Final answer:
The expected value of X is given by the integral of X multiplied by the probability density function over the interval.
For the given problem, the expected value of X is π/2.
Step-by-step explanation:
To find the expected value of a continuous random variable, such as the uniformly distributed random variable X over the interval [-π,π], we can use calculus.
The expected value of X, denoted as E(X) or µ, is given by the integral of X multiplied by the probability density function (PDF) over the interval.
In this case, since X is uniformly distributed, the PDF is constant over the interval and equal to 1/(b-a), where a is the lower bound (-π) and b is the upper bound (π).
So, the expected value is:
E(X) = µ = ∫(from -π to π) X * (1/(π-(-π))) dx
= (1/(2π)) ∫(from -π to π) X dx
= (1/2π) [X^2/2] (from -π to π)
= (1/2π) [(π^2 - (-π)^2)/2]
= (1/2π) [2π^2/2]
= π/2
Therefore, the expected value of X is π/2.