1. The problem states that $f(x) = kx^3$ for $0 < x < 1$ and $f(x) = 0$ elsewhere is a probability density function (p.d.f). We need to find the value of $k$. 2. For $f(x)$ to be a valid p.d.f, the total area under the curve must be 1. This means: $$\int_{-\infty}^{\infty} f(x) \, dx = 1$$ 3. Since $f(x) = 0$ outside $(0,1)$, the integral reduces to: $$\int_0^1 kx^3 \, dx = 1$$ 4. Compute the integral: $$k \int_0^1 x^3 \, dx = k \left[ \frac{x^4}{4} \right]_0^1 = k \times \frac{1}{4} = \frac{k}{4}$$ 5. Set the integral equal to 1: $$\frac{k}{4} = 1$$ 6. Solve for $k$: $$k = 4$$ Final answer: $k = 4$