The p.d.f of a continuous random variable X is `f(x) = (x^(2))/(3), - 1 lt x lt 2` 0 = otherwise Then the c.d.f of X is

Options

A. `(x^(3))/(9) +(1)/(9)`
B. `(x^(3))/(9) - (1)/(9)`
C. `(x^(2))/(4) +(1)/(4)`
D. `(1)/(9x^(3))+(1)/(9)`

Answer

To find the cumulative distribution function (c.d.f) of the continuous random variable \( X \) with the given probability density function (p.d.f) \( f(x) = \frac{x^2}{3} \) for \( -1 < x < 2 \) and \( f(x) = 0 \) otherwise, we will follow these steps:

### Step 1: Understand the Definition of Cumulative Distribution Function
The cumulative distribution function \( F(x) \) is defined as:
\[ F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t) \, dt \]

### Step 2: Identify the Intervals
Since \( f(x) = 0 \) for \( x \leq -1 \) and \( x \geq 2 \), we can break the calculation of \( F(x) \) into different intervals:
- For \( x < -1 \): \( F(x) = 0 \)
- For \( -1 \leq x < 2 \): \( F(x) = \int_{-1}^{x} f(t) \, dt \)
- For \( x \geq 2 \): \( F(x) = 1 \) (since the total area under the p.d.f must equal 1)

### Step 3: Calculate the CDF for the Interval \(-1 \leq x < 2\)
We will now calculate \( F(x) \) for the interval \( -1 \leq x < 2 \):
\[ F(x) = \int_{-1}^{x} f(t) \, dt = \int_{-1}^{x} \frac{t^2}{3} \, dt \]

### Step 4: Perform the Integration
Now we need to compute the integral:
\[ \int \frac{t^2}{3} \, dt = \frac{1}{3} \cdot \frac{t^3}{3} = \frac{t^3}{9} \]
Now, we will evaluate this from \(-1\) to \(x\):
\[ F(x) = \left[ \frac{t^3}{9} \right]_{-1}^{x} = \frac{x^3}{9} - \frac{(-1)^3}{9} \]
\[ F(x) = \frac{x^3}{9} + \frac{1}{9} = \frac{x^3 + 1}{9} \]

### Step 5: Combine All Parts to Form the Complete CDF
Now we can summarize the c.d.f \( F(x) \):
\[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{x^3 + 1}{9} & \text{for } -1 \leq x < 2 \\ 1 & \text{for } x \geq 2 \end{cases} \]

### Final Answer
Thus, the cumulative distribution function \( F(x) \) is:
\[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{x^3 + 1}{9} & \text{for } -1 \leq x < 2 \\ 1 & \text{for } x \geq 2 \end{cases} \]