Esercizi di analisi II

Esercizi di analisi II

+Esercizi di analisi II

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Esercizi di analisi II

Teoria dell’Analisi II

L’Analisi II è un ramo della matematica che si occupa dello studio delle funzioni di più variabili, delle serie e delle integrazioni multiple. Essa estende i concetti fondamentali dell’Analisi I, come limiti, continuità, derivazione e integrazione, a contesti più complessi.

Concetti Fondamentali

1. Limiti e Continuità in più variabili: Si studia il comportamento di funzioni di più variabili quando gli argomenti si avvicinano a un certo punto.

2. Derivate Parziali: Le derivate parziali di una funzione di più variabili descrivono la variazione della funzione rispetto a una sola variabile, mantenendo fisse le altre.

3. Gradiente e Derivate Direzionali: Il gradiente è un vettore che indica la direzione e il tasso di cambiamento massimo della funzione. Le derivate direzionali sono le derivate lungo una direzione specifica.

4. Integrali Doppio e Triplo: Si estendono i concetti di integrazione a funzioni di due o tre variabili, permettendo il calcolo di aree e volumi.

5. Teoremi di Stokes e Gauss: Questi teoremi collegano le integrali su superfici e volumi con le integrali sui bordi delle superfici.

Esercizi di Analisi II

Esercizio 1: Calcolo del Limite

Calcola il limite:

\lim_{(x,y) \to (0,0)} \frac{x^2 + y^2}{x^2 + y^2 + 1}.

$$\lim_{(x,y) \to (0,0)} \frac{x^2 + y^2}{x^2 + y^2 + 1}.$$

Soluzione:
Sostituendo il punto:

\lim_{(x,y) \to (0,0)} \frac{x^2 + y^2}{x^2 + y^2 + 1} = \frac{0 + 0}{0 + 0 + 1} = \frac{0}{1} = 0.

$$\lim_{(x,y) \to (0,0)} \frac{x^2 + y^2}{x^2 + y^2 + 1} = \frac{0 + 0}{0 + 0 + 1} = \frac{0}{1} = 0.$$

Quindi, il limite è 0.

Esercizio 2: Derivata Parziale

Calcola la derivata parziale della funzione
f(x,y) = x^2y + 3xy^2$f(x,y) = x^2y + 3xy^2$ rispetto a x$x$.

Soluzione:
Utilizziamo la regola della derivazione per calcolare la derivata parziale:

\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2y) + \frac{\partial}{\partial x}(3xy^2)

$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2y) + \frac{\partial}{\partial x}(3xy^2)$$

Calcoliamo ciascun termine:

• Derivata parziale di x^2y$x^2y$ rispetto a x$x$:

\frac{\partial}{\partial x}(x^2y) = 2xy.

$$\frac{\partial}{\partial x}(x^2y) = 2xy.$$

• Derivata parziale di 3xy^2$3xy^2$ rispetto a x$x$:

\frac{\partial}{\partial x}(3xy^2) = 3y^2.

$$\frac{\partial}{\partial x}(3xy^2) = 3y^2.$$

Quindi, la derivata parziale è:

\frac{\partial f}{\partial x} = 2xy + 3y^2.

$$\frac{\partial f}{\partial x} = 2xy + 3y^2.$$

Esercizio 3: Gradiente

Trova il gradiente della funzione
f(x,y) = x^3 - 3xy + y^3$f(x,y) = x^3 - 3xy + y^3$.

Soluzione:
Il gradiente è dato dal vettore delle derivate parziali:

\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right).

$$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right).$$

Calcoliamo le derivate parziali:

• \frac{\partial f}{\partial x} = 3x^2 - 3y 

  $$\frac{\partial f}{\partial x} = 3x^2 - 3y$$

• \frac{\partial f}{\partial y} = -3x + 3y^2 

  $$\frac{\partial f}{\partial y} = -3x + 3y^2$$

Quindi, il gradiente è:

\nabla f = (3x^2 - 3y, -3x + 3y^2).

$$\nabla f = (3x^2 - 3y, -3x + 3y^2).$$

Esercizio 4: Integrazione Doppia

Calcola l’integrale doppio della funzione
f(x,y) = xy$f(x,y) = xy$ sull’area delimitata dai punti (0,0), (1,0), (1,1), (0,1).

Soluzione:
L’integrale doppio si scrive come:

\iint_R xy \, dA,

$$\iint_R xy \, dA,$$

dove R è il quadrato con vertici in (0,0), (1,0), (1,1), (0,1).

L’integrale può essere calcolato come:

\int_0^1 \int_0^1 xy \, dy \, dx.

$$\int_0^1 \int_0^1 xy \, dy \, dx.$$

Calcoliamo prima l’integrale interno rispetto a y$y$:

\int_0^1 xy \, dy = x \left[ \frac{y^2}{2} \right]_0^1 = x \cdot \frac{1}{2} = \frac{x}{2}.

$$\int_0^1 xy \, dy = x \left[ \frac{y^2}{2} \right]_0^1 = x \cdot \frac{1}{2} = \frac{x}{2}.$$

Ora calcoliamo l’integrale esterno rispetto a x$x$:

\int_0^1 \frac{x}{2} \, dx = \frac{1}{2} \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.

$$\int_0^1 \frac{x}{2} \, dx = \frac{1}{2} \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.$$

Quindi, l’integrale doppio vale \frac{1}{4}$\frac{1}{4}$.

Esercizio Avanzato: Integrazione Tripla

Calcola l’integrale triplo della funzione

f(x,y,z) = xyz 

$$f(x,y,z) = xyz$$

sull’intervallo definito da 0 < x < 1$0 < x < 1$, 0 < y < 1 - x$0 < y < 1 - x$ e 0 < z < xy$0 < z < xy$.

Soluzione:
L’integrale triplo si scrive come:

\iiint_V xyz \, dV,

$$\iiint_V xyz \, dV,$$

dove V è il volume definito dai limiti dati.

Scriviamo l’integrale come:

\int_0^1 \int_0^{1-x} \int_0^{xy} xyz \, dz\, dy\, dx.

$$\int_0^1 \int_0^{1-x} \int_0^{xy} xyz \, dz\, dy\, dx.$$

Calcoliamo prima l’integrale interno rispetto a z$z$:

\int_0^{xy} xyz\, dz = xy \left[ \frac{z^2}{2} \right]_0^{xy} = xy \cdot \frac{(xy)^2}{2} = \frac{x^3y^3}{2}.

$$\int_0^{xy} xyz\, dz = xy \left[ \frac{z^2}{2} \right]_0^{xy} = xy \cdot \frac{(xy)^2}{2} = \frac{x^3y^3}{2}.$$

Ora calcoliamo l’integrale rispetto a y$y$ e poi a x$x$:

\int_0^{1-x} \frac{x^3y^3}{2}\, dy = \frac{x^3}{2}\left[\frac{y^4}{4}\right]_0^{1-x} = \frac{x^3(1-x)^4}{8}.

$$\int_0^{1-x} \frac{x^3y^3}{2}\, dy = \frac{x^3}{2}\left[\frac{y^4}{4}\right]_0^{1-x} = \frac{x^3(1-x)^4}{8}.$$

Infine calcoliamo l’integrale rispetto a x$x$:

\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx.

$$\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx.$$

Per calcolare l’integrale

\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx,

$$\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx,$$

iniziamo semplificando l’integrale:

\frac{1}{8} \int_0^{1} x^3(1-x)^4 \, dx.

$$\frac{1}{8} \int_0^{1} x^3(1-x)^4 \, dx.$$

Ora calcoliamo l’integrale

\int_0^{1} x^3(1-x)^4 \, dx.

$$\int_0^{1} x^3(1-x)^4 \, dx.$$

Passo 1: Espandere il termine (1-x)^4$(1-x)^4$

Utilizziamo il binomio di Newton per espandere (1-x)^4$(1-x)^4$:

(1-x)^4 = \sum_{k=0}^{4} \binom{4}{k} (-x)^k = 1 - 4x + 6x^2 - 4x^3 + x^4.

$$(1-x)^4 = \sum_{k=0}^{4} \binom{4}{k} (-x)^k = 1 - 4x + 6x^2 - 4x^3 + x^4.$$

Passo 2: Moltiplicare per x^3$x^3$

Ora moltiplichiamo x^3$x^3$ per l’espansione:

x^3(1-x)^4 = x^3(1 - 4x + 6x^2 - 4x^3 + x^4) = x^3 - 4x^4 + 6x^5 - 4x^6 + x^7.

$$x^3(1-x)^4 = x^3(1 - 4x + 6x^2 - 4x^3 + x^4) = x^3 - 4x^4 + 6x^5 - 4x^6 + x^7.$$

Passo 3: Calcolare l’integrale

Ora possiamo calcolare l’integrale:

\int_0^{1} (x^3 - 4x^4 + 6x^5 - 4x^6 + x^7) \, dx.

$$\int_0^{1} (x^3 - 4x^4 + 6x^5 - 4x^6 + x^7) \, dx.$$

Calcoliamo ogni termine separatamente:

1. Primo termine:

   \int_0^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^{1} = \frac{1}{4}.

   $$\int_0^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^{1} = \frac{1}{4}.$$

2. Secondo termine:

   \int_0^{1} -4x^4 \, dx = -4\left[ \frac{x^5}{5} \right]_0^{1} = -\frac{4}{5}.

   $$\int_0^{1} -4x^4 \, dx = -4\left[ \frac{x^5}{5} \right]_0^{1} = -\frac{4}{5}.$$

3. Terzo termine:

   \int_0^{1} 6x^5 \, dx = 6\left[ \frac{x^6}{6} \right]_0^{1} = 1.

   $$\int_0^{1} 6x^5 \, dx = 6\left[ \frac{x^6}{6} \right]_0^{1} = 1.$$

4. Quarto termine:

   \int_0^{1} -4x^6 \, dx = -4\left[ \frac{x^7}{7} \right]_0^{1} = -\frac{4}{7}.

   $$\int_0^{1} -4x^6 \, dx = -4\left[ \frac{x^7}{7} \right]_0^{1} = -\frac{4}{7}.$$

5. Quinto termine:

   \int_0^{1} x^7 \, dx = \left[ \frac{x^8}{8} \right]_0^{1} = \frac{1}{8}.

   $$\int_0^{1} x^7 \, dx = \left[ \frac{x^8}{8} \right]_0^{1} = \frac{1}{8}.$$

Passo 4: Sommare i risultati

Ora sommiamo i risultati ottenuti:

\int_0^{1} (x^3 - 4x^4 + 6x^5 - 4x^6 + x^7) \, dx = 
\frac{1}{4} - \frac{4}{5} + 1 - \frac{4}{7} + \frac{1}{8}.

$$\int_0^{1} (x^3 - 4x^4 + 6x^5 - 4x^6 + x^7) \, dx = \frac{1}{4} - \frac{4}{5} + 1 - \frac{4}{7} + \frac{1}{8}.$$

Passo 5: Trovare un denominatore comune

Il denominatore comune tra i numeri 4, 5, 7, 8$4, 5, 7, 8$ è 280$280$. Riscriviamo ciascun termine con il denominatore comune:

• \frac{1}{4} = \frac{70}{280},$\frac{1}{4} = \frac{70}{280},$
• -\frac{4}{5} = -\frac{224}{280},$-\frac{4}{5} = -\frac{224}{280},$
• 1 = \frac{280}{280},$1 = \frac{280}{280},$
• -\frac{4}{7} = -\frac{160}{280},$-\frac{4}{7} = -\frac{160}{280},$
• \frac{1}{8} = \frac{35}{280}.$\frac{1}{8} = \frac{35}{280}.$

Sommando i termini:

\frac{70 - 224 + 280 - 160 + 35}{280} = 
\frac{70 + 280 + 35 - (224 +160)}{280}
=
\frac{385 - 384}{280}
=
\frac{1}{280}.

$$\frac{70 - 224 + 280 - 160 + 35}{280} = \frac{70 + 280 + 35 - (224 +160)}{280} = \frac{385 - 384}{280} = \frac{1}{280}.$$

Passo Finale: Moltiplicare per \frac{1}{8}$\frac{1}{8}$

Ora moltiplichiamo il risultato finale per \frac{1}{8}$\frac{1}{8}$:

\int_0^{1} x^3(1-x)^4\, dx = 
\frac{1}{280}.

$$\int_0^{1} x^3(1-x)^4\, dx = \frac{1}{280}.$$

Pertanto,

\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx =
\frac{1/280}{8}= 
\frac{1}{2240}.

$$\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx = \frac{1/280}{8}= \frac{1}{2240}.$$

La risposta finale è quindi:

\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx = 
\boxed{\frac{1}{2240}}.

$$\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx = \boxed{\frac{1}{2240}}.$$

English version

Analysis Exercises II

Theory of Analysis II

Analysis II is a branch of mathematics that deals with the study of functions of several variables, series, and multiple integrations. It extends the fundamental concepts of Analysis I, such as limits, continuity, derivation, and integration, to more complex contexts.

Fundamental Concepts

1. Limits and Continuity in several variables: The behavior of functions of several variables is studied when the arguments approach a certain point.

2. Partial Derivatives: The partial derivatives of a function of several variables describe the variation of the function with respect to a single variable, keeping the others fixed.

3. Gradient and Directional Derivatives: The gradient is a vector that indicates the direction and the maximum rate of change of the function. Directional derivatives are the derivatives along a specific direction.

4. Double and Triple Integrals: The concepts of integration are extended to functions of two or three variables, allowing the calculation of areas and volumes.

5. Stokes and Gauss Theorems: These theorems connect the integrals on surfaces and volumes with the integrals on the boundaries of surfaces.

Exercises in Analysis II

Exercise 1: Calculating the Limit

Calculate the limit:

\lim_{(x,y) \to (0,0)} \frac{x^2 + y^2}{x^2 + y^2 + 1}.

$$\lim_{(x,y) \to (0,0)} \frac{x^2 + y^2}{x^2 + y^2 + 1}.$$

Solution:
Replacing the point:

\lim_{(x,y) \to (0,0)} \frac{x^2 + y^2}{x^2 + y^2 + 1} = \frac{0 + 0}{0 + 0 + 1} = \frac{0}{1} = 0.

$$\lim_{(x,y) \to (0,0)} \frac{x^2 + y^2}{x^2 + y^2 + 1} = \frac{0 + 0}{0 + 0 + 1} = \frac{0}{1} = 0.$$

So, the limit is 0.

Exercise 2: Partial Derivative

Calculate the partial derivative of the function
f(x,y) = x^2y + 3xy^2$f(x,y) = x^2y + 3xy^2$ with respect to x$x$.

Solution:
We use the differentiation rule to calculate the partial derivative:

\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2y) + \frac{\partial}{\partial x}(3xy^2)

$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2y) + \frac{\partial}{\partial x}(3xy^2)$$

We calculate each term:

• Partial derivative of x^2y$x^2y$ with respect to x$x$:

\frac{\partial}{\partial x}(x^2y) = 2xy.

$$\frac{\partial}{\partial x}(x^2y) = 2xy.$$

• Partial derivative of 3xy^2$3xy^2$ with respect to x$x$:

\frac{\partial}{\partial x}(3xy^2) = 3y^2.

$$\frac{\partial}{\partial x}(3xy^2) = 3y^2.$$

So, the partial derivative is:

\frac{\partial f}{\partial x} = 2xy + 3y^2.

$$\frac{\partial f}{\partial x} = 2xy + 3y^2.$$

Exercise 3: Gradient

Find the gradient of the function
f(x,y) = x^3 - 3xy + y^3$f(x,y) = x^3 - 3xy + y^3$.

Solution:
The gradient is given by the vector of partial derivatives:

\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right).

$$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right).$$

Let’s calculate the partial derivatives:

• \frac{\partial f}{\partial x} = 3x^2 - 3y 

  $$\frac{\partial f}{\partial x} = 3x^2 - 3y$$

• \frac{\partial f}{\partial y} = -3x + 3y^2 

  $$\frac{\partial f}{\partial y} = -3x + 3y^2$$

So, the gradient is:

\nabla f = (3x^2 - 3y, -3x + 3y^2).

$$\nabla f = (3x^2 - 3y, -3x + 3y^2).$$

Exercise 4: Double Integration

Calculate the double integral of the function
f(x,y) = xy$f(x,y) = xy$ over the area bounded by the points (0,0), (1,0), (1,1), (0,1).

Solution:
The double integral is written as:

\iint_R xy \, dA,

$$\iint_R xy \, dA,$$

where R is the square with vertices in (0,0), (1,0), (1,1), (0,1).

The integral can be calculated as:

\int_0^1 \int_0^1 xy \, dy \, dx.

$$\int_0^1 \int_0^1 xy \, dy \, dx.$$

First, let’s compute the internal integral with respect to y$y$:

\int_0^1 xy \, dy = x \left[ \frac{y^2}{2} \right]_0^1 = x \cdot \frac{1}{2} = \frac{x}{2}.

$$\int_0^1 xy \, dy = x \left[ \frac{y^2}{2} \right]_0^1 = x \cdot \frac{1}{2} = \frac{x}{2}.$$

Now, let’s compute the external integral with respect to x$x$:

\int_0^1 \frac{x}{2} \, dx = \frac{1}{2} \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.

$$\int_0^1 \frac{x}{2} \, dx = \frac{1}{2} \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.$$

So, the double integral is \frac{1}{4}$\frac{1}{4}$.

Advanced Exercise: Triple Integration

Compute the triple integral of the function

f(x,y,z) = xyz 

$$f(x,y,z) = xyz$$

over the interval defined by 0 < x < 1$0 < x < 1$, 0 < y < 1 - x$0 < y < 1 - x$ and 0 < z < xy$0 < z < xy$.

Solution:
The triple integral is written as:

\iiint_V xyz \, dV,

$$\iiint_V xyz \, dV,$$

where V is the volume defined by the given limits.

We write the integral as:

\int_0^1 \int_0^{1-x} \int_0^{xy} xyz \, dz\, dy\, dx.

$$\int_0^1 \int_0^{1-x} \int_0^{xy} xyz \, dz\, dy\, dx.$$

First, let’s compute the internal integral with respect to z$z$:

\int_0^{xy} xyz\, dz = xy \left[ \frac{z^2}{2} \right]_0^{xy} = xy \cdot \frac{(xy)^2}{2} = \frac{x^3y^3}{2}.

$$\int_0^{xy} xyz\, dz = xy \left[ \frac{z^2}{2} \right]_0^{xy} = xy \cdot \frac{(xy)^2}{2} = \frac{x^3y^3}{2}.$$

Now, let’s compute the integral with respect to y$y$ and then to x$x$:

\int_0^{1-x} \frac{x^3y^3}{2}\, dy = \frac{x^3}{2}\left[\frac{y^4}{4}\right]_0^{1-x} = \frac{x^3(1-x)^4}{8}.

$$\int_0^{1-x} \frac{x^3y^3}{2}\, dy = \frac{x^3}{2}\left[\frac{y^4}{4}\right]_0^{1-x} = \frac{x^3(1-x)^4}{8}.$$

Finally, we calculate the integral with respect to x$x$:

\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx.

$$\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx.$$

To calculate the integral

\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx,

$$\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx,$$

we begin by simplifying the integral:

\frac{1}{8} \int_0^{1} x^3(1-x)^4 \, dx.

$$\frac{1}{8} \int_0^{1} x^3(1-x)^4 \, dx.$$

Now, we calculate the integral

\int_0^{1} x^3(1-x)^4 \, dx.

$$\int_0^{1} x^3(1-x)^4 \, dx.$$

Step 1: Expand the term (1-x)^4$(1-x)^4$

We use Newton’s binomial to expand (1-x)^4$(1-x)^4$:

(1-x)^4 = \sum_{k=0}^{4} \binom{4}{k} (-x)^k = 1 - 4x + 6x^2 - 4x^3 + x^4.

$$(1-x)^4 = \sum_{k=0}^{4} \binom{4}{k} (-x)^k = 1 - 4x + 6x^2 - 4x^3 + x^4.$$

Step 2: Multiply by x^3$x^3$

Now we multiply x^3$x^3$ by the expansion:

x^3(1-x)^4 = x^3(1 - 4x + 6x^2 - 4x^3 + x^4) = x^3 - 4x^4 + 6x^5 - 4x^6 + x^7.

$$x^3(1-x)^4 = x^3(1 - 4x + 6x^2 - 4x^3 + x^4) = x^3 - 4x^4 + 6x^5 - 4x^6 + x^7.$$

Step 3: Calculate the integral

Now we can calculate the integral:

\int_0^{1} (x^3 - 4x^4 + 6x^5 - 4x^6 + x^7) \, dx.

$$\int_0^{1} (x^3 - 4x^4 + 6x^5 - 4x^6 + x^7) \, dx.$$

Let’s calculate each term separately:

1. First term:

\int_0^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^{1} = \frac{1}{4}.

$$\int_0^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^{1} = \frac{1}{4}.$$

2. Second term:

\int_0^{1} -4x^4 \, dx = -4\left[ \frac{x^5}{5} \right]_0^{1} = -\frac{4}{5}.

$$\int_0^{1} -4x^4 \, dx = -4\left[ \frac{x^5}{5} \right]_0^{1} = -\frac{4}{5}.$$

3. Third term:

\int_0^{1} 6x^5 \, dx = 6\left[ \frac{x^6}{6} \right]_0^{1} = 1.

$$\int_0^{1} 6x^5 \, dx = 6\left[ \frac{x^6}{6} \right]_0^{1} = 1.$$

4. Fourth term:

\int_0^{1} -4x^6 \, dx = -4\left[ \frac{x^7}{7} \right]_0^{1} = -\frac{4}{7}.

$$\int_0^{1} -4x^6 \, dx = -4\left[ \frac{x^7}{7} \right]_0^{1} = -\frac{4}{7}.$$

5. Fifth term:

\int_0^{1} x^7 \, dx = \left[ \frac{x^8}{8} \right]_0^{1} = \frac{1}{8}.

$$\int_0^{1} x^7 \, dx = \left[ \frac{x^8}{8} \right]_0^{1} = \frac{1}{8}.$$

Step 4: Add the results

Now let’s add the results obtained:

\int_0^{1} (x^3 - 4x^4 + 6x^5 - 4x^6 + x^7) \, dx =
\frac{1}{4} - \frac{4}{5} + 1 - \frac{4}{7} + \frac{1}{8}.

$$\int_0^{1} (x^3 - 4x^4 + 6x^5 - 4x^6 + x^7) \, dx = \frac{1}{4} - \frac{4}{5} + 1 - \frac{4}{7} + \frac{1}{8}.$$

Step 5: Find a common denominator

The common denominator between the numbers 4, 5, 7, 8$4, 5, 7, 8$ is 280$280$. Let’s rewrite each term with the common denominator:

• \frac{1}{4} = \frac{70}{280},$\frac{1}{4} = \frac{70}{280},$
• -\frac{4}{5} = -\frac{224}{280},$-\frac{4}{5} = -\frac{224}{280},$
• 1 = \frac{280}{280},$1 = \frac{280}{280},$
• -\frac{4}{7} = -\frac{160}{280},$-\frac{4}{7} = -\frac{160}{280},$
• \frac{1}{8} = \frac{35}{280}.$\frac{1}{8} = \frac{35}{280}.$

Adding the terms:

\frac{70 - 224 + 280 - 160 + 35}{280} =
\frac{70 + 280 + 35 - (224 +160)}{280}
=
\frac{385 - 384}{280}
=
\frac{1}{280}.

$$\frac{70 - 224 + 280 - 160 + 35}{280} = \frac{70 + 280 + 35 - (224 +160)}{280} = \frac{385 - 384}{280} = \frac{1}{280}.$$

Final Step: Multiply by \frac{1}{8}$\frac{1}{8}$

Now we multiply the final result by \frac{1}{8}$\frac{1}{8}$:

\int_0^{1} x^3(1-x)^4\, dx =
\frac{1}{280}.

$$\int_0^{1} x^3(1-x)^4\, dx = \frac{1}{280}.$$

Therefore,

\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx =
\frac{1/280}{8}=
\frac{1}{2240}.

$$\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx = \frac{1/280}{8}= \frac{1}{2240}.$$

The final answer is then:

\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx =
\boxed{\frac{1}{2240}}.

$$\int_0^{1} \frac{x^3(1-x)^4}{8}\, dx = \boxed{\frac{1}{2240}}.$$