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Lezione 24

Integrali impropri

Analizziamo il primo caso.

Definizione

: [a, +∞) [a,

Sia f limitata ed integrabile su M] per ogni M a.

>

R

−→ [a, +∞)

Si dice che f è integrabile in senso generalizzato su se esiste

finito il limite M

Z (x)dx.

lim f

M→+∞ a

In tal caso si scrive M

Z Z

∞ (x)dx = (x)dx.

f lim f

M→+∞

a a (−∞,

Analogamente si definisce l’integrabilità su b]. dsm

Marco Castellani (L’Aquila) Matematica Generale Lezione 24 9 / 15

Lezione 24

Integrali impropri

Esempio

Calcolare i seguenti due integrali impropri

Z Z

∞ ∞

1 1

e

dx dx

2

x x

1 1

Per quanto riguarda il primo si ha

M

Z Z

∞ 1 1 x=M = =

= = lim ln M

dx lim dx lim ln x ∞.

x x M→∞

M→∞ M→∞ x=1

1 1

Quindi l’area sottesa è infinita. Invece per il secondo limite si ha

x=M

M

Z

Z ∞ 1 1

1 = = = + =

−2

dx lim x dx lim lim 1 1

− −

2 x M

x M→∞ M→∞ M→∞

1

1 x=1 dsm

e l’area in questo caso è finita!

Marco Castellani (L’Aquila) Matematica Generale Lezione 24 10 / 15

Lezione 24

Integrali impropri

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Quali analogie ci sono tra la serie armonica e le aree sottese dalle

1 1

(x) = =

funzioni f e g(x) ?

2

x x dsm

Marco Castellani (L’Aquila) Matematica Generale Lezione 24 11 / 15

Lezione 24

Integrali impropri

Analizziamo il secondo caso.

Definizione

: (a,

Sia f b] illimitata in un intorno destro di a ed integrabile su

R

[a + b] per ogni 0. Si dice che f è integrabile in senso

ε, ε >

(a,

generalizzato su b] se esiste finito il limite

b

Z (x)dx.

f

lim

ε→0 + a+ε

In tal caso si scrive b

b Z

Z (x)dx.

(x)dx = f

f lim

ε→0 + a+ε

a

Se f è illimitata in un intorno sinistro di b si ragiona sugli intervalli del

dsm

[a,

tipo b ε].

Marco Castellani (L’Aquila) Matematica Generale Lezione 24 12 / 15

Lezione 24

Integrali impropri

Esempio

Calcolare i seguenti due integrali impropri

1 1

Z Z

1 1

dx dx

e √

x x

0 0

Per quanto riguarda il primo si ha

1

1 Z

Z 1

1 x=1 = =

= = lim ln

dx lim dx lim ln x ε

− ∞.

x x ε→0

ε→0 ε→0 +

+ + x=ε

ε

0

Quindi l’area sottesa è infinita. Invece per il secondo limite si ha

1 1

Z Z

1 √ √

x=1 = =

=

= −1/2 lim 2 2

x 2

x dx lim 2

dx lim ε

√ −

x=ε

x ε→0

ε→0

ε→0 +

+

+ ε

0 dsm

e l’area in questo caso è finita!

Marco Castellani (L’Aquila) Matematica Generale Lezione 24 13 / 15

Lezione 24

Integrali impropri ·

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1

=

Come mai g(x) sottende un’area finita mentre l’area sottesa da

√ x

1

(x) =

f è infinita?

x dsm

Marco Castellani (L’Aquila) Matematica Generale Lezione 24 14 / 15


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AUTORE

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PUBBLICATO

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DESCRIZIONE DISPENSA

Calcolo degli integrali definiti: legame con gli integrali indefiniti e teorema fondamentale del calcolo integrale. Integrali impropri: integrale di una funzione definito su un intervallo illimitato; integrale di una funzione illimitata definito su un intervallo limitato. Alcuni esempi fondamentali per le funzioni potenza ad esponente reale [math]f(x)=x^\alpha,\ \alpha\in\mathbb{R}[/math].


DETTAGLI
Corso di laurea: Corso di laurea in economia e amministrazione delle imprese
SSD:
Università: L'Aquila - Univaq
A.A.: 2011-2012

I contenuti di questa pagina costituiscono rielaborazioni personali del Publisher Atreyu di informazioni apprese con la frequenza delle lezioni di Matematica Generale e studio autonomo di eventuali libri di riferimento in preparazione dell'esame finale o della tesi. Non devono intendersi come materiale ufficiale dell'università L'Aquila - Univaq o del prof Castellani Marco.

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