Given:
- Probability density function: f(x) = 1/2 |x|e-|x|

To find:
- Variance of the random variable X

Solution:
The variance of a random variable X is given by the formula:

Var(X) = E(X^2) - [E(X)]^2

where E(X) is the expected value of X.

Step 1: Calculate the expected value E(X)
To calculate the expected value E(X), we integrate X multiplied by the probability density function f(x) over the entire range of X.

E(X) = ∫(x * f(x)) dx

Step 1.1: Determine the range of X
Since the probability density function is defined as f(x) = 1/2 |x|e-|x|, we can see that it is symmetric about the y-axis. Therefore, the range of X is from -∞ to +∞.

Step 1.2: Calculate the expected value E(X)
E(X) = ∫(x * f(x)) dx
= ∫(x * 1/2 |x|e-|x|) dx

To calculate this integral, we can split it into two parts based on the sign of x:
1. For x > 0, the integral becomes:
∫(x * 1/2 xe-x) dx

2. For x < 0,="" the="" integral="" />
∫(x * 1/2 (-x)e-(-x)) dx
= ∫(-x * 1/2 xe-x) dx

Simplifying the integrals, we get:
E(X) = ∫(x * 1/2 xe-x) dx + ∫(-x * 1/2 xe-x) dx
= ∫(x^2 * 1/2 e-x) dx

Now we can integrate by parts. Let u = x^2/2 and dv = e-x dx.
Then du = x dx and v = -e-x.

Using the integration by parts formula:
∫udv = uv - ∫vdu

We can apply this formula to the integral, which gives:
E(X) = [(-x^2/2)e-x] - ∫(-e-x * x) dx
= [(-x^2/2)e-x] + ∫(xe-x) dx
= [(-x^2/2)e-x] + [-xe-x - ∫(-e-x) dx]
= [(-x^2/2)e-x] + [-xe-x + e-x] + C
= e-x (-x^2/2 - x + 1) + C

Step 1.3: Calculate the definite integral
To find the expected value E(X), we need to calculate the definite integral of the expression obtained in the previous step over the range of X.

E(X) = ∫[e-x (-x^2/2 - x + 1)] dx
= ∫e-x (-x^2/2 - x + 1) dx

Since the range of X is from -∞ to +∞,