The mean value (expected value) of a continuous random variable $X$ with probability density function (PDF) $f(x)$ is given by:

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

In this case, the random variable $X$ has the PDF:

$f(x) = 0.75(1 - x^2) \quad \text{for } -1 \leq x \leq 1$

To find the mean, we need to evaluate:

$E[X] = \int_{-1}^{1} x \cdot 0.75(1 - x^2) \, dx$

Let's compute this integral step-by-step.

1. Set up the integral:

$E[X] = 0.75 \int_{-1}^{1} x (1 - x^2) \, dx$

2. Expand the integrand:

$E[X] = 0.75 \int_{-1}^{1} (x - x^3) \, dx$

3. Split the integral:

$E[X] = 0.75 \left( \int_{-1}^{1} x \, dx - \int_{-1}^{1} x^3 \, dx \right)$

4. Evaluate each integral separately:

   • For $\int_{-1}^{1} x \, dx$:

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

   • For $\int_{-1}^{1} x^3 \, dx$:

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

5. Combine the results:

$E[X] = 0.75 \left( 0 - 0 \right) = 0$

Thus, the mean value of $x$ is:

$E[X] = 0$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How would the variance of $x$ be calculated for the given PDF?
2. What is the median value of $x$ for this distribution?
3. How do you find the mode of the given density function?
4. If we change the interval to $[-2, 2]$, what would be the normalization constant for the PDF?
5. How does the mean value change if the PDF is modified to $f(x) = c(1 - x^2)$ where $c$ is a new constant?

Tip: The mean value for a symmetric distribution around 0 is often 0, as the positive and negative contributions cancel out.