VARIANCE OF THE VARIABLE x DISTRIBUTED ACCORDING T EXPONENTIAL LAW f(x)= ) 2e^-2x,&xgeqslant 0 0,&xlt 0 IS Select one: 1/2 -1/2 (Ln2)/2 1/4 -1/4

To find the variance of the random variable X, we need to first find the mean (expected value) of X, denoted as μ, and then use the formula for variance:Var(X) = E(X^2) - [E(X)]^2where E(X^2) is the expected value of X^2 and [E(X)]^2 is the square of the mean.Given the probability density function (pdf) of X:f(x) = { 2e^(-2x), x ≥ 00, x < 0we can find the mean by integrating the product of x and f(x) over the entire range of X:μ = ∫(x * f(x)) dxIntegrating from 0 to ∞:μ = ∫(x * 2e^(-2x)) dxLet u = -2x, then du = -2 dx, and when x = 0, u = 0, and when x = ∞, u = -∞.μ = ∫(u/2 * 2e^u) duIntegrating by parts:μ = [u * e^u]_(-∞)^0 - ∫(e^u) duμ = 0 - [e^u]_(-∞)^0μ = -1Now, we can find E(X^2) by integrating the product of x^2 and f(x entire range of X:E(X^2) = ∫(x^2 * f(x)) dxIntegrating from 0 to ∞:E(X^2) = ∫(x^2 * 2e^(-2x)) dxLet u = -2x, then du = -2 dx, and when x = 0, u = 0, and when x = ∞, u = -∞.E(X^2) = ∫(u^2/4 * 2e^u) duIntegrating by parts:E(X^2) = [u^2/8 * e^u]_(-∞)^0 - ∫(u/4 * e^u) duE(X^2) = 0 - [u^2/8 * e^u]_(-∞)^0 + ∫(u/4 * e^u) duE(X^2) = 0 - [u^2/8 * e^u]_(-∞)^0 + [u/8 * e^u]_(-∞)^0E(X^2) = 0 - 0 + 1/8E(X^2) = 1/8Now, we can find the variance using the formula:Var(X) = E(X^2) - [E(X)]^2Var(X) = 1/8 - (-1)^2Var(X) = 1/8 - 1Var = -7/8Therefore, the variance of the random variable X is -7/8.