
)
|
p
′
[
∫
a
b
|
g
(
x
)
|
p
′
d
x
]
p
′
/
p
′
d
x
≤
1
p
+
1
p
′
=
1
,
{\displaystyle \int _{a}^{b}{\frac {|f(x)|}{[\int _{a}^{b}|f(x)|^{p}\;dx]^{1/p}}}{\frac {|g(x)|}{[\int _{a}^{b}|g(x)|^{p'}\;dx]^{1/p'}}}\;dx\leq {\frac {1}{p}}\int _{a}^{b}{\frac {|f(x)|^{p}}{[\int _{a}^{b}|f(x)|^{p}\;dx]^{p/p}}}\;dx+{\frac {1}{p'}}\int _{a}^{b}{\frac {|g(x)|^{p'}}{[\int _{a}^{b}|g(x)|^{p'}\;dx]^{p'/p'}}}\;dx\leq {\frac {1}{p}}+{\frac {1}{p'}}=1,}
1
[
∫
a
b
|
f
(
x
)
|
p
d
x
]
1
/
p
1
[
∫
a
b
|
g
(
x
)
|
p
′
d
x
]
1
/
p
′
∫
a
b
|
f
(
x
)
|
|
g
(
x
)
|
d
x
≤
1
p
∫
a
b
|
f
(
x
)
|
p
d
x
[
∫
a
b
|
f
(
x
)
|
p
d
x
]
+
1
p
′
∫
a
b
|
g
(
x
)
|
p
′
d
x
[
∫
a
b
|
g
(
x
)
|
p
′
d
x
]
≤
1
p
+
1
p
′
=
1
,
{\displaystyle {\frac {1}{[\int _{a}^{b}|f(x)|^{p}\;dx]^{1/p}}}{\frac {1}{[\int _{a}^{b}|g(x)|^{p'}\;dx]^{1/p'}}}\int _{a}^{b}|f(x)||g(x)|\;dx\leq {\frac {1}{p}}{\frac {\int _{a}^{b}|f(x)|^{p}\;dx}{[\int _{a}^{b}|f(x)|^{p}\;dx]}}+{\frac {1}{p'}}{\frac {\int _{a}^{b}|g(x)|^{p'}\;dx}{[\int _{a}^{b}|g(x)|^{p'}\;dx]}}\leq {\frac {1}{p}}+{\frac {1}{p'}}=1,}
1
[
∫
a
b
|
f
(
x
)
|
p
d
x
]
1
/
p
1
[
∫
a
b
|
g
(
x
)
|
p
′
d
x
]
1
/
p
′
∫
a
b
|
f
(
x
)
|
|
g
(
x
)
|
d
x
≤
1
p
⋅
1
+
1
p
′
⋅
1
=
1
p
+
1
p
′
=
1
,
{\displaystyle {\frac {1}{[\int _{a}^{b}|f(x)|^{p}\;dx]^{1/p}}}{\frac {1}{[\int _{a}^{b}|g(x)|^{p'}\;dx]^{1/p'}}}\int _{a}^{b}|f(x)||g(x)|\;dx\leq {\frac {1}{p}}\cdot 1+{\frac {1}{p'}}\cdot 1={\frac {1}{p}}+{\frac {1}{p'}}=1,}
∫
a
b
|
f
(
x
)
|
|
g
(
x
)
|
d
x
≤
[
∫
a
b
|
f
(
x
)
|
p
d
x
]
1
/
p
[
∫
a
b
|
g
(
x
)
|
p
′
d
x
]
1
/
p
′
.
{\displaystyle \int _{a}^{b}|f(x)||g(x)|\;dx\leq [\int _{a}^{b}|f(x)|^{p}\;dx]^{1/p}[\int _{a}^{b}|g(x)|^{p'}\;dx]^{1/p'}.}
Pastaba. Atskiru atveju, kai p=p'=2, Helderio nelygybė integralams virsta šitokia:
|
∫
a
b
f
(
x
)
g
(
x
)
d
x
|
≤
∫
a
b
|
f
(
x
)
|
2
d
x
∫
a
b
|
g
(
x
)
|
2
d
x
;
(
10.39
)
{\displaystyle \left|\int _{a}^{b}f(x)g(x)\;dx\right|\leq {\sqrt {\int _{a}^{b}|f(x)|^{2}\;dx}}{\sqrt {\int _{a}^{b}|g(x)|^{2}\;dx}};\quad (10.39)}
ji vadinama Koši—Buniakovskio nelygybe integralams .
keisti
Su bet kokiomis neneigiamomis ir segmente [a; b] integruojamomis funkcijomis f(x) ir g(x) ir bet kokiu skaičiumi p >1 teisinga šitokia nelygybė:
(
∫
a
b
[
f
(
x
)
+
g
(
x
)
]
p
d
x
)
1
/
p
≤
(
∫
a
b
[
f
(
x
)
]
p
d
x
)
1
/
p
+
(
∫
a
b
[
g
(
x
)
]
p
d
x
)
1
/
p
;
(
10.40
)
{\displaystyle \left(\int _{a}^{b}[f(x)+g(x)]^{p}\;dx\right)^{1/p}\leq \left(\int _{a}^{b}[f(x)]^{p}\;dx\right)^{1/p}+\left(\int _{a}^{b}[g(x)]^{p}\;dx\right)^{1/p};\quad (10.40)}
ji vadinama Minkovskio nelygybe integralams . Norint ją įrodyti, reikia pradėti nuo formulės
∫
a
b
[
f
(
x
)
+
g
(
x
)
]
p
d
x
=
∫
a
b
f
(
x
)
[
f
(
x
)
+
g
(
x
)
]
p
−
1
d
x
+
∫
a
b
g
(
x
)
[
f
(
x
)
+
g
(
x
)
]
p
−
1
d
x
{\displaystyle \int _{a}^{b}[f(x)+g(x)]^{p}\;dx=\int _{a}^{b}f(x)[f(x)+g(x)]^{p-1}\;dx+\int _{a}^{b}g(x)[f(x)+g(x)]^{p-1}\;dx}
ir integralams, esantiems dešinėje tos formulės pusėje, taikyti Helderio nelygybę. Įrodymo detales paliekame skaitytojui.
Kadangi
(
p
−
1
)
p
′
=
p
{\displaystyle (p-1)p'=p}
1
p
′
=
p
−
1
p
,
{\displaystyle {\frac {1}{p'}}={\frac {p-1}{p}},}
∫
a
b
[
f
(
x
)
+
g
(
x
)
]
p
d
x
≤
[
∫
a
b
[
f
(
x
)
]
p
d
x
]
1
p
[
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
(
p
−
1
)
p
′
d
x
]
1
p
′
+
[
∫
a
b
[
g
(
x
)
]
p
d
x
]
1
p
[
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
(
p
−
1
)
p
′
d
x
]
1
p
′
=
{\displaystyle \int _{a}^{b}[f(x)+g(x)]^{p}\;dx\leq \left[\int _{a}^{b}[f(x)]^{p}\;dx\right]^{1 \over p}\left[\int _{a}^{b}(f(x)+g(x))^{(p-1)p'}\;dx\right]^{1 \over p'}+\left[\int _{a}^{b}[g(x)]^{p}\;dx\right]^{1 \over p}\left[\int _{a}^{b}(f(x)+g(x))^{(p-1)p'}\;dx\right]^{1 \over p'}=}
=
(
[
∫
a
b
[
f
(
x
)
]
p
d
x
]
1
p
+
[
∫
a
b
[
g
(
x
)
]
p
d
x
]
1
p
)
[
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
p
d
x
]
p
−
1
p
.
{\displaystyle =\left(\left[\int _{a}^{b}[f(x)]^{p}\;dx\right]^{1 \over p}+\left[\int _{a}^{b}[g(x)]^{p}\;dx\right]^{1 \over p}\right)\left[\int _{a}^{b}(f(x)+g(x))^{p}\;dx\right]^{p-1 \over p}.}
Padaliję paskutinės nelygybės abi puses iš
[
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
p
d
x
]
p
−
1
p
,
{\displaystyle \left[\int _{a}^{b}(f(x)+g(x))^{p}\;dx\right]^{p-1 \over p},}
[
∫
a
b
[
f
(
x
)
+
g
(
x
)
]
p
d
x
]
[
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
p
d
x
]
−
p
−
1
p
≤
[
∫
a
b
[
f
(
x
)
]
p
d
x
]
1
p
+
[
∫
a
b
[
g
(
x
)
]
p
d
x
]
1
p
,
{\displaystyle \left[\int _{a}^{b}[f(x)+g(x)]^{p}\;dx\right]\left[\int _{a}^{b}(f(x)+g(x))^{p}\;dx\right]^{-{\frac {p-1}{p}}}\leq \left[\int _{a}^{b}[f(x)]^{p}\;dx\right]^{1 \over p}+\left[\int _{a}^{b}[g(x)]^{p}\;dx\right]^{1 \over p},}
[
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
p
d
x
]
1
−
p
−
1
p
≤
[
∫
a
b
[
f
(
x
)
]
p
d
x
]
1
p
+
[
∫
a
b
[
g
(
x
)
]
p
d
x
]
1
p
,
{\displaystyle \left[\int _{a}^{b}(f(x)+g(x))^{p}\;dx\right]^{1-{\frac {p-1}{p}}}\leq \left[\int _{a}^{b}[f(x)]^{p}\;dx\right]^{1 \over p}+\left[\int _{a}^{b}[g(x)]^{p}\;dx\right]^{1 \over p},}
[
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
p
d
x
]
p
−
(
p
−
1
)
p
≤
[
∫
a
b
[
f
(
x
)
]
p
d
x
]
1
p
+
[
∫
a
b
[
g
(
x
)
]
p
d
x
]
1
p
,
{\displaystyle \left[\int _{a}^{b}(f(x)+g(x))^{p}\;dx\right]^{\frac {p-(p-1)}{p}}\leq \left[\int _{a}^{b}[f(x)]^{p}\;dx\right]^{1 \over p}+\left[\int _{a}^{b}[g(x)]^{p}\;dx\right]^{1 \over p},}
[
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
p
d
x
]
1
p
≤
[
∫
a
b
[
f
(
x
)
]
p
d
x
]
1
p
+
[
∫
a
b
[
g
(
x
)
]
p
d
x
]
1
p
.
{\displaystyle \left[\int _{a}^{b}(f(x)+g(x))^{p}\;dx\right]^{\frac {1}{p}}\leq \left[\int _{a}^{b}[f(x)]^{p}\;dx\right]^{1 \over p}+\left[\int _{a}^{b}[g(x)]^{p}\;dx\right]^{1 \over p}.}
Indukcijos metodu iš (10.40) nelygybės galima gauti nelygybę, kuri tinka n neneigiamų ir segmente [a; b] integruojamų funkcijų
f
1
(
x
)
,
f
2
(
x
)
,
.
.
.
,
f
n
(
x
)
:
{\displaystyle f_{1}(x),\;f_{2}(x),\;...,\;f_{n}(x):}
(
∫
a
b
[
f
1
(
x
)
+
f
2
(
x
)
+
.
.
.
+
f
n
(
x
)
]
p
d
x
)
1
/
p
≤
(
∫
a
b
[
f
1
(
x
)
]
p
d
x
)
1
/
p
+
(
∫
a
b
[
f
2
(
x
)
]
p
d
x
)
1
/
p
+
.
.
.
+
(
∫
a
b
[
f
n
(
x
)
]
p
d
x
)
1
/
p
.
{\displaystyle \left(\int _{a}^{b}[f_{1}(x)+f_{2}(x)+...+f_{n}(x)]^{p}\;dx\right)^{1/p}\leq \left(\int _{a}^{b}[f_{1}(x)]^{p}\;dx\right)^{1/p}+\left(\int _{a}^{b}[f_{2}(x)]^{p}\;dx\right)^{1/p}+...+\left(\int _{a}^{b}[f_{n}(x)]^{p}\;dx\right)^{1/p}.}
keisti
3
∘
.
{\displaystyle 3^{\circ }.}
f
(
x
)
⋅
g
(
x
)
{\displaystyle f(x)\cdot g(x)}
∫
a
b
[
f
(
x
)
±
g
(
x
)
]
d
x
=
∫
a
b
f
(
x
)
d
x
±
∫
a
b
g
(
x
)
d
x
.
(
10.8
)
{\displaystyle \int _{a}^{b}[f(x)\pm g(x)]\;dx=\int _{a}^{b}f(x)\;dx\pm \int _{a}^{b}g(x)\;dx.\quad (10.8)}
Pirmiausia įrodykime, kad funkcija
f
(
x
)
±
g
(
x
)
{\displaystyle f(x)\pm g(x)}
ξ
i
,
{\displaystyle \xi _{i},}
∑
i
=
1
n
[
f
(
ξ
i
)
±
g
(
ξ
i
)
]
Δ
x
i
=
∑
i
=
1
n
f
(
ξ
i
)
Δ
x
i
±
∑
i
=
1
n
g
(
ξ
i
)
Δ
x
i
.
{\displaystyle \sum _{i=1}^{n}[f(\xi _{i})\pm g(\xi _{i})]\Delta x_{i}=\sum _{i=1}^{n}f(\xi _{i})\Delta x_{i}\pm \sum _{i=1}^{n}g(\xi _{i})\Delta x_{i}.}
Kai egzistuoja dešinės pusės riba, egzistuoja ir kairės pusės riba. Todėl funkcija
f
(
x
)
±
g
(
x
)
{\displaystyle f(x)\pm g(x)}
Dabar įrodysime, kad integruojamų funkcijų sandauga yra integruojama funckija. Kadangi funkcijos f(x) ir g(x) integruojamos segmente [a; b], tai jos tame segmente yra aprėžtos, taigi
|
f
(
x
)
|
≤
A
,
{\displaystyle |f(x)|\leq A,\;}
|
g
(
x
)
|
≤
B
.
{\displaystyle |g(x)|\leq B.}
[
x
i
−
1
;
x
i
]
{\displaystyle [x_{i-1};\;x_{i}]}
f
(
x
″
)
g
(
x
″
)
−
f
(
x
′
)
g
(
x
′
)
=
[
f
(
x
″
)
−
f
(
x
′
)
]
g
(
x
″
)
+
[
g
(
x
″
)
−
g
(
x
′
)
]
f
(
x
′
)
.
{\displaystyle f(x'')g(x'')-f(x')g(x')=[f(x'')-f(x')]g(x'')+[g(x'')-g(x')]f(x').}
Kadangi
|
f
(
x
″
)
g
(
x
″
)
−
f
(
x
′
)
g
(
x
′
)
|
≤
ω
i
,
{\displaystyle |f(x'')g(x'')-f(x')g(x')|\leq \omega _{i},\;\;}
|
f
(
x
″
)
−
f
(
x
′
)
|
≤
w
i
,
{\displaystyle |f(x'')-f(x')|\leq w_{i},\;\;}
|
g
(
x
″
)
−
g
(
x
′
)
|
≤
w
¯
i
,
{\displaystyle |g(x'')-g(x')|\leq {\overline {w}}_{i},}
kai
ω
i
,
w
i
,
w
¯
i
{\displaystyle \omega _{i},\;w_{i},\;{\overline {w}}_{i}}
f
(
x
)
g
(
x
)
,
{\displaystyle f(x)g(x),}
[
x
i
−
1
;
x
i
]
,
{\displaystyle [x_{i-1};\;x_{i}],}
ω
i
≤
B
w
i
+
A
w
¯
i
.
{\displaystyle \omega _{i}\leq Bw_{i}+A{\overline {w}}_{i}.}
Todėl
∑
i
=
1
n
ω
i
Δ
x
i
≤
B
∑
i
=
1
n
w
i
Δ
x
i
+
A
∑
i
=
1
n
w
¯
i
Δ
x
i
.
{\displaystyle \sum _{i=1}^{n}\omega _{i}\Delta x_{i}\leq B\sum _{i=1}^{n}w_{i}\Delta x_{i}+A\sum _{i=1}^{n}{\overline {w}}_{i}\Delta x_{i}.}
Kadangi f(x) ir g(x) yra integruojamos segmente [a; b], tai bet kokį duotąjį
ε
>
0
{\displaystyle \varepsilon >0}
T , kad
∑
i
=
1
n
w
i
Δ
x
i
<
ε
2
B
{\displaystyle \sum _{i=1}^{n}w_{i}\Delta x_{i}<{\frac {\varepsilon }{2B}}}
∑
i
=
1
n
w
¯
i
Δ
x
i
<
ε
2
A
.
{\displaystyle \sum _{i=1}^{n}{\overline {w}}_{i}\Delta x_{i}<{\frac {\varepsilon }{2A}}.}
S
−
s
=
∑
i
=
1
n
ω
i
Δ
x
i
<
B
ε
2
B
+
A
ε
2
A
=
ε
.
{\displaystyle S-s=\sum _{i=1}^{n}\omega _{i}\Delta x_{i}<B\;{\frac {\varepsilon }{2B}}+A\;{\frac {\varepsilon }{2A}}=\varepsilon .}
Taigi integruojamų funkcijų sandauga yra integruojama funkcija. _________________
* Toje tapatybėje taškus x' ir x" galima pasirinkti taip, kad kairė pusė kiek norima mažai skirtųsi nuo
ω
i
.
{\displaystyle \omega _{i}.}
4
∘
.
{\displaystyle 4^{\circ }.}
c
f
(
x
)
{\displaystyle cf(x)\;}
c
=
const
{\displaystyle c={\text{const}}}
∫
a
b
c
f
(
x
)
d
x
=
c
∫
a
b
f
(
x
)
d
x
.
(
10.9
)
{\displaystyle \int _{a}^{b}cf(x)\;dx=c\int _{a}^{b}f(x)\;dx.\quad (10.9)}
Iš tikrųjų funkcijų f(x) ir c f(x) integralinės sumos skiriasi pastoviu daugikliu c . Todėl funkcija cf(x) integruojama ir teisinga (10.9) formulė.
Šiame skirsnyje nurodysime, kaip įvertinami apibrėžtiniai integralai, kurių pointegralinės funkcijos tenkina vienokias ar kitokias sąlygas.
1
∘
.
{\displaystyle 1^{\circ }.}
Tarkime, kad integruojama segmente [a; b] funkcija
f
(
x
)
{\displaystyle f(x)}
∫
a
b
f
(
x
)
d
x
≥
0.
{\displaystyle \int _{a}^{b}f(x)\;dx\geq 0.}
Iš tikrųjų kiekviena tokios funkcijos integralinė suma yra neneigiama. Todėl integralinių sumų riba
I
=
∫
a
b
f
(
x
)
d
x
{\displaystyle I=\int _{a}^{b}f(x)\;dx}
1 pastaba. Jeigu
f
(
x
)
{\displaystyle f(x)}
[a; b] ir
f
(
x
)
≥
m
,
{\displaystyle f(x)\geq m,}
∫
a
b
f
(
x
)
d
x
≥
m
(
b
−
a
)
.
{\displaystyle \int _{a}^{b}f(x)\;dx\geq m(b-a).}
Iš tikrųjų funkcija
f
(
x
)
−
m
{\displaystyle f(x)-m}
∫
a
b
[
f
(
x
)
−
m
]
d
x
≥
0.
{\displaystyle \int _{a}^{b}[f(x)-m]\;dx\geq 0.}
∫
a
b
f
(
x
)
d
x
≥
∫
a
b
m
d
x
=
m
∫
a
b
d
x
=
m
(
b
−
a
)
.
{\displaystyle \int _{a}^{b}f(x)\;dx\geq \int _{a}^{b}m\;dx=m\int _{a}^{b}dx=m(b-a).}
2
∘
.
{\displaystyle 2^{\circ }.}
Jei funkcija
f
(
x
)
{\displaystyle f(x)}
[a; b], tai
∫
a
b
f
(
x
)
d
x
≥
c
>
0.
{\displaystyle \int _{a}^{b}f(x)\;dx\geq c>0.}
Kadangi funkcija
f
(
x
)
{\displaystyle f(x)}
ξ
,
{\displaystyle \xi ,}
f
(
ξ
)
=
2
k
>
0.
{\displaystyle f(\xi )=2k>0.}
ξ
,
{\displaystyle \xi ,}
f
(
x
)
{\displaystyle f(x)}
∫
p
q
f
(
x
)
d
x
≥
k
(
q
−
p
)
>
0.
{\displaystyle \int _{p}^{q}f(x)\;dx\geq k(q-p)>0.}
Pagal apibrėžtinių integralų savybę
∫
a
b
f
(
x
)
d
x
=
∫
a
c
f
(
x
)
d
x
+
∫
c
b
f
(
x
)
d
x
,
{\displaystyle \int _{a}^{b}f(x)\;dx=\int _{a}^{c}f(x)\;dx+\int _{c}^{b}f(x)\;dx,}
∫
a
b
f
(
x
)
d
x
=
∫
a
p
f
(
x
)
d
x
+
∫
p
q
f
(
x
)
d
x
+
∫
q
b
f
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}f(x)\;dx=\int _{a}^{p}f(x)\;dx+\int _{p}^{q}f(x)\;dx+\int _{q}^{b}f(x)\;dx.}
Kadangi
f
(
x
)
≥
0
{\displaystyle f(x)\geq 0}
∫
p
q
f
(
x
)
d
x
≥
c
>
0
{\displaystyle \int _{p}^{q}f(x)\;dx\geq c>0\;}
c
=
k
(
q
−
p
)
{\displaystyle c=k(q-p)}
∫
a
b
f
(
x
)
d
x
≥
c
>
0.
{\displaystyle \int _{a}^{b}f(x)\;dx\geq c>0.}
3
∘
.
{\displaystyle 3^{\circ }.}
Jei funkcijos
f
(
x
)
{\displaystyle f(x)}
g
(
x
)
{\displaystyle g(x)}
[a; b] ir
f
(
x
)
≥
g
(
x
)
{\displaystyle f(x)\geq g(x)}
∫
a
b
f
(
x
)
d
x
≥
∫
a
b
g
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}f(x)\;dx\geq \int _{a}^{b}g(x)\;dx.}
Iš tikrųjų funkcija
f
(
x
)
−
g
(
x
)
≥
0
{\displaystyle f(x)-g(x)\geq 0}
1
∘
{\displaystyle 1^{\circ }}
2 pastaba. Jei funkcija
f
(
x
)
{\displaystyle f(x)}
[a; b], tai funkcija
|
f
(
x
)
|
{\displaystyle |f(x)|}
|
∫
a
b
f
(
x
)
d
x
|
≤
∫
a
b
|
f
(
x
)
|
d
x
.
{\displaystyle \left|\int _{a}^{b}f(x)\;dx\right|\leq \int _{a}^{b}|f(x)|\;dx.}
Iš pradžių įrodykime, kad integruojamos funkcijos f(x) modulis |f(x)| yra integruojamas. Pažymėkime
M
i
{\displaystyle M_{i}}
m
i
{\displaystyle m_{i}}
[
x
i
−
1
;
x
i
]
{\displaystyle [x_{i-1};x_{i}]}
M
i
′
{\displaystyle M_{i}'}
m
i
′
{\displaystyle m_{i}'}
M
i
′
−
m
i
′
≤
M
i
−
m
i
{\displaystyle M_{i}'-m_{i}'\leq M_{i}-m_{i}\;}
M
i
{\displaystyle M_{i}}
m
i
{\displaystyle m_{i}}
M
i
{\displaystyle M_{i}}
m
i
{\displaystyle m_{i}}
M
i
>
0
,
{\displaystyle M_{i}>0,\;}
m
i
≤
0
{\displaystyle m_{i}\leq 0\;}
M
i
−
m
i
=
−
|
M
i
|
−
(
−
|
m
i
|
)
=
−
|
M
i
|
+
|
m
i
|
>
0
{\displaystyle M_{i}-m_{i}=-|M_{i}|-(-|m_{i}|)=-|M_{i}|+|m_{i}|>0}
M
i
′
−
m
i
′
=
|
M
i
|
−
|
m
i
|
<
0
,
{\displaystyle M_{i}'-m_{i}'=|M_{i}|-|m_{i}|<0,}
M
i
′
−
m
i
′
≤
M
i
−
m
i
{\displaystyle M_{i}'-m_{i}'\leq M_{i}-m_{i}}
S
′
−
s
′
≤
S
−
s
.
{\displaystyle S'-s'\leq S-s.}
S
−
s
<
ε
,
{\displaystyle S-s<\varepsilon ,}
S
′
−
s
′
<
ε
,
{\displaystyle S'-s'<\varepsilon ,}
Dabar įrodysime mus dominantį įvertį. Kadangi
−
|
f
(
x
)
|
≤
f
(
x
)
≤
|
f
(
x
)
|
,
{\displaystyle -|f(x)|\leq f(x)\leq |f(x)|,}
−
∫
a
b
|
f
(
x
)
|
d
x
≤
∫
a
b
f
(
x
)
d
x
≤
∫
a
b
|
f
(
x
)
|
d
x
.
{\displaystyle -\int _{a}^{b}|f(x)|\;dx\leq \int _{a}^{b}f(x)\;dx\leq \int _{a}^{b}|f(x)|\;dx.}
Tai ir reiškia, kad
|
∫
a
b
f
(
x
)
d
x
|
≤
∫
a
b
|
f
(
x
)
|
d
x
.
{\displaystyle \left|\int _{a}^{b}f(x)\;dx\right|\leq \int _{a}^{b}|f(x)|\;dx.}
________________________
* Iš to, kad funkcija |f(x)| yra integruojama, dar neišplaukia f(x) integruojamumas. Pavyzdžiui, funkcija
f
(
x
)
=
{
1
,
kai
x
racionalusis
,
−
1
,
kai
x
iracionalusis
,
{\displaystyle f(x)={\begin{cases}1,\;\;{\text{kai}}\;x\;{\text{racionalusis}},&\\-1,\;\;{\text{kai}}\;x\;{\text{iracionalusis}},&\end{cases}}}
|
f
(
x
)
|
=
1
{\displaystyle |f(x)|=1}
4
∘
.
{\displaystyle 4^{\circ }.}
Sakykime, funkcijos
f
(
x
)
{\displaystyle f(x)}
g
(
x
)
{\displaystyle g(x)}
[a; b] ir
g
(
x
)
≥
0.
{\displaystyle g(x)\geq 0.}
f
(
x
)
{\displaystyle f(x)}
[a; b], tai
m
∫
a
b
g
(
x
)
d
x
≤
∫
a
b
f
(
x
)
g
(
x
)
d
x
≤
M
∫
a
b
g
(
x
)
d
x
.
(
10.11
)
{\displaystyle m\int _{a}^{b}g(x)\;dx\leq \int _{a}^{b}f(x)g(x)\;dx\leq M\int _{a}^{b}g(x)\;dx.\quad (10.11)}
(10.11) formulė išplaukia iš to, kad visiems x iš segmento [a; b] yra teisingos nelygybės
m
g
(
x
)
≤
f
(
x
)
g
(
x
)
≤
M
g
(
x
)
{\displaystyle mg(x)\leq f(x)g(x)\leq Mg(x)\;}
3
∘
{\displaystyle 3^{\circ }}
4
∘
{\displaystyle 4^{\circ }}
Tarkime, kad funkcija
f
(
x
)
{\displaystyle f(x)}
[a; b], o m ir M yra jos tikslieji rėžiai tame segmente. Tada yra toks skaičius
μ
,
{\displaystyle \mu ,}
m
≤
μ
≤
M
,
{\displaystyle m\leq \mu \leq M,}
∫
a
b
f
(
x
)
d
x
=
μ
(
b
−
a
)
.
(
10.12
)
{\displaystyle \int _{a}^{b}f(x)\;dx=\mu (b-a).\quad (10.12)}
Paėmę
g
(
x
)
=
1
{\displaystyle g(x)=1}
∫
a
b
1
⋅
d
x
=
b
−
a
,
{\displaystyle \int _{a}^{b}1\cdot dx=b-a,}
m
(
b
−
a
)
≤
∫
a
b
f
(
x
)
d
x
≤
M
(
b
−
a
)
.
(
10.12.1
)
{\displaystyle m(b-a)\leq \int _{a}^{b}f(x)\;dx\leq M(b-a).\quad (10.12.1)}
Skaičių
1
b
−
a
∫
a
b
f
(
x
)
d
x
{\displaystyle {\frac {1}{b-a}}\int _{a}^{b}f(x)\;dx}
μ
,
{\displaystyle \mu ,}
8.5 teorema. Sakykime, funkcija
f
(
x
)
{\displaystyle f(x)}
f
(
a
)
{\displaystyle f(a)}
f
(
b
)
,
{\displaystyle f(b),}
f
(
a
)
<
0
,
f
(
b
)
>
0
{\displaystyle f(a)<0,\;f(b)>0}
[a; b] viduje yra taškas
ξ
,
{\displaystyle \xi ,}
.8.6 teorema. Sakykime, funkcija
f
(
x
)
{\displaystyle f(x)}
[a; b] ir
f
(
a
)
=
A
,
{\displaystyle f(a)=A,\;}
f
(
b
)
=
B
.
{\displaystyle f(b)=B.}
[a; b] yra toks taškas
ξ
,
{\displaystyle \xi ,}
f
(
ξ
)
=
C
.
{\displaystyle f(\xi )=C.}
Įrodymas. Užtenka išnagrinėti atvejį, kai
A
≠
B
,
{\displaystyle A\neq B,}
C nesutampa nei su skaičiumi A , nei su skaičiumi B . Konkretumo dėlei tarkime, kad A<B ir A<C<B. Sudarykime funkciją
ϕ
(
x
)
=
f
(
x
)
−
C
.
{\displaystyle \phi (x)=f(x)-C.}
ϕ
(
a
)
=
f
(
a
)
−
C
=
A
−
C
<
0
,
ϕ
(
b
)
=
f
(
b
)
−
C
=
B
−
C
>
0.
{\displaystyle \phi (a)=f(a)-C=A-C<0,\quad \phi (b)=f(b)-C=B-C>0.}
Pagal 8.5 teoremą segmento [a; b] viduje yra toks taškas
ξ
,
{\displaystyle \xi ,}
ϕ
(
ξ
)
=
f
(
ξ
)
−
C
=
0.
{\displaystyle \phi (\xi )=f(\xi )-C=0.}
f
(
ξ
)
=
C
.
{\displaystyle f(\xi )=C.}
8.7 (pirmoji Vejerštraso) teorema. Jei funkcija
f
(
x
)
{\displaystyle f(x)}
[a; b], tai ji tame segmente yra aprėžta .[Pavyzdžiui, funkcija f(x)=1/x pusintervalyje (0; 10] yra tolydi, bet segmente [0; 10] nėra tolydi (sąlygoje pasakyta segmentas, o ne pusintervalis), nes dalyba iš nulio negalima. O jei imti reikšmes artimas nuliui, kaip 0.0001, tai tom reikšmėm artėjant prie nulio, funkcijos f(x)=1/x reikšmės artės prie begalybės ir tokia funkcija nėra aprėžta. Be to, ši funkcija (f(x)=1/x) nėra tolygiai tolydi pusintervalyje (0; 10], nes su labai mažu skirtumu
δ
x
{\displaystyle \delta x}
Ox ašies yra labai didelis skirtumas
δ
y
{\displaystyle \delta y}
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
(
−
∞
;
∞
)
,
{\displaystyle (-\infty ;\;\infty ),}
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
x reikšmės) pasikeitimas yra 0.1, tai
1000000.1
2
−
1000000
2
=
1
,
000
,
000
,
200
,
000.01
−
1
,
000
,
000
,
000
,
000
=
200000.01
,
{\displaystyle 1000000.1^{2}-1000000^{2}=1,000,000,200,000.01-1,000,000,000,000=200000.01,}
δ
y
{\displaystyle \delta y}
[
0
;
∞
)
{\displaystyle [0;\;\infty )}
f
(
x
)
=
x
,
{\displaystyle f(x)={\sqrt {x}},}
1000000.1
−
1000000
≈
1000.00004999999875
−
1000
=
4.999999875
⋅
10
−
5
=
{\displaystyle {\sqrt {1000000.1}}-{\sqrt {1000000}}\approx 1000.00004999999875-1000=4.999999875\cdot 10^{-5}=}
Pastebėsime, kad
10.1
⋅
10.1
=
10.1
(
10
+
0.1
)
=
10.1
⋅
10
+
10.1
⋅
0.1
=
101
+
1.01
=
102.01.
{\displaystyle 10.1\cdot 10.1=10.1(10+0.1)=10.1\cdot 10+10.1\cdot 0.1=101+1.01=102.01.}
8.8 (antroji Vejerštraso) teorema. Jei funkcija
f
(
x
)
{\displaystyle f(x)}
[a; b], tai tame segmente pasiekia savo tikslųjį viršutinį ir tikslųjį apatinį rėžį (t. y. segmente [a; b] yra tokie taškai
x
1
{\displaystyle x_{1}}
x
2
,
{\displaystyle x_{2},}
f
(
x
1
)
=
M
,
{\displaystyle f(x_{1})=M,\;}
f
(
x
2
)
=
m
{\displaystyle f(x_{2})=m}
Jei funkcija
f
(
x
)
{\displaystyle f(x)}
tolydi segmente [a; b], tai egzistuoja tokie segmento taškai p ir q , kad
f
(
p
)
=
m
{\displaystyle f(p)=m}
f
(
q
)
=
M
{\displaystyle f(q)=M\;}
ξ
,
{\displaystyle \xi ,}
f
(
ξ
)
=
μ
.
{\displaystyle f(\xi )=\mu .}
∫
a
b
f
(
x
)
d
x
=
f
(
ξ
)
(
b
−
a
)
.
(
10.13
)
{\displaystyle \int _{a}^{b}f(x)\;dx=f(\xi )(b-a).\quad (10.13)}
Ji vadinama pirmąja vidurinės reikšmės formule . Įrodysime šitokį teiginį. Tarkime, kad funkcijos
f
(
x
)
{\displaystyle f(x)}
g
(
x
)
{\displaystyle g(x)}
[a; b], o m ir M yra
f
(
x
)
{\displaystyle f(x)}
[a; b]. Be to, tarkime, kad funkcija
g
(
x
)
≥
0
{\displaystyle g(x)\geq 0\;}
arba
g
(
x
)
≤
0
{\displaystyle g(x)\leq 0}
visame segmente [a; b]. Tada yra toks skaičius
μ
,
{\displaystyle \mu ,}
m
≤
μ
≤
M
,
{\displaystyle m\leq \mu \leq M,}
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
μ
∫
a
b
g
(
x
)
d
x
.
(
10.14
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=\mu \int _{a}^{b}g(x)\;dx.\quad (10.14)}
Skyrium imant, jeigu
f
(
x
)
{\displaystyle f(x)}
[a; b], tai jame yra toks skaičius
ξ
,
{\displaystyle \xi ,}
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
f
(
ξ
)
∫
a
b
g
(
x
)
d
x
.
(
10.15
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=f(\xi )\int _{a}^{b}g(x)\;dx.\quad (10.15)}
(10.15) formulė vadinama pirmąja apibendrinta vidurinės reikšmės formule .
Įrodysime (10.14) formulę. Jeigu
∫
a
b
g
(
x
)
d
x
=
0
,
{\displaystyle \int _{a}^{b}g(x)\;dx=0,}
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
0.
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=0.}
μ
{\displaystyle \mu }
∫
a
b
g
(
x
)
d
x
>
0
,
{\displaystyle \int _{a}^{b}g(x)\;dx>0,}
∫
a
b
g
(
x
)
d
x
,
{\displaystyle \int _{a}^{b}g(x)\;dx,}
m
≤
∫
a
b
f
(
x
)
g
(
x
)
d
x
∫
a
b
g
(
x
)
d
x
≤
M
.
{\displaystyle m\leq {\frac {\int _{a}^{b}f(x)g(x)\;dx}{\int _{a}^{b}g(x)\;dx}}\leq M.}
Pažymėję skaičių
∫
a
b
f
(
x
)
g
(
x
)
d
x
∫
a
b
g
(
x
)
d
x
{\displaystyle {\frac {\int _{a}^{b}f(x)g(x)\;dx}{\int _{a}^{b}g(x)\;dx}}}
μ
,
{\displaystyle \mu ,}
Jeigu f(x) yra tolydi segmente [a; b], tai, kad ir koks būtų skaičius
μ
{\displaystyle \mu }
m ir M , tame segmente yra toks taškas
ξ
,
{\displaystyle \xi ,}
f
(
ξ
)
=
μ
,
{\displaystyle f(\xi )=\mu ,}
4 pastaba. Jei funkcija f(x) nėra tolydi, tai (10.15) formulė, apskritai kalbant, nėra teisinga.Teisingas šitoks teiginys. Jei segmente [a; b] funkcija
f
(
x
)
{\displaystyle f(x)}
f
(
x
)
{\displaystyle f(x)}
ξ
,
{\displaystyle \xi ,}
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
+
g
(
b
)
∫
ξ
b
f
(
x
)
d
x
.
(
10.16
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx+g(b)\int _{\xi }^{b}f(x)\;dx.\quad (10.16)}
(10.16) formulė vadinama antąja vidurinės reikšmės , arba Bonė *, formule. Suformuluotas teiginys įrodytas šio skyriaus 2 priede. [Paraboloido pastaba. Tame antro priedo įrodyme yra kažkas teisingo, bet bendrai paėmus (10.16) formulė yra neteisinga.]
* Bonė (1819–1892) – prancūzų matematikas.
keisti
Dar kartą suformuluosime 6 paragrafo 4 skirsnio teiginį.
Jei segmente [a; b] funkcija
g
(
x
)
{\displaystyle g(x)}
f
(
x
)
{\displaystyle f(x)}
ξ
,
{\displaystyle \xi ,}
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
+
g
(
b
)
∫
ξ
b
f
(
x
)
d
x
.
(
10.16
)
∗
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx+g(b)\int _{\xi }^{b}f(x)\;dx.\quad (10.16)^{*}}
* Patogumo dėlei paliekame ankstesnį formulės numerį.
Iš pradžių įrodysime pagalbinį teiginį.
Abelio** lema. Tarkime, kad
v
1
≥
v
2
≥
.
.
.
≥
v
n
≥
0
{\displaystyle v_{1}\geq v_{2}\geq ...\geq v_{n}\geq 0\;}
u
1
,
u
2
,
.
.
.
,
u
n
{\displaystyle \;u_{1},\;u_{2},\;...,\;u_{n}}
S
i
=
u
1
+
u
2
+
.
.
.
+
u
i
{\displaystyle S_{i}=u_{1}+u_{2}+...+u_{i}}
i
{\displaystyle i}
S
i
{\displaystyle S_{i}}
v
i
u
1
+
v
2
u
2
+
.
.
.
+
v
n
u
n
{\displaystyle v_{i}u_{1}+v_{2}u_{2}+...+v_{n}u_{n}}
A
v
1
{\displaystyle Av_{1}}
B
v
1
.
{\displaystyle Bv_{1}.}
** N. H. Abelis (1802–1829) – norvegų matematikas.Įrodymas. Turime
u
1
=
S
1
,
{\displaystyle u_{1}=S_{1},\;}
u
i
=
S
i
−
S
i
−
1
.
{\displaystyle u_{i}=S_{i}-S_{i-1}.}
v
i
u
1
+
v
2
u
2
+
.
.
.
+
v
n
u
n
=
v
1
S
1
+
v
2
(
S
2
−
S
1
)
+
v
3
(
S
3
−
S
2
)
+
.
.
.
+
v
n
(
S
n
−
S
n
−
1
)
=
{\displaystyle v_{i}u_{1}+v_{2}u_{2}+...+v_{n}u_{n}=v_{1}S_{1}+v_{2}(S_{2}-S_{1})+v_{3}(S_{3}-S_{2})+...+v_{n}(S_{n}-S_{n-1})=}
=
S
1
(
v
1
−
v
2
)
+
S
2
(
v
2
−
v
3
)
+
S
3
(
v
3
−
v
4
)
+
.
.
.
+
S
n
−
1
(
v
n
−
1
−
v
n
)
+
S
n
v
n
.
{\displaystyle =S_{1}(v_{1}-v_{2})+S_{2}(v_{2}-v_{3})+S_{3}(v_{3}-v_{4})+...+S_{n-1}(v_{n-1}-v_{n})+S_{n}v_{n}.}
Kadangi
v
i
≥
0
{\displaystyle v_{i}\geq 0}
v
i
−
v
i
+
1
≥
0
,
{\displaystyle v_{i}-v_{i+1}\geq 0,}
S
i
{\displaystyle S_{i}}
A , o po to B , gausime nelygybes
A
[
(
v
1
−
v
2
)
+
(
v
2
−
v
3
)
+
(
v
3
−
v
4
)
+
.
.
.
+
(
v
n
−
1
−
v
n
)
+
v
n
]
≤
S
1
(
v
1
−
v
2
)
+
S
2
(
v
2
−
v
3
)
+
S
3
(
v
3
−
v
4
)
+
.
.
.
+
S
n
−
1
(
v
n
−
1
−
v
n
)
+
S
n
v
n
≤
{\displaystyle A[(v_{1}-v_{2})+(v_{2}-v_{3})+(v_{3}-v_{4})+...+(v_{n-1}-v_{n})+v_{n}]\leq S_{1}(v_{1}-v_{2})+S_{2}(v_{2}-v_{3})+S_{3}(v_{3}-v_{4})+...+S_{n-1}(v_{n-1}-v_{n})+S_{n}v_{n}\leq }
≤
B
[
(
v
1
−
v
2
)
+
(
v
2
−
v
3
)
+
(
v
3
−
v
4
)
+
.
.
.
+
(
v
n
−
1
−
v
n
)
+
v
n
]
.
{\displaystyle \leq B[(v_{1}-v_{2})+(v_{2}-v_{3})+(v_{3}-v_{4})+...+(v_{n-1}-v_{n})+v_{n}].}
Pastebėję, kad reiškiniai laužtiniuose skliaustuose lygūs
v
1
,
{\displaystyle v_{1},}
A
v
1
≤
S
1
(
v
1
−
v
2
)
+
S
2
(
v
2
−
v
3
)
+
S
3
(
v
3
−
v
4
)
+
.
.
.
+
S
n
−
1
(
v
n
−
1
−
v
n
)
+
S
n
v
n
≤
B
v
1
.
{\displaystyle Av_{1}\leq S_{1}(v_{1}-v_{2})+S_{2}(v_{2}-v_{3})+S_{3}(v_{3}-v_{4})+...+S_{n-1}(v_{n-1}-v_{n})+S_{n}v_{n}\leq Bv_{1}.}
Lema įrodyta.
Pastaba. Abelio lemos įrodyme pritaikėme sumos
∑
k
=
1
n
v
k
u
k
{\displaystyle \sum _{k=1}^{n}v_{k}u_{k}}
6 paragrafo 4 skirsnio teiginio įrodymas. Tarkime, kad funkcija g(x) nedidėja segmente [a; b] ir yra jame neneigiama. Kadangi funkcija
f
(
x
)
g
(
x
)
{\displaystyle f(x)g(x)}
3
∘
{\displaystyle 3^{\circ }}
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
lim
Δ
→
0
∑
i
=
1
n
f
(
x
i
−
1
)
g
(
x
i
−
1
)
Δ
x
i
;
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=\lim _{\Delta \to 0}\sum _{i=1}^{n}f(x_{i-1})g(x_{i-1})\;\Delta x_{i};\;}
Δ
=
max
Δ
x
i
.
{\displaystyle \Delta ={\text{max}}\;\Delta x_{i}.}
Funkcijos f(x) tiksliuosius rėžius segmente
[
x
i
−
1
;
x
1
]
{\displaystyle [x_{i-1};\;x_{1}]}
M
i
{\displaystyle M_{i}}
m
i
.
{\displaystyle m_{i}.}
∑
i
=
1
n
m
i
g
(
x
i
−
1
)
Δ
x
i
≤
∑
i
=
1
n
f
(
x
i
−
1
)
g
(
x
i
−
1
)
Δ
x
i
≤
∑
i
=
1
n
M
i
g
(
x
i
−
1
)
Δ
x
i
.
(
10.41
)
{\displaystyle \sum _{i=1}^{n}m_{i}g(x_{i-1})\;\Delta x_{i}\leq \sum _{i=1}^{n}f(x_{i-1})g(x_{i-1})\;\Delta x_{i}\leq \sum _{i=1}^{n}M_{i}g(x_{i-1})\;\Delta x_{i}.\quad (10.41)}
Bet g(x) nedidėja segmente [a; b], todėl skirtumas
∑
i
=
1
n
M
i
g
(
x
i
−
1
)
Δ
x
i
−
∑
i
=
1
n
m
i
g
(
x
i
−
1
)
Δ
x
i
=
∑
i
=
1
n
(
M
i
−
m
i
)
g
(
x
i
−
1
)
Δ
x
i
{\displaystyle \sum _{i=1}^{n}M_{i}g(x_{i-1})\;\Delta x_{i}-\sum _{i=1}^{n}m_{i}g(x_{i-1})\;\Delta x_{i}=\sum _{i=1}^{n}(M_{i}-m_{i})g(x_{i-1})\;\Delta x_{i}}
nėra didesnis už skaičių
∑
i
=
1
n
(
M
i
−
m
i
)
g
(
a
)
Δ
x
i
=
g
(
a
)
∑
i
=
1
n
(
M
i
−
m
i
)
Δ
x
i
.
{\displaystyle \sum _{i=1}^{n}(M_{i}-m_{i})g(a)\;\Delta x_{i}=g(a)\sum _{i=1}^{n}(M_{i}-m_{i})\;\Delta x_{i}.\;}
∑
i
=
1
n
(
M
i
−
m
i
)
Δ
x
i
=
∑
i
=
1
n
ω
Δ
x
i
{\displaystyle \sum _{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}=\sum _{i=1}^{n}\omega \;\Delta x_{i}\;}
Δ
→
0.
{\displaystyle \Delta \to 0.}
kad ir kokie būtų skaičiai
μ
i
,
{\displaystyle \mu _{i},}
m
i
≤
μ
i
≤
M
i
,
{\displaystyle m_{i}\leq \mu _{i}\leq M_{i},}
∑
i
=
1
n
m
i
g
(
x
i
−
1
)
Δ
x
i
,
∑
i
=
1
n
μ
i
g
(
x
i
−
1
)
Δ
x
i
,
∑
i
=
1
n
M
i
g
(
x
i
−
1
)
Δ
x
i
{\displaystyle \sum _{i=1}^{n}m_{i}g(x_{i-1})\;\Delta x_{i},\quad \sum _{i=1}^{n}\mu _{i}g(x_{i-1})\;\Delta x_{i},\quad \sum _{i=1}^{n}M_{i}g(x_{i-1})\;\Delta x_{i}}
riba, kad
Δ
→
0
,
{\displaystyle \Delta \to 0,}
∫
a
b
f
(
x
)
g
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx.}
μ
i
,
{\displaystyle \mu _{i},\;}
m
i
≤
μ
i
≤
M
i
,
{\displaystyle m_{i}\leq \mu _{i}\leq M_{i},}
∫
x
i
−
1
x
i
f
(
x
)
d
x
=
μ
i
Δ
x
i
.
{\displaystyle \int _{x_{i-1}}^{x_{i}}f(x)\;dx=\mu _{i}\;\Delta x_{i}.}
Kadangi funkcija
F
(
x
)
=
∫
a
x
f
(
t
)
d
t
{\displaystyle F(x)=\int _{a}^{x}f(t)\;dt\;}
x gali būti bet kokia reikšmė iš segmento [a; b]) tolydi segmente [a; b], tai skaičius
S
i
=
∑
k
=
1
i
μ
k
Δ
x
k
=
∫
a
x
i
f
(
t
)
d
t
{\displaystyle S_{i}=\sum _{k=1}^{i}\mu _{k}\;\Delta x_{k}=\int _{a}^{x_{i}}f(t)\;dt}
m ir tiksliojo viršutinio rėžio M segmente [a; b].
[Paraboloido pataisymas .
Kadangi funkcija
F
(
x
)
=
∫
a
x
f
(
t
)
d
t
{\displaystyle F(x)=\int _{a}^{x}f(t)\;dt\;}
x gali būti bet kokia reikšmė iš segmento [a; b]) tolydi segmente [a; b], tai skaičius
S
i
=
∑
k
=
1
i
μ
k
Δ
x
k
=
∫
a
x
i
f
(
t
)
d
t
{\displaystyle S_{i}=\sum _{k=1}^{i}\mu _{k}\;\Delta x_{k}=\int _{a}^{x_{i}}f(t)\;dt}
m padauginto iš
(
x
i
−
a
)
{\displaystyle (x_{i}-a)}
M padauginto iš
(
x
i
−
a
)
{\displaystyle (x_{i}-a)}
Taip gaunama pagal (10.12.1) formulę, t. y.
m
(
x
i
−
a
)
≤
∫
a
x
i
f
(
t
)
d
t
≤
M
(
x
i
−
a
)
,
{\displaystyle m(x_{i}-a)\leq \int _{a}^{x_{i}}f(t)\;dt\leq M(x_{i}-a),\;}
m
(
b
−
a
)
≤
∫
a
b
f
(
x
)
d
x
≤
M
(
b
−
a
)
.
(
10.12.1
)
{\displaystyle m(b-a)\leq \int _{a}^{b}f(x)\;dx\leq M(b-a).\quad (10.12.1)}
Imkime
v
1
=
g
(
a
)
,
v
2
=
g
(
x
1
)
,
v
3
=
g
(
x
2
)
,
.
.
.
,
v
n
=
g
(
x
n
−
1
)
,
{\displaystyle v_{1}=g(a),\;v_{2}=g(x_{1}),\;v_{3}=g(x_{2}),\;...,\;v_{n}=g(x_{n-1}),\;}
u
1
=
μ
1
Δ
x
1
,
u
2
=
μ
2
Δ
x
2
,
.
.
.
,
u
n
=
μ
n
Δ
x
n
.
{\displaystyle u_{1}=\mu _{1}\Delta x_{1},\;u_{2}=\mu _{2}\Delta x_{2},\;...,\;u_{n}=\mu _{n}\Delta x_{n}.}
Kadangi
v
1
≥
v
2
≥
.
.
.
≥
v
n
≥
0
{\displaystyle v_{1}\geq v_{2}\geq ...\geq v_{n}\geq 0\;}
S
i
=
∑
k
=
1
i
u
k
{\displaystyle S_{i}=\sum _{k=1}^{i}u_{k}\;}
m ir M , tai remiantis Abelio lema, suma
∑
i
=
1
n
g
(
x
i
−
1
)
μ
i
Δ
x
1
{\displaystyle \sum _{i=1}^{n}g(x_{i-1})\mu _{i}\;\Delta x_{1}\;}
g
(
x
1
−
1
)
=
g
(
x
0
)
=
g
(
a
)
{\displaystyle g(x_{1-1})=g(x_{0})=g(a)}
[Paraboloido pataisymas .
Kadangi
v
1
≥
v
2
≥
.
.
.
≥
v
n
≥
0
{\displaystyle v_{1}\geq v_{2}\geq ...\geq v_{n}\geq 0\;}
S
i
=
∑
k
=
1
i
u
k
=
∑
k
=
1
i
μ
k
Δ
x
k
{\displaystyle S_{i}=\sum _{k=1}^{i}u_{k}=\sum _{k=1}^{i}\mu _{k}\Delta x_{k}\;}
m
(
x
i
−
a
)
{\displaystyle m(x_{i}-a)}
M
(
x
i
−
a
)
,
{\displaystyle M(x_{i}-a),}
∑
i
=
1
n
g
(
x
i
−
1
)
μ
i
Δ
x
1
{\displaystyle \sum _{i=1}^{n}g(x_{i-1})\mu _{i}\;\Delta x_{1}\;}
m
(
x
n
−
a
)
g
(
a
)
{\displaystyle m(x_{n}-a)g(a)}
M
(
x
n
−
a
)
g
(
a
)
{\displaystyle M(x_{n}-a)g(a)\;}
g
(
x
1
−
1
)
=
g
(
x
0
)
=
g
(
a
)
{\displaystyle g(x_{1-1})=g(x_{0})=g(a)}
Bet tada ir tos sumos riba, kai
Δ
→
0
,
{\displaystyle \Delta \to 0,}
m
g
(
a
)
{\displaystyle mg(a)}
M
g
(
a
)
,
{\displaystyle Mg(a),}
g
(
a
)
m
≤
∫
a
b
f
(
x
)
g
(
x
)
d
x
≤
g
(
a
)
M
.
{\displaystyle g(a)m\leq \int _{a}^{b}f(x)g(x)\;dx\leq g(a)M.}
[Paraboloido pataisymas .
Bet tada ir tos sumos riba, kai
Δ
→
0
,
{\displaystyle \Delta \to 0,}
m
(
b
−
a
)
g
(
a
)
{\displaystyle m(b-a)g(a)}
M
(
b
−
a
)
g
(
a
)
,
{\displaystyle M(b-a)g(a),}
g
(
a
)
m
(
b
−
a
)
≤
∫
a
b
f
(
x
)
g
(
x
)
d
x
≤
g
(
a
)
M
(
b
−
a
)
.
{\displaystyle g(a)m(b-a)\leq \int _{a}^{b}f(x)g(x)\;dx\leq g(a)M(b-a).}
Tolydžioji funkcija
F
(
x
)
=
∫
a
x
f
(
t
)
d
t
{\displaystyle F(x)=\int _{a}^{x}f(t)\;dt\;}
A , esančią tarp jos tiksliųjų rėžių m ir M , t. y. yra toks taškas
ξ
,
{\displaystyle \xi ,}
F
(
ξ
)
=
∫
a
ξ
f
(
t
)
d
t
=
A
=
∫
a
b
f
(
x
)
g
(
x
)
d
x
g
(
a
)
.
{\displaystyle F(\xi )=\int _{a}^{\xi }f(t)\;dt\;=A={\frac {\int _{a}^{b}f(x)g(x)\;dx}{g(a)}}.}
Todėl
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
.
(
10.42
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx.\quad (10.42)}
[Paraboloido pataisymas .
Tolydžioji funkcija
F
(
x
)
=
∫
a
x
f
(
t
)
d
t
{\displaystyle F(x)=\int _{a}^{x}f(t)\;dt\;}
A , esančią tarp reikšmių
m
(
x
−
a
)
{\displaystyle m(x-a)}
M
(
x
−
a
)
,
{\displaystyle M(x-a),}
ξ
,
{\displaystyle \xi ,}
(
x
−
a
)
=
1
,
{\displaystyle (x-a)=1,}
F
(
ξ
)
=
∫
a
ξ
f
(
t
)
d
t
=
A
=
∫
a
b
f
(
x
)
g
(
x
)
d
x
(
x
−
a
)
g
(
a
)
.
{\displaystyle F(\xi )=\int _{a}^{\xi }f(t)\;dt\;=A={\frac {\int _{a}^{b}f(x)g(x)\;dx}{(x-a)g(a)}}.}
Bet yra toks skaičius A ,
A
=
∫
a
b
f
(
x
)
g
(
x
)
d
x
(
b
−
a
)
g
(
a
)
,
{\displaystyle A={\frac {\int _{a}^{b}f(x)g(x)\;dx}{(b-a)g(a)}},}
kad iš
g
(
a
)
m
(
b
−
a
)
≤
∫
a
b
f
(
x
)
g
(
x
)
d
x
≤
g
(
a
)
M
(
b
−
a
)
{\displaystyle g(a)m(b-a)\leq \int _{a}^{b}f(x)g(x)\;dx\leq g(a)M(b-a)}
gauname
g
(
a
)
A
(
b
−
a
)
=
g
(
a
)
(
b
−
a
)
∫
a
b
f
(
x
)
g
(
x
)
d
x
(
b
−
a
)
g
(
a
)
=
∫
a
b
f
(
x
)
g
(
x
)
d
x
.
{\displaystyle g(a)A(b-a)=g(a)(b-a){\frac {\int _{a}^{b}f(x)g(x)\;dx}{(b-a)g(a)}}=\int _{a}^{b}f(x)g(x)\;dx.}
Bet visgi gali būti, kad kai (b-a)=1, tai tolydžioji funkcija
F
(
x
)
=
∫
a
x
f
(
t
)
d
t
{\displaystyle F(x)=\int _{a}^{x}f(t)\;dt\;}
A , esančią tarp reikšmių
m
(
x
−
a
)
{\displaystyle m(x-a)}
M
(
x
−
a
)
,
{\displaystyle M(x-a),}
ξ
,
{\displaystyle \xi ,}
(
x
−
a
)
=
1
,
{\displaystyle (x-a)=1,}
F
(
ξ
)
=
∫
a
ξ
f
(
t
)
d
t
=
A
=
∫
a
b
f
(
x
)
g
(
x
)
d
x
(
x
−
a
)
g
(
a
)
.
{\displaystyle F(\xi )=\int _{a}^{\xi }f(t)\;dt\;=A={\frac {\int _{a}^{b}f(x)g(x)\;dx}{(x-a)g(a)}}.}
Čia
b
−
a
=
x
−
a
=
1.
{\displaystyle b-a=x-a=1.}
Tada
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
(
b
−
a
)
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
;
(
10.42
Paraboloido
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=(b-a)g(a)\int _{a}^{\xi }f(x)\;dx;\quad (10.42\;\;{\text{Paraboloido}})}
čia
b
−
a
=
1
,
a
<
ξ
<
b
.
{\displaystyle b-a=1,\;\;a<\xi <b.}
Jei nedidėjanti funkcija g(x) įgyja ir neigiamų reikšmių, tai funkcija
h
(
x
)
=
g
(
x
)
−
g
(
b
)
{\displaystyle h(x)=g(x)-g(b)}
[Jei segmente [a; b] funkcija g(x)<0 ir yra nedidėjanti, tai [įprastai] g(a)>g(b) (bet jei g(x) funkcija yra horizontali tiesi linija, tada gali būti ir taip
g
(
a
)
≥
g
(
b
)
,
{\displaystyle g(a)\geq g(b),}
h
(
x
)
=
g
(
x
)
−
g
(
b
)
>
0
{\displaystyle h(x)=g(x)-g(b)>0}
∫
a
b
f
(
x
)
[
g
(
x
)
−
g
(
b
)
]
d
x
=
[
g
(
a
)
−
g
(
b
)
]
∫
a
ξ
f
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}f(x)[g(x)-g(b)]\;dx=[g(a)-g(b)]\int _{a}^{\xi }f(x)\;dx.}
[Taip neteisingai:
∫
a
b
f
(
x
)
[
g
(
x
)
−
g
(
b
)
]
d
x
=
∫
a
b
f
(
x
)
g
(
x
)
d
x
−
∫
a
b
f
(
x
)
g
(
b
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
−
g
(
b
)
∫
a
b
f
(
x
)
d
x
,
{\displaystyle \int _{a}^{b}f(x)[g(x)-g(b)]\;dx=\int _{a}^{b}f(x)g(x)\;dx-\int _{a}^{b}f(x)g(b)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx-g(b)\int _{a}^{b}f(x)\;dx,}
nes
h
(
x
)
=
g
(
x
)
−
g
(
b
)
{\displaystyle h(x)=g(x)-g(b)}
∫
a
b
f
(
x
)
h
(
x
)
d
x
=
h
(
a
)
∫
a
ξ
f
(
x
)
d
x
=
[
g
(
a
)
−
g
(
b
)
]
∫
a
ξ
f
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}f(x)h(x)\;dx=h(a)\int _{a}^{\xi }f(x)\;dx=[g(a)-g(b)]\int _{a}^{\xi }f(x)\;dx.}
Iš čia po nesudetingų pertvarkymų gauname (10.16) formulę.
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
+
g
(
b
)
∫
ξ
b
f
(
x
)
d
x
.
(
10.16
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx+g(b)\int _{\xi }^{b}f(x)\;dx.\quad (10.16)}
[Bet pagal (10.42) formulę
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
.
(
10.42
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx.\quad (10.42)}
Vadinasi,
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
+
g
(
b
)
∫
ξ
b
f
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
,
(
10.
D
)
{\displaystyle g(a)\int _{a}^{\xi }f(x)\;dx+g(b)\int _{\xi }^{b}f(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx,\quad (10.D)}
g
(
b
)
∫
ξ
b
f
(
x
)
d
x
=
0.
{\displaystyle g(b)\int _{\xi }^{b}f(x)\;dx=0.}
Toks samprotavimas neteisingas, nes kairėje (10.D) lygybės pusėje
ξ
{\displaystyle \xi }
h
(
x
)
=
g
(
x
)
−
g
(
b
)
,
{\displaystyle h(x)=g(x)-g(b),}
ξ
{\displaystyle \xi }
ξ
{\displaystyle \xi }
[
∫
a
b
f
(
x
)
[
g
(
x
)
−
g
(
b
)
]
d
x
=
[
g
(
a
)
−
g
(
b
)
]
∫
a
ξ
f
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
−
g
(
b
)
∫
a
ξ
f
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
+
g
(
b
)
∫
ξ
a
f
(
x
)
d
x
,
{\displaystyle \int _{a}^{b}f(x)[g(x)-g(b)]\;dx=[g(a)-g(b)]\int _{a}^{\xi }f(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx-g(b)\int _{a}^{\xi }f(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx+g(b)\int _{\xi }^{a}f(x)\;dx,}
∫
a
b
f
(
x
)
[
g
(
x
)
−
g
(
b
)
]
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
+
g
(
b
)
∫
ξ
a
f
(
x
)
d
x
,
{\displaystyle \int _{a}^{b}f(x)[g(x)-g(b)]\;dx=g(a)\int _{a}^{\xi }f(x)\;dx+g(b)\int _{\xi }^{a}f(x)\;dx,}
∫
a
b
f
(
x
)
g
(
x
)
d
x
−
g
(
b
)
∫
a
b
f
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
+
g
(
b
)
∫
ξ
a
f
(
x
)
d
x
,
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx-g(b)\int _{a}^{b}f(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx+g(b)\int _{\xi }^{a}f(x)\;dx,}
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
+
g
(
b
)
∫
ξ
a
f
(
x
)
d
x
+
g
(
b
)
∫
a
b
f
(
x
)
d
x
,
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx+g(b)\int _{\xi }^{a}f(x)\;dx+g(b)\int _{a}^{b}f(x)\;dx,}
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
+
g
(
b
)
(
∫
ξ
a
f
(
x
)
d
x
+
∫
a
b
f
(
x
)
d
x
)
,
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx+g(b)(\int _{\xi }^{a}f(x)\;dx+\int _{a}^{b}f(x)\;dx),}
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
+
g
(
b
)
∫
ξ
b
f
(
x
)
d
x
.
(
10.16
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx+g(b)\int _{\xi }^{b}f(x)\;dx.\quad (10.16)}
Vadinasi, (10.16) formulė teisinga, remiantis išvesta (10.C) formule.]
[ Šiaip, pagalvojus, taip turėtų būti teisingai :Tolydžioji funkcija
F
(
x
)
=
∫
a
x
f
(
t
)
d
t
{\displaystyle F(x)=\int _{a}^{x}f(t)\;dt\;}
A , esančią tarp reikšmių
m
(
x
−
a
)
{\displaystyle m(x-a)}
M
(
x
−
a
)
,
{\displaystyle M(x-a),}
ξ
,
{\displaystyle \xi ,}
A
1
(
ξ
−
a
)
=
∫
a
ξ
f
(
t
)
d
t
,
{\displaystyle A_{1}(\xi -a)=\int _{a}^{\xi }f(t)\;dt,}
A
2
(
b
−
a
)
=
∫
a
b
f
(
t
)
d
t
.
{\displaystyle A_{2}(b-a)=\int _{a}^{b}f(t)\;dt.}
[pagal
∫
a
b
f
(
x
)
d
x
=
μ
(
b
−
a
)
.
(
10.12
)
{\displaystyle \int _{a}^{b}f(x)\;dx=\mu (b-a).\quad (10.12)}
ir
m
(
b
−
a
)
≤
∫
a
b
f
(
x
)
d
x
≤
M
(
b
−
a
)
.
(
10.12.1
)
{\displaystyle m(b-a)\leq \int _{a}^{b}f(x)\;dx\leq M(b-a).\quad (10.12.1)}
Ir iš
g
(
a
)
m
(
b
−
a
)
≤
∫
a
b
f
(
x
)
g
(
x
)
d
x
≤
g
(
a
)
M
(
b
−
a
)
,
{\displaystyle g(a)m(b-a)\leq \int _{a}^{b}f(x)g(x)\;dx\leq g(a)M(b-a),}
ir
m
(
b
−
a
)
≤
A
2
(
b
−
a
)
≤
M
(
b
−
a
)
,
{\displaystyle m(b-a)\leq A_{2}(b-a)\leq M(b-a),}
(taip neteisingai:
m
(
b
−
a
)
≤
A
1
(
ξ
−
a
)
≤
M
(
b
−
a
)
{\displaystyle m(b-a)\leq A_{1}(\xi -a)\leq M(b-a)}
gaunasi
g
(
a
)
m
(
b
−
a
)
≤
A
2
g
(
a
)
(
b
−
a
)
≤
g
(
a
)
M
(
b
−
a
)
,
{\displaystyle g(a)m(b-a)\leq A_{2}g(a)(b-a)\leq g(a)M(b-a),}
t. y.
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
A
2
g
(
ξ
)
(
b
−
a
)
=
g
(
ξ
)
∫
a
b
f
(
t
)
d
t
=
[
g
(
ξ
)
∫
a
b
f
(
x
)
d
x
]
.
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=A_{2}g(\xi )(b-a)=g(\xi )\int _{a}^{b}f(t)\;dt=[g(\xi )\int _{a}^{b}f(x)\;dx].}
Gavome, kad kai
ξ
{\displaystyle \xi }
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
g
(
ξ
)
∫
a
b
f
(
x
)
d
x
.
(
10.
B
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=g(\xi )\int _{a}^{b}f(x)\;dx.\quad (10.B)}
T. y. gavome (10.15) formulę ir nieko daugiau. (Ir tai dar nežinoma ar
∫
a
b
f
(
x
)
g
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx}
A
2
g
(
a
)
(
b
−
a
)
=
g
(
a
)
∫
a
b
f
(
t
)
d
t
{\displaystyle A_{2}g(a)(b-a)=g(a)\int _{a}^{b}f(t)\;dt}
Bet (10.B) formulėje
g
(
a
)
>
g
(
ξ
)
{\displaystyle g(a)>g(\xi )}
∫
a
b
f
(
x
)
d
x
>
∫
a
ξ
f
(
x
)
d
x
,
{\displaystyle \int _{a}^{b}f(x)\;dx>\int _{a}^{\xi }f(x)\;dx,}
Todėl galima rasti tokį
ξ
,
{\displaystyle \xi ,}
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
.
(
10.
C
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx.\quad (10.C)}
(10.42) formulė įrodyta.
Taip pat turėtų būti teisinga ir tai:
F
(
ξ
)
=
∫
a
ξ
f
(
t
)
d
t
=
A
=
∫
a
b
f
(
x
)
g
(
x
)
d
x
(
ξ
−
a
)
g
(
a
)
.
{\displaystyle F(\xi )=\int _{a}^{\xi }f(t)\;dt\;=A={\frac {\int _{a}^{b}f(x)g(x)\;dx}{(\xi -a)g(a)}}.}
T. y.
g
(
a
)
(
ξ
−
a
)
∫
a
ξ
f
(
t
)
d
t
=
∫
a
b
f
(
x
)
g
(
x
)
d
x
{\displaystyle g(a)(\xi -a)\int _{a}^{\xi }f(t)\;dt=\int _{a}^{b}f(x)g(x)\;dx}
arba
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
g
(
a
)
(
ξ
−
a
)
∫
a
ξ
f
(
x
)
d
x
,
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=g(a)(\xi -a)\int _{a}^{\xi }f(x)\;dx,}
a
<
ξ
<
b
.
{\displaystyle a<\xi <b.}
ξ
{\displaystyle \xi }
∫
a
b
f
(
x
)
g
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx.}
Turint galvoje (10.14) ir (10.15) formules:
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
μ
∫
a
b
g
(
x
)
d
x
.
(
10.14
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=\mu \int _{a}^{b}g(x)\;dx.\quad (10.14)}
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
f
(
ξ
)
∫
a
b
g
(
x
)
d
x
.
(
10.15
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=f(\xi )\int _{a}^{b}g(x)\;dx.\quad (10.15)}
Ir turint galvoje, kad
∫
a
b
f
(
x
)
g
(
x
)
d
x
≤
g
(
a
)
(
b
−
a
)
∫
a
b
f
(
x
)
d
x
,
(
10.
A
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx\leq g(a)(b-a)\int _{a}^{b}f(x)\;dx,\quad (10.A)}
nes pagal (10.15) formulę, atsižvelgiant į tai, kad g(a)>g(b), gauname:
∫
a
b
f
(
x
)
g
(
x
)
d
x
≤
g
(
a
)
∫
a
b
f
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx\leq g(a)\int _{a}^{b}f(x)\;dx.}
Ir jeigu (b-a)>1, tai (10.A) formulė garantuotai teisinga.
Vadinasi, (10.42) formulė teisinga, kai
b
−
a
≥
1
,
a
<
ξ
<
b
.
{\displaystyle b-a\geq 1,\;\;a<\xi <b.}
∫
a
b
f
(
x
)
g
(
x
)
d
x
=
g
(
a
)
∫
a
ξ
f
(
x
)
d
x
.
(
10.42
)
{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx.\quad (10.42)}
]
![{\displaystyle \sum _{i=1}^{n}a_{i}b_{i}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}\left[\sum _{i=1}^{n}b_{i}^{p'}\right]^{1 \over p'},\quad (10.27)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4b7363721773efd93cb9865a6357a50e55a427e)
![{\displaystyle A_{i}={\frac {a_{i}}{[\sum _{i=1}^{n}a_{i}^{p}]^{1 \over p}}}\;,\quad B_{i}={\frac {b_{i}}{[\sum _{i=1}^{n}b_{i}^{p'}]^{1 \over p'}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b39117121a3be5405937c1fe7f981fd88c548dbf)
![{\displaystyle \sum _{i=1}^{n}A_{i}^{p}=\sum _{i=1}^{n}\left({\frac {a_{i}}{[\sum _{i=1}^{n}a_{i}^{p}]^{1 \over p}}}\right)^{p}=\sum _{i=1}^{n}{\frac {a_{i}^{p}}{[\sum _{i=1}^{n}a_{i}^{p}]^{p \over p}}}=\sum _{i=1}^{n}{\frac {a_{i}^{p}}{[\sum _{i=1}^{n}a_{i}^{p}]}}={\frac {\sum _{i=1}^{n}a_{i}^{p}}{[\sum _{i=1}^{n}a_{i}^{p}]}}=1.\quad ({\text{Paraboloido}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/46610190a23d38e1da79562d9846bf2d030fd4a2)
![{\displaystyle \sum _{i=1}^{n}A_{i}B_{i}={\frac {\sum _{i=1}^{n}a_{i}b_{i}}{[\sum _{i=1}^{n}a_{i}^{p}]^{1 \over p}[\sum _{i=1}^{n}b_{i}^{p'}]^{1 \over p'}}}\leq 1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f7e2fae005fb64d1a905d737d86defe4e0aa79e)
![{\displaystyle \left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{1 \over p}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p},\quad (10.31)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16584dcaf45b2db41eb2c489446b91f01aa4ea37)
![{\displaystyle a_{i}(a_{i}+b_{i})^{3-1}+b_{i}(a_{i}+b_{i})^{3-1}=a_{i}(a_{i}^{2}+2a_{i}b_{i}+b_{i}^{2})+b_{i}(a_{i}^{2}+2a_{i}b_{i}+b_{i}^{2})=[a_{i}^{3}+2a_{i}^{2}b_{i}+a_{i}b_{i}^{2}]+[a_{i}^{2}b_{i}+2a_{i}b_{i}^{2}+b_{i}^{3}]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/207fec28c69d92df9f8f05694271a0b3485541dc)
![{\displaystyle \sum _{i=1}^{n}(a_{i}+b_{i})^{p}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}\left[\sum _{i=1}^{n}(a_{i}+b_{i})^{(p-1)p'}\right]^{1 \over p'}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p}\left[\sum _{i=1}^{n}(a_{i}+b_{i})^{(p-1)p'}\right]^{1 \over p'}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2f52c87a3402ebf486bdd9000d4460668718337)
![{\displaystyle =\left(\left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p}\right)\left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{p-1 \over p}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8fba8ea4f83097a82e3c0bb31eb0d947247be271)
![{\displaystyle \left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{p-1 \over p},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40c164549e4c30fd2f40c4e01e08806ba4f7a937)
![{\displaystyle \sum _{i=1}^{n}(a_{i}+b_{i})^{p}\leq \left(\left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p}\right)\left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{p-1 \over p},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76acb043a14853bbe7583d89bc29d5165797a389)
![{\displaystyle \left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]\left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{-{\frac {p-1}{p}}}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/919768eb8f909c269fc1042c8535a35a6c8bf2cd)
![{\displaystyle \left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{1-{\frac {p-1}{p}}}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c405341118a8dcb0e742168a013e30d4433d2dcb)
![{\displaystyle \left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{\frac {p-(p-1)}{p}}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18d4c330fc5970b1af643c2c196fb38786b2434d)
![{\displaystyle \left[\sum _{i=1}^{n}(a_{i}+b_{i})^{p}\right]^{\frac {1}{p}}\leq \left[\sum _{i=1}^{n}a_{i}^{p}\right]^{1 \over p}+\left[\sum _{i=1}^{n}b_{i}^{p}\right]^{1 \over p}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a888b1a008676650ff61d3c77269925654d4edc4)
![{\displaystyle |f(x)|^{r}|f(x)|^{r-[r]};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ce23447b3bc85559ae1ce6562520a2646eb57cd)
![{\displaystyle |f(x)|^{[r]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c285bdfa43376f4f0a512f4e97012e0c2cc5ea9f)
![{\displaystyle |f(x)|^{r-[r]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27a9bff7dbf1fc663b09d492995b27db261babca)
![{\displaystyle S-s\leq \left[\sum _{i=1}^{n}(M_{i}^{1/p}-m_{i}^{1/p})^{p}\Delta x_{i}\right]^{1/p}\left[\sum _{i=1}^{n}\Delta x_{i}\right]^{1/p'}.\quad (10.34)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb5d13cb9cddb53101058668a971f2b9719a4a9a)
![{\displaystyle \left[1-\left({\frac {m_{i}}{M_{i}}}\right)^{1/p}\right]^{p}\leq 1-{\frac {m_{i}}{M_{i}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8425414d204427991b5bfc9d6dbe8447ecca04b)
![{\displaystyle S-s\leq \left[\sum _{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}\right]^{1/p}(b-a)^{1/p'}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4314e5378fe1dbb92feffd8a0330867e0ea62c23)
![{\displaystyle S-s\leq \left[\sum _{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}\right]^{1/p}(b-a)^{\frac {p-1}{p}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/526fede8ac819b48bb7438f527260cbc5659fb6c)
^{p-1}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4148238c3d2a564ee01b97741163f245135ef22)
^{p-1}<\varepsilon ^{p}(b-a)^{p-1}<\varepsilon ^{p},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80c4aea061b2d30fa8648ad36dfe0f362b1b54b0)
![{\displaystyle \left|\int _{a}^{b}f(x)g(x)\;dx\right|\leq \left[\int _{a}^{b}|f(x)|^{p}\;dx\right]^{1/p}\left[\int _{a}^{b}|g(x)|^{p'}\;dx\right]^{1/p'},\quad (10.36)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef4685672505619955f7143c17f286b191d00ccf)
![{\displaystyle A(x)={\frac {|f(x)|}{[\int _{a}^{b}|f(x)|^{p}\;dx]^{1/p}}},\quad B(x)={\frac {|g(x)|}{[\int _{a}^{b}|g(x)|^{p'}\;dx]^{1/p'}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b08d3933cd1a5bab7e4dcf2736ad0529c492ceee)
![{\displaystyle \left[\int _{a}^{b}|f(x)|^{p}\;dx\right]^{1/p}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/042a8487c6da18990cf06d2cd1601e885c27e799)
![{\displaystyle \left[\int _{a}^{b}|g(x)|^{p'}\;dx\right]^{1/p'}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b978e58b127f7e12bcaff52ef4175bbafe1e9f25)
![{\displaystyle \int _{a}^{b}|f(x)||g(x)|\;dx\leq \left[\int _{a}^{b}|f(x)|^{p}\;dx\right]^{1/p}\left[\int _{a}^{b}|g(x)|^{p'}\;dx\right]^{1/p'}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8b6a5f9735b7b82bd7fdb19f146e6c016814150)
![{\displaystyle \int _{a}^{b}{\frac {|f(x)|}{[\int _{a}^{b}|f(x)|^{p}\;dx]^{1/p}}}{\frac {|g(x)|}{[\int _{a}^{b}|g(x)|^{p'}\;dx]^{1/p'}}}\;dx\leq {\frac {1}{p}}\int _{a}^{b}{\frac {|f(x)|^{p}}{[\int _{a}^{b}|f(x)|^{p}\;dx]^{p/p}}}\;dx+{\frac {1}{p'}}\int _{a}^{b}{\frac {|g(x)|^{p'}}{[\int _{a}^{b}|g(x)|^{p'}\;dx]^{p'/p'}}}\;dx\leq {\frac {1}{p}}+{\frac {1}{p'}}=1,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e24101bad252cea5150d30394838fceb85592f85)
![{\displaystyle {\frac {1}{[\int _{a}^{b}|f(x)|^{p}\;dx]^{1/p}}}{\frac {1}{[\int _{a}^{b}|g(x)|^{p'}\;dx]^{1/p'}}}\int _{a}^{b}|f(x)||g(x)|\;dx\leq {\frac {1}{p}}{\frac {\int _{a}^{b}|f(x)|^{p}\;dx}{[\int _{a}^{b}|f(x)|^{p}\;dx]}}+{\frac {1}{p'}}{\frac {\int _{a}^{b}|g(x)|^{p'}\;dx}{[\int _{a}^{b}|g(x)|^{p'}\;dx]}}\leq {\frac {1}{p}}+{\frac {1}{p'}}=1,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/639e86ed9edb567d3d7c5e847df016d532facf74)
![{\displaystyle {\frac {1}{[\int _{a}^{b}|f(x)|^{p}\;dx]^{1/p}}}{\frac {1}{[\int _{a}^{b}|g(x)|^{p'}\;dx]^{1/p'}}}\int _{a}^{b}|f(x)||g(x)|\;dx\leq {\frac {1}{p}}\cdot 1+{\frac {1}{p'}}\cdot 1={\frac {1}{p}}+{\frac {1}{p'}}=1,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/930b7ef192b465ea7682a2839cbd2ff69980f4b0)
![{\displaystyle \int _{a}^{b}|f(x)||g(x)|\;dx\leq [\int _{a}^{b}|f(x)|^{p}\;dx]^{1/p}[\int _{a}^{b}|g(x)|^{p'}\;dx]^{1/p'}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f433ab2f66603a3f6bf2c12fb330e3b7efaa5433)
![{\displaystyle \left(\int _{a}^{b}[f(x)+g(x)]^{p}\;dx\right)^{1/p}\leq \left(\int _{a}^{b}[f(x)]^{p}\;dx\right)^{1/p}+\left(\int _{a}^{b}[g(x)]^{p}\;dx\right)^{1/p};\quad (10.40)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a330e9b5afed30084dbe70cc13c64ea671006426)
![{\displaystyle \int _{a}^{b}[f(x)+g(x)]^{p}\;dx=\int _{a}^{b}f(x)[f(x)+g(x)]^{p-1}\;dx+\int _{a}^{b}g(x)[f(x)+g(x)]^{p-1}\;dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d2a5a91a538098064b8763e8b0e82e3fc4882a0)
![{\displaystyle \int _{a}^{b}[f(x)+g(x)]^{p}\;dx\leq \left[\int _{a}^{b}[f(x)]^{p}\;dx\right]^{1 \over p}\left[\int _{a}^{b}(f(x)+g(x))^{(p-1)p'}\;dx\right]^{1 \over p'}+\left[\int _{a}^{b}[g(x)]^{p}\;dx\right]^{1 \over p}\left[\int _{a}^{b}(f(x)+g(x))^{(p-1)p'}\;dx\right]^{1 \over p'}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/292dde8996fc575586de4719b0f07cdada68df4f)
![{\displaystyle =\left(\left[\int _{a}^{b}[f(x)]^{p}\;dx\right]^{1 \over p}+\left[\int _{a}^{b}[g(x)]^{p}\;dx\right]^{1 \over p}\right)\left[\int _{a}^{b}(f(x)+g(x))^{p}\;dx\right]^{p-1 \over p}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3c2d8b9fb870b828d012c1e1c3dc06b09275705)
![{\displaystyle \left[\int _{a}^{b}(f(x)+g(x))^{p}\;dx\right]^{p-1 \over p},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c14f003dfdd23626ebf266cf547c26c0258a467)
![{\displaystyle \left[\int _{a}^{b}[f(x)+g(x)]^{p}\;dx\right]\left[\int _{a}^{b}(f(x)+g(x))^{p}\;dx\right]^{-{\frac {p-1}{p}}}\leq \left[\int _{a}^{b}[f(x)]^{p}\;dx\right]^{1 \over p}+\left[\int _{a}^{b}[g(x)]^{p}\;dx\right]^{1 \over p},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e6a9f861004f7bbdbd350183cbae8193144e539)
![{\displaystyle \left[\int _{a}^{b}(f(x)+g(x))^{p}\;dx\right]^{1-{\frac {p-1}{p}}}\leq \left[\int _{a}^{b}[f(x)]^{p}\;dx\right]^{1 \over p}+\left[\int _{a}^{b}[g(x)]^{p}\;dx\right]^{1 \over p},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f73cf995fd50d8fbe4555b5f6d5d1a182dba98f)
![{\displaystyle \left[\int _{a}^{b}(f(x)+g(x))^{p}\;dx\right]^{\frac {p-(p-1)}{p}}\leq \left[\int _{a}^{b}[f(x)]^{p}\;dx\right]^{1 \over p}+\left[\int _{a}^{b}[g(x)]^{p}\;dx\right]^{1 \over p},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8da210ddf63eed8f8bc83173306f9832ba8eada0)
![{\displaystyle \left[\int _{a}^{b}(f(x)+g(x))^{p}\;dx\right]^{\frac {1}{p}}\leq \left[\int _{a}^{b}[f(x)]^{p}\;dx\right]^{1 \over p}+\left[\int _{a}^{b}[g(x)]^{p}\;dx\right]^{1 \over p}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66e81fa5f98f8809714e161c0789f858a4cddbf7)
![{\displaystyle \left(\int _{a}^{b}[f_{1}(x)+f_{2}(x)+...+f_{n}(x)]^{p}\;dx\right)^{1/p}\leq \left(\int _{a}^{b}[f_{1}(x)]^{p}\;dx\right)^{1/p}+\left(\int _{a}^{b}[f_{2}(x)]^{p}\;dx\right)^{1/p}+...+\left(\int _{a}^{b}[f_{n}(x)]^{p}\;dx\right)^{1/p}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c8c4a1e05349e6f8446c921001b251001072f8e)
![{\displaystyle \int _{a}^{b}[f(x)\pm g(x)]\;dx=\int _{a}^{b}f(x)\;dx\pm \int _{a}^{b}g(x)\;dx.\quad (10.8)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/736c3220874733e4e6545cb58384cb76ec84504b)
![{\displaystyle \sum _{i=1}^{n}[f(\xi _{i})\pm g(\xi _{i})]\Delta x_{i}=\sum _{i=1}^{n}f(\xi _{i})\Delta x_{i}\pm \sum _{i=1}^{n}g(\xi _{i})\Delta x_{i}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bb4203b76119f1e2177ee44bb86f19f7f5f9107)
![{\displaystyle [x_{i-1};\;x_{i}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9026a925b74f3b5ba4efe91e547aa7bbb28380a)
![{\displaystyle f(x'')g(x'')-f(x')g(x')=[f(x'')-f(x')]g(x'')+[g(x'')-g(x')]f(x').}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d44b21db938689b856d7b423186826638a09f72c)
![{\displaystyle [x_{i-1};\;x_{i}],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98f95060e0c30ae748c619513ae630be0533a9b3)
![{\displaystyle \int _{a}^{b}[f(x)-m]\;dx\geq 0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53b0071d22ea6609084712db486929be251e5e8d)
![{\displaystyle [x_{i-1};x_{i}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a1c62e61054537b8a6952bf19f623388c38b138)
![{\displaystyle A[(v_{1}-v_{2})+(v_{2}-v_{3})+(v_{3}-v_{4})+...+(v_{n-1}-v_{n})+v_{n}]\leq S_{1}(v_{1}-v_{2})+S_{2}(v_{2}-v_{3})+S_{3}(v_{3}-v_{4})+...+S_{n-1}(v_{n-1}-v_{n})+S_{n}v_{n}\leq }](https://wikimedia.org/api/rest_v1/media/math/render/svg/fad1072e31fd1ea4298729c3a6c0736320de5e93)
![{\displaystyle \leq B[(v_{1}-v_{2})+(v_{2}-v_{3})+(v_{3}-v_{4})+...+(v_{n-1}-v_{n})+v_{n}].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0546290d1a69f5ec70d6d2b8644c24a08384a4d1)
![{\displaystyle [x_{i-1};\;x_{1}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2b104c8b3c7fa5dfe3e757cd2359096e3bb552b)
![{\displaystyle \int _{a}^{b}f(x)[g(x)-g(b)]\;dx=[g(a)-g(b)]\int _{a}^{\xi }f(x)\;dx.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90fb94755934f6dc6147fb5881e2446da2763332)
![{\displaystyle \int _{a}^{b}f(x)[g(x)-g(b)]\;dx=\int _{a}^{b}f(x)g(x)\;dx-\int _{a}^{b}f(x)g(b)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx-g(b)\int _{a}^{b}f(x)\;dx,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0bf3f4a70fa5cc9c2aaddf1dc7484483ecbcbcca)
![{\displaystyle \int _{a}^{b}f(x)h(x)\;dx=h(a)\int _{a}^{\xi }f(x)\;dx=[g(a)-g(b)]\int _{a}^{\xi }f(x)\;dx.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/148febd4d5ce640f42aa9974cda56903afd6cb55)
![{\displaystyle \int _{a}^{b}f(x)[g(x)-g(b)]\;dx=[g(a)-g(b)]\int _{a}^{\xi }f(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx-g(b)\int _{a}^{\xi }f(x)\;dx=g(a)\int _{a}^{\xi }f(x)\;dx+g(b)\int _{\xi }^{a}f(x)\;dx,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a2d4cd6c7b38554c8ec3f32e60cb04b39e03e22)
![{\displaystyle \int _{a}^{b}f(x)[g(x)-g(b)]\;dx=g(a)\int _{a}^{\xi }f(x)\;dx+g(b)\int _{\xi }^{a}f(x)\;dx,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9da1a5aa17821fd1ad52c099276020d0cea7658f)
![{\displaystyle \int _{a}^{b}f(x)g(x)\;dx=A_{2}g(\xi )(b-a)=g(\xi )\int _{a}^{b}f(t)\;dt=[g(\xi )\int _{a}^{b}f(x)\;dx].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e2499b34de4f218e94ff6a9cd229f4422aa453c)
