Since the PDF is a constant c within the support and 0 outside, we can split the integral into two parts:
$E(X) = ∫_{-2}^{2} x * c dx + ∫_{99.5}^{100.5} x * c dx$
Now, we can factor out the constant c and evaluate the integrals:
E(X) = c * [x^2/2]_{-2}^{2} + c * [x^2/2]_{99.5}^{100.5}
E(X) = c * (4/2 - 4/2) + c * (10100.25/2 - 9900.25/2)
E(X) = c * (100)
To find the value of c, we use the fact that the total probability over the support must equal 1:
1 = ∫_{-2}^{2} c dx + ∫_{99.5}^{100.5} c dx
1 = c * (4 + 1)
c = 1/5
Therefore, the expected value of X is:
E(X) = (1/5) * (100)
E(X) = 20
So the answer is 20.