Esercizi sugli integrali impropri

Esercizi sugli integrali impropri

+Esercizi sugli integrali impropri

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Esercizi sugli integrali impropri

Teoria degli Integrali Impropri

Definizione

Un integrale improprio è un integrale che presenta una delle seguenti caratteristiche:

1. Limiti Infiniti: L’integrale ha almeno uno dei limiti di integrazione che tende a infinito. Ad esempio:

   \int_{a}^{\infty} f(x) \, dx 

   $$\int_{a}^{\infty} f(x) \, dx$$

2. Discontinuità: La funzione integranda presenta discontinuità in un punto all’interno dell’intervallo di integrazione. Ad esempio:

   \int_{a}^{b} f(x) \, dx 

   $$\int_{a}^{b} f(x) \, dx$$

dove f(x)$f(x)$ ha una discontinuità in c \in (a, b)$c \in (a, b)$.

Calcolo degli Integrali Impropri

Per calcolare un integrale improprio, si utilizza il limite. Ad esempio, per un integrale con limite infinito:

\int_{a}^{\infty} f(x) \, dx = \lim_{b \to \infty} \int_{a}^{b} f(x) \, dx 

$$\int_{a}^{\infty} f(x) \, dx = \lim_{b \to \infty} \int_{a}^{b} f(x) \, dx$$

Per un integrale con discontinuità:

\int_{a}^{b} f(x) \, dx = \lim_{\epsilon \to 0} \left( \int_{a}^{c - \epsilon} f(x) \, dx + \int_{c + \epsilon}^{b} f(x) \, dx \right)

$$\int_{a}^{b} f(x) \, dx = \lim_{\epsilon \to 0} \left( \int_{a}^{c - \epsilon} f(x) \, dx + \int_{c + \epsilon}^{b} f(x) \, dx \right)$$

Esercizi sugli Integrali Impropri

Esercizio 1: Limite Infinito

Calcola l’integrale improprio:

I = \int_{1}^{\infty} \frac{1}{x^2} \, dx 

$$I = \int_{1}^{\infty} \frac{1}{x^2} \, dx$$

Soluzione:
Scriviamo l’integrale come limite:

I = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^2} \, dx 

$$I = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^2} \, dx$$

Calcoliamo l’integrale definito:

\int \frac{1}{x^2} \, dx = -\frac{1}{x}

$$\int \frac{1}{x^2} \, dx = -\frac{1}{x}$$

Quindi,

I = \lim_{b \to \infty} \left[ -\frac{1}{x} \right]_{1}^{b} = \lim_{b \to \infty} (-\frac{1}{b} + 1) = 0 + 1 = 1

$$I = \lim_{b \to \infty} \left[ -\frac{1}{x} \right]_{1}^{b} = \lim_{b \to \infty} (-\frac{1}{b} + 1) = 0 + 1 = 1$$

L’integrale converge a 1.

Esercizio 2: Discontinuità

Calcola l’integrale improprio:

I = \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx 

$$I = \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx$$

Soluzione:
Poiché la funzione ha una discontinuità in x = 0$x = 0$, scriviamo l’integrale come limite:

I = \lim_{\epsilon \to 0^+} \int_{\epsilon}^{1} x^{-\frac{1}{2}}\, dx 

$$I = \lim_{\epsilon \to 0^+} \int_{\epsilon}^{1} x^{-\frac{1}{2}}\, dx$$

Calcoliamo l’integrale definito:

\int x^{-\frac{1}{2}}\, dx = 2x^{\frac{1}{2}}

$$\int x^{-\frac{1}{2}}\, dx = 2x^{\frac{1}{2}}$$

Quindi,

I = \lim_{\epsilon \to 0^+} [2x^{\frac{1}{2}}]_{\epsilon}^{1} = 2(1 - (\sqrt{\epsilon})) = 2(1 - 0) = 2

$$I = \lim_{\epsilon \to 0^+} [2x^{\frac{1}{2}}]_{\epsilon}^{1} = 2(1 - (\sqrt{\epsilon})) = 2(1 - 0) = 2$$

L’integrale converge a 2.

Esercizio 3: Limite Infinito con Funzione Esponenziale

Calcola l’integrale improprio:

I = \int_{0}^{\infty} e^{-x} \, dx 

$$I = \int_{0}^{\infty} e^{-x} \, dx$$

Soluzione:
Scriviamo l’integrale come limite:

I = \lim_{b \to \infty} \int_{0}^{b} e^{-x} \, dx 

$$I = \lim_{b \to \infty} \int_{0}^{b} e^{-x} \, dx$$

Calcoliamo l’integrale definito:

\int e^{-x}\, dx = -e^{-x}

$$\int e^{-x}\, dx = -e^{-x}$$

Quindi,

I = \lim_{b \to \infty} [-e^{-x}]_{0}^{b} = -e^{-b} + e^{0}

$$I = \lim_{b \to \infty} [-e^{-x}]_{0}^{b} = -e^{-b} + e^{0}$$

Poiché e^{-b}$e^{-b}$ tende a 0 quando b$b$ tende a infinito:

I = 0 + 1 = 1

$$I = 0 + 1 = 1$$

L’integrale converge a 1.

Esercizio 4: Confronto di Integrali Impropri

Determina se gli integrali seguenti convergono o divergono:

a) I_1 = \int_{1}^{\infty} e^{-x^2}\, dx$I_1 = \int_{1}^{\infty} e^{-x^2}\, dx$
b) I_2 = \int_{0}^{\infty} x\, e^{-x}\, dx$I_2 = \int_{0}^{\infty} x\, e^{-x}\, dx$

Soluzione:
a) Per l’integrale I_1$I_1$ possiamo notare che la funzione e^{-x^2}$e^{-x^2}$ decresce molto rapidamente. Possiamo confrontare con una funzione che sappiamo converge. Infatti,

I_1 < I_3 = C

$$I_1 < I_3 = C$$

per qualche costante C. Dunque, converge.

b) Per l’integrale I_2$I_2$ possiamo calcolare direttamente:

Utilizziamo il limite:

I_2 = \lim_{b\to\infty}\int_0^b x e^{-x}\,dx

$$I_2 = \lim_{b\to\infty}\int_0^b x e^{-x}\,dx$$

Utilizzando integrazione per parti con u=x$u=x$ e dv=e^{-x}\,dx$dv=e^{-x}\,dx$, otteniamo:

• du=dx$du=dx$
• v=-e^{-x}$v=-e^{-x}$

Quindi,

I_2=\lim_{b\to\infty}\left[-xe^{-x}\bigg|_0^b+\int_0^be^{-x}\,dx\right]
=0+[-e^{-x}]_0^b=0+(-(-e^{0}))=1

$$I_2=\lim_{b\to\infty}\left[-xe^{-x}\bigg|_0^b+\int_0^be^{-x}\,dx\right] =0+[-e^{-x}]_0^b=0+(-(-e^{0}))=1$$

L’integrale converge a 1.

English version

Improper Integral Exercises

Improper Integral Theory

Definition

An improper integral is an integral that has one of the following characteristics:

1. Infinite Limits: The integral has at least one of the limits of integration that tends to infinity. For example:

\int_{a}^{\infty} f(x) \, dx

$$\int_{a}^{\infty} f(x) \, dx$$

2. Discontinuity: The integrand function has discontinuity at a point within the interval of integration. For example:

\int_{a}^{b} f(x) \, dx

$$\int_{a}^{b} f(x) \, dx$$

where f(x)$f(x)$ has a discontinuity at c \in (a, b)$c \in (a, b)$.

Improper Integral Calculation

To calculate an improper integral, the limit is used. For example, for an integral with infinite limit:

\int_{a}^{\infty} f(x) \, dx = \lim_{b \to \infty} \int_{a}^{b} f(x) \, dx

$$\int_{a}^{\infty} f(x) \, dx = \lim_{b \to \infty} \int_{a}^{b} f(x) \, dx$$

For an integral with discontinuity:

\int_{a}^{b} f(x) \, dx = \lim_{\epsilon \to 0} \left( \int_{a}^{c - \epsilon} f(x) \, dx + \int_{c + \epsilon}^{b} f(x) \, dx \right)

$$\int_{a}^{b} f(x) \, dx = \lim_{\epsilon \to 0} \left( \int_{a}^{c - \epsilon} f(x) \, dx + \int_{c + \epsilon}^{b} f(x) \, dx \right)$$

Exercises on Improper Integrals

Exercise 1: Infinite Limit

Calculate the improper integral:

I = \int_{1}^{\infty} \frac{1}{x^2} \, dx

$$I = \int_{1}^{\infty} \frac{1}{x^2} \, dx$$

Solution:
Let’s write the integral as a limit:

I = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^2} \, dx

$$I = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^2} \, dx$$

Let’s calculate the definite integral:

\int \frac{1}{x^2} \, dx = -\frac{1}{x}

$$\int \frac{1}{x^2} \, dx = -\frac{1}{x}$$

So,

I = \lim_{b \to \infty} \left[ -\frac{1}{x} \right]_{1}^{b} = \lim_{b \to \infty} (-\frac{1}{b} + 1) = 0 + 1 = 1

$$I = \lim_{b \to \infty} \left[ -\frac{1}{x} \right]_{1}^{b} = \lim_{b \to \infty} (-\frac{1}{b} + 1) = 0 + 1 = 1$$

The integral converges to 1.

Exercise 2: Discontinuity

Calculate the improper integral:

I = \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx

$$I = \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx$$

Solution:
Since the function has a discontinuity at x = 0$x = 0$, we write the integral as a limit:

I = \lim_{\epsilon \to 0^+} \int_{\epsilon}^{1} x^{-\frac{1}{2}}\, dx

$$I = \lim_{\epsilon \to 0^+} \int_{\epsilon}^{1} x^{-\frac{1}{2}}\, dx$$

Calculate the definite integral:

\int x^{-\frac{1}{2}}\, dx = 2x^{\frac{1}{2}}

$$\int x^{-\frac{1}{2}}\, dx = 2x^{\frac{1}{2}}$$

So,

I = \lim_{\epsilon \to 0^+} [2x^{\frac{1}{2}}]_{\epsilon}^{1} = 2(1 - (\sqrt{\epsilon})) = 2(1 - 0) = 2

$$I = \lim_{\epsilon \to 0^+} [2x^{\frac{1}{2}}]_{\epsilon}^{1} = 2(1 - (\sqrt{\epsilon})) = 2(1 - 0) = 2$$

The integral converges to 2.

Exercise 3: Infinite Limit with Exponential Function

Calculate the improper integral:

I = \int_{0}^{\infty} e^{-x} \, dx

$$I = \int_{0}^{\infty} e^{-x} \, dx$$

Solution:
Let’s write the integral as a limit:

I = \lim_{b \to \infty} \int_{0}^{b} e^{-x} \, dx

$$I = \lim_{b \to \infty} \int_{0}^{b} e^{-x} \, dx$$

Let’s calculate the definite integral:

\int e^{-x}\, dx = -e^{-x}

$$\int e^{-x}\, dx = -e^{-x}$$

So,

I = \lim_{b \to \infty} [-e^{-x}]_{0}^{b} = -e^{-b} + e^{0}

$$I = \lim_{b \to \infty} [-e^{-x}]_{0}^{b} = -e^{-b} + e^{0}$$

Since e^{-b}$e^{-b}$ tends to 0 when b$b$ tends to infinity:

I = 0 + 1 = 1

$$I = 0 + 1 = 1$$

The integral converges to 1.

Exercise 4: Comparison of Improper Integrals

Determine whether the following integrals converge or diverge:

a) I_1 = \int_{1}^{\infty} e^{-x^2}\, dx$I_1 = \int_{1}^{\infty} e^{-x^2}\, dx$
b) I_2 = \int_{0}^{\infty} x\, e^{-x}\, dx$I_2 = \int_{0}^{\infty} x\, e^{-x}\, dx$

Solution:
a) For the integral I_1$I_1$ we can notice that the function e^{-x^2}$e^{-x^2}$ decreases very rapidly. We can compare with a function that we know converges. In fact,

I_1 < I_3 = C

$$I_1 < I_3 = C$$

for some constant C. Therefore, converges.

b) For the integral I_2$I_2$ we can directly calculate:

We use the limit:

I_2 = \lim_{b\to\infty}\int_0^b x e^{-x}\,dx

$$I_2 = \lim_{b\to\infty}\int_0^b x e^{-x}\,dx$$

Using integration by parts with u=x$u=x$ and dv=e^{-x}\,dx$dv=e^{-x}\,dx$, we get:

• du=dx$du=dx$
• v=-e^{-x}$v=-e^{-x}$

So,

I_2=\lim_{b\to\infty}\left[-xe^{-x}\bigg|_0^b+\int_0^be^{-x}\,dx\right]
=0+[-e^{-x}]_0^b=0+(-(-e^{0}))=1

$$I_2=\lim_{b\to\infty}\left[-xe^{-x}\bigg|_0^b+\int_0^be^{-x}\,dx\right] =0+[-e^{-x}]_0^b=0+(-(-e^{0}))=1$$

The integral converges to 1.