If the p.d.f. of a c.r.v. X is
`f(x)={(kx^2 (1-x^3)","0lexle1,),(0",",elsewhere):}`
then :k…

Options

A. 6
B. 5
C. 4
D. 3

Answer

To find the value of \( k \) in the given probability density function (p.d.f.) of a continuous random variable \( X \), we start with the following p.d.f.:

\[ f(x) = \begin{cases} kx^2(1-x^3) & \text{for } 0 \leq x \leq 1 \\ 0 & \text{elsewhere} \end{cases} \]

### Step 1: Set up the integral of the p.d.f.

The total probability must equal 1. Therefore, we need to integrate the p.d.f. over its entire range (from 0 to 1 in this case):

\[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \]

Since \( f(x) = 0 \) outside the interval [0, 1], we can simplify our integral:

\[ \int_{0}^{1} kx^2(1-x^3) \, dx = 1 \]

### Step 2: Expand the integrand

Next, we expand the integrand \( kx^2(1-x^3) \):

\[ kx^2(1-x^3) = kx^2 - kx^5 \]

### Step 3: Set up the integral

Now, we can set up the integral:

\[ \int_{0}^{1} (kx^2 - kx^5) \, dx = 1 \]

### Step 4: Integrate term by term

We can separate the integral:

\[ k \int_{0}^{1} x^2 \, dx - k \int_{0}^{1} x^5 \, dx = 1 \]

Calculating each integral:

1. For \( \int_{0}^{1} x^2 \, dx \):

\[ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \]

2. For \( \int_{0}^{1} x^5 \, dx \):

\[ \int_{0}^{1} x^5 \, dx = \left[ \frac{x^6}{6} \right]_{0}^{1} = \frac{1^6}{6} - \frac{0^6}{6} = \frac{1}{6} \]

### Step 5: Substitute back into the equation

Substituting these results back into our equation gives:

\[ k \left( \frac{1}{3} - \frac{1}{6} \right) = 1 \]

### Step 6: Simplify the equation

Now simplify the expression inside the parentheses:

\[ \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6} \]

So we have:

\[ k \cdot \frac{1}{6} = 1 \]

### Step 7: Solve for \( k \)

Now, solve for \( k \):

\[ k = 1 \cdot 6 = 6 \]

Thus, the value of \( k \) is:

\[ \boxed{6} \]