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To solve the problem in the image, we are tasked with finding the expected value \( E(X) \) of a given probability density function (PDF).

### Step-by-Step Solution:

The probability density function \( f(x) \) is given as:

\[
f(x) = \begin{cases} 
\frac{3}{4}(1 - x^2) & \text{for } x \in [-1, 1] \\
0 & \text{otherwise}
\end{cases}
\]

The expected value \( E(X) \) of a continuous random variable \( X \) is calculated as:

\[
E(X) = \int_{-\infty}^{\infty} x f(x) \, dx
\]

Since \( f(x) = 0 \) outside the interval \([-1, 1]\), we only need to integrate within the range \( x \in [-1, 1] \):

\[
E(X) = \int_{-1}^{1} x f(x) \, dx = \int_{-1}^{1} x \cdot \frac{3}{4} (1 - x^2) \, dx
\]

### Step 1: Set up the integral

\[
E(X) = \frac{3}{4} \int_{-1}^{1} x (1 - x^2) \, dx
\]

### Step 2: Expand the integrand

\[
E(X) = \frac{3}{4} \int_{-1}^{1} (x - x^3) \, dx
\]

### Step 3: Break the integral into two parts

\[
E(X) = \frac{3}{4} \left( \int_{-1}^{1} x \, dx - \int_{-1}^{1} x^3 \, dx \right)
\]

### Step 4: Solve each integral

#### First integral:

\[
\int_{-1}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{1} = \frac{1^2}{2} - \frac{(-1)^2}{2} = 0
\]

#### Second integral:

\[
\int_{-1}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{-1}^{1} = \frac{1^4}{4} - \frac{(-1)^4}{4} = 0
\]

### Step 5: Combine the results

Since both integrals evaluate to zero, the expected value is:

\[
E(X) = \frac{3}{4} \times (0 - 0) = 0
\]

### Final Answer:

\[
E(X) = 0
\]

Let me know if you would like further clarifications or have additional questions!

---

Here are 5 follow-up questions you may find interesting:

1. What is the variance \( \text{Var}(X) \) for the given density function?
2. How do you interpret the result \( E(X) = 0 \) in the context of this distribution?
3. Could we calculate the median of \( X \) and how does it differ from the mean?
4. What is the cumulative distribution function (CDF) for this probability density?
5. How would the expected value change if the limits of integration were different?

**Tip:** The symmetry of the distribution plays a critical role in simplifying the calculation of the expected value. Symmetric distributions around zero often have \( E(X) = 0 \).

Berechnen Sie den Erwartungswert E(x) für die gegebene Dichtefunktion f(x) = \frac{3}{4}(1 - x^2) für x \in [-1, 1].

Learn how to calculate the expected value E(X) for a continuous random variable X given its probability density function f(x) = 3/4(1 - x^2) for x in the range [-1, 1]. This problem involves core concepts in probability theory and integration.

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