1. **State the problem:** We are given a function $$f(x) = A(4x - 2x^2)$$ defined on the interval $$0 \leq x \leq 2$$. We need to find the value of $$A$$ such that $$f(x)$$ is a probability density function (pdf). 2. **Recall the property of a pdf:** The total area under the pdf curve over its domain must equal 1. Mathematically, $$\int_0^2 f(x) \, dx = 1$$ 3. **Set up the integral:** $$\int_0^2 A(4x - 2x^2) \, dx = 1$$ 4. **Factor out the constant $$A$$:** $$A \int_0^2 (4x - 2x^2) \, dx = 1$$ 5. **Calculate the integral:** $$\int_0^2 4x \, dx = \left[2x^2\right]_0^2 = 2(2)^2 - 0 = 8$$ $$\int_0^2 2x^2 \, dx = \left[\frac{2x^3}{3}\right]_0^2 = \frac{2(2)^3}{3} - 0 = \frac{16}{3}$$ 6. **Combine the integrals:** $$\int_0^2 (4x - 2x^2) \, dx = 8 - \frac{16}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}$$ 7. **Substitute back:** $$A \times \frac{8}{3} = 1$$ 8. **Solve for $$A$$:** $$A = \frac{1}{\frac{8}{3}} = \frac{3}{8}$$ **Final answer:** $$A = \frac{3}{8}$$