Analisi matematica > Calcolo degli integrali di Riemann
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Algebra degli integrali di Riemann
modifica
Siano
f
,
g
∈
R
[
a
,
b
]
,
λ
∈
R
{\displaystyle f,g\in {\mathcal {R}}_{[a,b]},\ \lambda \in \mathbb {R} }
∫
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))dx=\int _{a}^{b}f(x)dx+\int _{a}^{b}g(x)dx}
cioè l'integrale di una somma è uguale alla somma degli integrali. Infatti, siano
σ
′
{\displaystyle \sigma '}
σ
″
{\displaystyle \sigma ''}
[
a
,
b
]
{\displaystyle [a,b]}
σ
=
σ
′
∪
σ
″
{\displaystyle \sigma =\sigma '\cup \sigma ''}
σ
{\displaystyle \sigma }
∫
a
_
b
(
f
(
x
)
+
g
(
x
)
)
d
x
≤
S
(
f
+
g
,
σ
)
≤
S
(
f
,
σ
)
+
S
(
g
,
σ
)
≤
S
(
f
,
σ
′
)
+
s
(
g
,
σ
″
)
{\displaystyle \int _{a}^{\_b}(f(x)+g(x))dx\leq S(f+g,\sigma )\leq S(f,\sigma )+S(g,\sigma )\leq S(f,\sigma ')+s(g,\sigma '')}
Dunque
∫
a
_
b
(
f
(
x
)
+
g
(
x
)
)
d
x
≤
∫
a
_
b
f
(
x
)
d
x
+
∫
a
_
b
g
(
x
)
d
x
=
∫
a
b
f
(
x
)
d
x
+
∫
a
b
g
(
x
)
d
x
{\displaystyle \int _{a}^{\_b}(f(x)+g(x))dx\leq \int _{a}^{\_b}f(x)dx+\int _{a}^{\_b}g(x)dx=\int _{a}^{b}f(x)dx+\int _{a}^{b}g(x)dx}
D'altra parte
∫
_
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))dx\geq \int _{a}^{b}f(x)dx+\int _{a}^{b}g(x)dx}
In conclusione
∫
a
_
b
(
f
(
x
)
+
g
(
x
)
)
d
x
≤
∫
a
b
f
(
x
)
d
x
+
∫
a
b
g
(
x
)
d
x
≤
∫
_
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)dx+\int _{a}^{b}g(x)dx\leq \int _{\_a}^{b}(f(x)+g(x))dx\leq \int _{a}^{\_b}(f(x)+g(x))dx}
Ma il primo e l'ultimo membro sono uguali! Dunque tutti questi
≤
{\displaystyle \leq }
∫
a
_
b
(
f
(
x
)
+
g
(
x
)
)
d
x
=
∫
_
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))dx=\int _{\_a}^{b}(f(x)+g(x))dx=\int _{a}^{b}f(x)dx+\int _{a}^{b}g(x)dx}
Moltiplicazione di un integrale per un numero reale
modifica
∫
a
b
λ
f
(
x
)
d
x
=
λ
∫
a
b
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}\lambda f(x)dx=\lambda \int _{a}^{b}f(x)dx}
Infatti, se
λ
≠
0
{\displaystyle \lambda \neq 0}
λ
>
0
{\displaystyle \lambda >0}
λ
S
(
f
,
σ
)
=
S
(
λ
f
,
σ
)
{\displaystyle \lambda S(f,\sigma )=S(\lambda f,\sigma )}
λ
s
(
f
,
σ
)
=
s
(
λ
f
,
σ
)
{\displaystyle \lambda s(f,\sigma )=s(\lambda f,\sigma )}
Se
λ
<
0
{\displaystyle \lambda <0}
λ
S
(
f
,
σ
)
=
s
(
λ
,
σ
)
{\displaystyle \lambda S(f,\sigma )=s(\lambda ,\sigma )}
λ
s
(
f
,
σ
)
=
S
(
λ
f
,
σ
)
{\displaystyle \lambda s(f,\sigma )=S(\lambda f,\sigma )}
Supponiamo
f
(
x
)
≤
g
(
x
)
,
∀
x
∈
[
a
,
b
]
{\displaystyle f(x)\leq g(x),\ \forall x\in [a,b]}
∫
a
b
f
(
x
)
d
x
≤
∫
a
b
g
(
x
)
{\displaystyle \int _{a}^{b}f(x)dx\leq \int _{a}^{b}g(x)}
Infatti
f
(
x
)
≤
g
(
x
)
,
∀
x
∈
[
a
,
b
]
⇒
S
(
f
,
σ
)
≤
S
(
g
,
σ
)
,
∀
σ
∈
Ω
[
a
,
b
]
{\displaystyle f(x)\leq g(x),\ \forall x\in [a,b]\Rightarrow S(f,\sigma )\leq S(g,\sigma ),\ \forall \sigma \in \Omega _{[a,b]}}
∫
a
b
f
(
x
)
d
x
≤
∫
a
b
g
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x)dx\leq \int _{a}^{b}g(x)dx}
Valore assoluto di un integrale
modifica
Sia
|
f
|
∈
R
[
a
,
b
]
{\displaystyle |f|\in {\mathcal {R}}_{[a,b]}}
|
∫
a
b
f
(
x
)
f
x
|
≤
∫
a
b
|
f
(
x
)
|
d
x
{\displaystyle \left|\int _{a}^{b}f(x)fx\right|\leq \int _{a}^{b}|f(x)|dx}
Infatti, prendiamo un
ε
>
0
{\displaystyle \varepsilon >0}
σ
=
{
x
0
,
…
,
x
n
}
{\displaystyle \sigma =\{x_{0},\dots ,x_{n}\}}
S
(
f
,
σ
)
−
s
(
f
,
σ
)
<
ε
{\displaystyle S(f,\sigma )-s(f,\sigma )<\varepsilon }
S
(
|
f
|
,
σ
)
−
s
(
|
f
|
,
σ
)
=
∑
i
=
1
n
(
sup
I
i
|
f
|
−
inf
I
i
|
f
|
)
(
x
i
+
1
−
x
i
)
≤
{\displaystyle S(|f|,\sigma )-s(|f|,\sigma )=\sum _{i=1}^{n}\left(\sup _{I_{i}}|f|-\inf _{I_{i}}|f|\right)(x_{i+1}-x_{i})\leq }
∑
i
=
1
n
(
sup
I
i
f
−
inf
I
i
f
)
(
x
i
+
1
−
x
i
)
≤
S
(
f
,
σ
)
−
s
(
f
,
σ
)
<
ε
{\displaystyle \sum _{i=1}^{n}\left(\sup _{I_{i}}f-\inf _{I_{i}}f\right)(x_{i+1}-x_{i})\leq S(f,\sigma )-s(f,\sigma )<\varepsilon }
Per il Teorema di Riemann abbiamo provato che
|
f
|
{\displaystyle |f|}
[
a
,
b
]
{\displaystyle [a,b]}
±
f
≤
|
f
|
{\displaystyle \pm f\leq |f|}
±
∫
a
b
f
(
x
)
d
x
≤
∫
a
b
|
f
(
x
)
|
d
x
{\displaystyle \pm \int _{a}^{b}f(x)dx\leq \int _{a}^{b}|f(x)|dx}
Dunque
|
∫
a
b
f
(
x
)
d
x
|
=
max
{
∫
a
b
f
(
x
)
d
x
,
−
∫
a
b
f
(
x
)
d
x
}
≤
∫
a
b
|
f
(
x
)
|
d
x
{\displaystyle \left|\int _{a}^{b}f(x)dx\right|=\max \left\{\int _{a}^{b}f(x)dx,-\int _{a}^{b}f(x)dx\right\}\leq \int _{a}^{b}|f(x)|dx}
modifica
Sia
x
0
∈
[
a
,
b
]
{\displaystyle x_{0}\in [a,b]}
h
≠
0
{\displaystyle h\neq 0}
x
o
+
h
∈
[
a
,
b
]
{\displaystyle x_{o}+h\in [a,b]}
I
f
(
x
0
+
h
)
−
I
f
(
x
0
)
=
∫
a
x
0
+
h
f
(
t
)
d
t
−
∫
a
x
0
f
(
t
)
d
t
=
∫
x
0
x
0
+
h
f
(
t
)
d
t
=
μ
(
h
)
h
{\displaystyle I_{f}(x_{0}+h)-I_{f}(x_{0})=\int _{a}^{x_{0}+h}f(t)dt-\int _{a}^{x_{0}}f(t)dt=\int _{x_{0}}^{x_{0}+h}f(t)dt=\mu (h)h}
ove
μ
(
h
)
=
1
h
∫
x
0
x
0
+
h
f
(
t
)
d
t
{\displaystyle \mu (h)={\frac {1}{h}}\int _{x_{0}}^{x_{0}+h}f(t)dt}
inf
[
x
0
,
x
0
+
h
]
f
≤
μ
(
h
)
≤
sup
[
x
0
,
x
0
+
h
]
f
{\displaystyle \inf _{[x_{0},x_{0}+h]}f\leq \mu (h)\leq \sup _{[x_{0},x_{0}+h]}f}
lim
h
→
0
(
I
f
(
x
0
+
h
)
−
I
f
(
x
0
)
)
=
0
{\displaystyle \lim _{h\to 0}(I_{f}(x_{0}+h)-I_{f}(x_{0}))=0}
Quindi
I
f
{\displaystyle I_{f}}
2. Sia
x
0
{\displaystyle x_{0}}
f
{\displaystyle f}
ε
{\displaystyle \varepsilon }
δ
{\displaystyle \delta }
|
t
−
x
0
|
<
δ
,
∀
t
∈
[
a
,
b
]
{\displaystyle |t-x_{0}|<\delta ,\ \ \forall t\in [a,b]}
f
(
x
0
)
−
ε
<
f
(
t
)
<
f
(
x
0
)
+
ε
{\displaystyle f(x_{0})-\varepsilon <f(t)<f(x_{0})+\varepsilon }
dunque
Nota: problema con la dimostrazione. chiedere al prof.
modifica
Scomponiamo questo integrale come somma di più integrali. Abbiamo:
I
(
x
)
=
∫
x
0
x
f
(
t
)
d
t
=
∫
x
0
a
f
(
t
)
d
t
+
∫
a
x
f
(
t
)
d
t
=
∫
a
x
0
f
(
t
)
d
t
+
∫
a
x
f
(
t
)
d
t
{\displaystyle I(x)=\int _{x_{0}}^{x}f(t)dt=\int _{x_{0}}^{a}f(t)dt+\int _{a}^{x}f(t)dt=\int _{a}^{x_{0}}f(t)dt+\int _{a}^{x}f(t)dt}
Ora, notiamo che
∫
a
x
0
f
(
t
)
d
t
{\displaystyle \int _{a}^{x_{0}}f(t)dt}
I
{\displaystyle I}
∫
a
x
0
f
(
t
)
d
t
=
c
{\displaystyle \int _{a}^{x_{0}}f(t)dt=c}
Adesso, abbiamo che
(
c
+
∫
a
x
f
(
t
)
d
t
)
′
=
(
c
+
I
f
(
x
)
)
′
=
f
(
x
)
{\displaystyle \left(c+\int _{a}^{x}f(t)dt\right)'=\left(c+I_{f}(x)\right)'=f(x)}
.
◻
{\displaystyle \Box }
Sia
f
∈
C
1
(
[
a
,
b
]
,
R
)
{\displaystyle f\in {\mathcal {C}}^{1}([a,b],\mathbb {R} )}
∫
a
b
f
′
(
x
)
d
x
=
f
(
b
)
−
f
(
a
)
{\displaystyle \int _{a}^{b}f'(x)dx=f(b)-f(a)}
Innanzitutto, essendo
f
′
(
x
)
{\displaystyle f'(x)}
[
a
,
b
]
{\displaystyle [a,b]}
f
{\displaystyle f}
f
′
{\displaystyle f'}
∫
a
b
f
′
(
x
)
d
x
=
[
f
(
x
)
]
a
b
=
f
(
b
)
−
f
(
a
)
{\displaystyle \int _{a}^{b}f'(x)dx=[f(x)]_{a}^{b}=f(b)-f(a)}
◻
{\displaystyle \Box }
modifica
Definiamo il concetto di primitiva di una funzione. Consideriamo una funzione
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\to \mathbb {R} }
F
:
[
a
,
b
]
→
R
{\displaystyle F:[a,b]\to \mathbb {R} }
F
′
(
x
)
=
f
(
x
)
{\displaystyle F'(x)=f(x)}
primitiva di
f
{\displaystyle f}
Per il precedente corollario, sappiamo ora una cosa molto importante: ogni funzione continua ha almeno una primitiva, cioè la sua funzione integrale!
Enunciamo ora il seguente, fondamentale, Teorema:
Per ipotesi
f
{\displaystyle f}
∀
ε
>
0
∃
σ
∈
Ω
[
a
,
b
]
:
S
(
f
,
σ
)
−
s
(
f
,
σ
)
<
ε
{\displaystyle \forall \varepsilon >0\exists \sigma \in \Omega _{[a,b]}\ :\ S(f,\sigma )-s(f,\sigma )<\varepsilon }
F
(
b
)
−
F
(
a
)
{\displaystyle F(b)-F(a)}
F
(
b
)
−
F
(
a
)
=
F
(
x
n
)
−
F
(
x
0
)
=
∑
i
=
1
n
(
F
(
x
i
)
−
F
(
x
i
−
1
)
)
{\displaystyle F(b)-F(a)=F(x_{n})-F(x_{0})=\sum _{i=1}^{n}(F(x_{i})-F(x_{i-1}))}
Per il Teorema del valor medio, sappiamo che esiste un
c
i
∈
]
x
i
−
1
,
x
i
[
:
F
(
x
i
)
−
F
(
x
i
−
1
)
=
F
′
(
c
i
)
(
x
i
−
x
i
−
1
)
{\displaystyle c_{i}\in ]x_{i-1},x_{i}[\ :\ F(x_{i})-F(x_{i-1})=F'(c_{i})(x_{i}-x_{i-1})}
cioè
f
(
c
i
)
(
x
i
−
x
i
−
1
)
{\displaystyle f(c_{i})(x_{i}-x_{i-1})}
perché
F
{\displaystyle F}
f
{\displaystyle f}
Mettendo insieme le due espressioni, otteniamo:
F
(
b
)
−
F
(
a
)
=
∑
i
=
1
n
f
(
c
i
)
(
x
i
−
x
i
−
1
)
≤
∑
i
=
1
n
sup
[
x
i
−
1
,
x
i
]
f
⋅
(
x
i
−
x
i
−
1
)
{\displaystyle F(b)-F(a)=\sum _{i=1}^{n}f(c_{i})(x_{i}-x_{i-1})\leq \sum _{i=1}^{n}\sup _{[x_{i-1},x_{i}]}f\cdot (x_{i}-x_{i-1})}
=
S
(
f
,
σ
)
<
s
(
f
,
σ
)
+
ε
≤
∫
a
b
f
(
x
)
d
x
+
ε
{\displaystyle =S(f,\sigma )<s(f,\sigma )+\varepsilon \leq \int _{a}^{b}f(x)dx+\varepsilon }
F
(
b
)
−
F
(
a
)
=
∑
i
=
1
n
f
(
c
i
)
(
x
i
−
x
i
−
1
)
≥
∑
i
=
1
n
inf
[
x
i
−
1
,
x
i
]
f
⋅
(
x
i
−
x
i
−
1
)
{\displaystyle F(b)-F(a)=\sum _{i=1}^{n}f(c_{i})(x_{i}-x_{i-1})\geq \sum _{i=1}^{n}\inf _{[x_{i-1},x_{i}]}f\cdot (x_{i}-x_{i-1})}
=
s
(
f
,
σ
)
>
S
(
f
,
σ
)
−
ε
≥
∫
a
b
f
(
x
)
d
x
−
ε
{\displaystyle =s(f,\sigma )>S(f,\sigma )-\varepsilon \geq \int _{a}^{b}f(x)dx-\varepsilon }
Dunque,
F
(
b
)
−
F
(
a
)
=
∫
a
b
f
(
x
)
d
x
{\displaystyle F(b)-F(a)=\int _{a}^{b}f(x)dx}
◻
{\displaystyle \Box }
non è condizione necessaria per l'integrabilità e nemmeno sufficiente. Infatti esistono funzioni che non hanno primitiva ma sono integrabili, così come esistono funzioni che hanno primitiva ma non sono integrabili secondo Riemann.
modifica
![{\displaystyle f,g\in {\mathcal {R}}_{[a,b]},\ \lambda \in \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/cdf17726ef11b123a8d2d05e88db01f27b9c629f)
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
![{\displaystyle f(x)\leq g(x),\ \forall x\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3525e2dc18f1715e82c990ab196a38f7a112caff)
![{\displaystyle f(x)\leq g(x),\ \forall x\in [a,b]\Rightarrow S(f,\sigma )\leq S(g,\sigma ),\ \forall \sigma \in \Omega _{[a,b]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69976ec31ffeaa52e34e7f6debdff8ffe8f1f1fe)
![{\displaystyle |f|\in {\mathcal {R}}_{[a,b]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e60a70bd8894479d899b10385ff8a4d67a3c1699)
![{\displaystyle \inf _{[a,b]}f\leq \mu \leq \sup _{[a,b]}f}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca206145e5405d786ac6c7cb7c91a18f92ad63e8)
![{\displaystyle f\in {\mathcal {R}}_{[a,b]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eeed0ef8d5d09dd9658616a878f8f554554f329a)
![{\displaystyle I_{f}:[a,b]\to \mathbb {R} ,\ \ I_{f}(x)=\int _{a}^{x}f(t)dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53a875db5af4d0eda04c3833bf6c57c79cf2fed0)
![{\displaystyle x_{0}\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06636653315ee7c3b5dc9bdb6ac3fb8cccadc145)
![{\displaystyle x_{o}+h\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6276327a26686192958fb9e39e349a7e98d1f745)
![{\displaystyle \inf _{[x_{0},x_{0}+h]}f\leq \mu (h)\leq \sup _{[x_{0},x_{0}+h]}f}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6064fa2eb12d7dbe1500663dd2c0733e892bb0cb)
![{\displaystyle |t-x_{0}|<\delta ,\ \ \forall t\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/232ac84ffed5b471205a1ef00323705bc5deea68)
Nota:
![{\displaystyle x\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/026357b404ee584c475579fb2302a4e9881b8cce)
![{\displaystyle f\in {\mathcal {C}}^{1}([a,b],\mathbb {R} )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/286bec294de0b424ee6bccc3e9dcf11c4e859b1a)
![{\displaystyle \int _{a}^{b}f'(x)dx=[f(x)]_{a}^{b}=f(b)-f(a)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7022b65df990ccfa6b58d0b76ce1099040be1e16)
![{\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 \int _{a}^{b}f(x)dx=[F(x)]_{a}^{b}=F(b)-F(a)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f35df72551cd47d6fd7dfff2e19a83ee8f56d1a)
![{\displaystyle \forall \varepsilon >0\exists \sigma \in \Omega _{[a,b]}\ :\ S(f,\sigma )-s(f,\sigma )<\varepsilon }](https://wikimedia.org/api/rest_v1/media/math/render/svg/19085841e259ce06a364958f0efdc6df6513fcab)
![{\displaystyle c_{i}\in ]x_{i-1},x_{i}[\ :\ F(x_{i})-F(x_{i-1})=F'(c_{i})(x_{i}-x_{i-1})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7188bd6d96b63fc95d3ba6a3c54c162b1b6d1334)
![{\displaystyle F(b)-F(a)=\sum _{i=1}^{n}f(c_{i})(x_{i}-x_{i-1})\leq \sum _{i=1}^{n}\sup _{[x_{i-1},x_{i}]}f\cdot (x_{i}-x_{i-1})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d380e9d19281eed0abf9c093e69a1a4d0ffa3e4)
![{\displaystyle F(b)-F(a)=\sum _{i=1}^{n}f(c_{i})(x_{i}-x_{i-1})\geq \sum _{i=1}^{n}\inf _{[x_{i-1},x_{i}]}f\cdot (x_{i}-x_{i-1})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ad7ec77237833d9d884569120d4f247d5d1e24e)
![{\displaystyle f\in {\mathcal {C}}([a,b],\mathbb {R} )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/574d77a18b0a63187362202b74f55b9039f9b499)
![{\displaystyle g\in {\mathcal {C}}^{1}([a,b],\mathbb {R} )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67e52fb699c56111b87d3d3ae3be10ff2fa0b52c)
![{\displaystyle \int _{a}^{b}f(x)g(x)dx=[F(x)g(x)]_{a}^{b}-\int _{a}^{b}F(x)g'(x)dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e83ad5eaf7676241440812a7b37e175750572fb)
![{\displaystyle Fg\in {\mathcal {C}}^{1}([a,b],\mathbb {R} )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfc75d8667a010c50a7155c013fdf878db30e0a2)
![{\displaystyle [F(x)g(x)]_{a}^{b}=\int _{a}^{b}(F(x)g(x))'dx=\int _{a}^{b}(F'(x)g(x)+F(x)g'(x))dx=\int _{a}^{b}f(x)g(x)dx+\int _{a}^{b}F(x)g'(x)dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf5d143ed35563261f4547d0ad826c39a0b961f3)
![{\displaystyle \varphi :[\alpha ,\beta ]\to [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/311229a90ba552f7eb0b8b3287b8e3a80a371773)
