Question 1192148
**a) Find the value of k**

* **Condition for a valid probability density function (PDF):** 
    * The total area under the PDF curve must be equal to 1. 
    * Mathematically, this means: ∫[from 0 to ∞] f(x) dx = 1

* **Set up the integral:**
    * ∫[from 0 to ∞] kxe^(-x^2) dx = 1

* **Solve the integral:**
    * Let u = -x^2 
    * du = -2x dx 
    * dx = -du / (2x)

    * Substitute: ∫[from 0 to ∞] kxe^u * (-du / (2x)) 
                      = -k/2 ∫[from 0 to -∞] e^u du 
                      = -k/2 [e^u] [from 0 to -∞]
                      = -k/2 [0 - 1] 
                      = k/2

* **Solve for k:**
    * k/2 = 1
    * k = 2

**Therefore, the value of k is 2.**

**b) Find the distribution function of X (cumulative density function)**

* **Definition:** The cumulative distribution function (CDF), F(x), gives the probability that the random variable X takes on a value less than or equal to x.

* **Calculate F(x):**
    * For x ≤ 0: F(x) = P(X ≤ x) = 0 (since f(x) = 0 for x ≤ 0)
    * For x > 0: 
        * F(x) = P(X ≤ x) = ∫[from 0 to x] f(t) dt 
                    = ∫[from 0 to x] 2te^(-t^2) dt 

        * Let u = -t^2 
        * du = -2t dt 
        * dt = -du / (2t)

        * Substitute: ∫[from 0 to -x^2] 2te^u * (-du / (2t)) 
                          = -∫[from 0 to -x^2] e^u du
                          = -[e^u] [from 0 to -x^2]
                          = -(e^(-x^2) - e^0)
                          = 1 - e^(-x^2)

* **Combine the results:**

   * F(x) = 
       * 0, for x ≤ 0
       * 1 - e^(-x^2), for x > 0

**Therefore, the distribution function of X is:**

   F(x) = { 
       0, for x ≤ 0
       1 - e^(-x^2), for x > 0 
   }