The area bounded by the curve $\mathrm{y}=4 x-x^2$ and X - axis in square units, is $\qquad$

The area bounded by $y = 4x - x^2$ and the X-axis is determined by finding the points of intersection. Setting $y = 0$ yields $4x - x^2 = 0$ or $x(4 - x) = 0$, so $x = 0$ and $x = 4$ serve as the limits of integration.

Between these limits, the parabola is above the X-axis since $y$ is positive. The area is thus given by the definite integral $\int_{0}^{4} (4x - x^2) dx$.

Evaluating the integral:

$\left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4} = \left(32 - \frac{64}{3}\right) - 0 = \frac{32}{3}$.

This matches option A.

Answer: A