Answer is: option3

Solution:

Find the points where the graph intersects the x-axis

Set \( f(x) = 0 \):

\[ 4x - x^3 = 0 \Rightarrow x(4 - x^2) = 0 \Rightarrow x = 0 \quad \text{or} \quad x^2 = 4 \Rightarrow x = \pm 2 \]

So, the curve intersects the x-axis at \( x = -2, 0, 2 \). But since we only want the area in the first quadrant, we consider the region between \( x = 0 \) and \( x = 2 \), where \( f(x) \geq 0 \).

\[ \text{Area} = \int_0^2 (4x - x^3) \, dx \]

\[ \int_0^2 (4x - x^3) \, dx = \int_0^2 4x \, dx - \int_0^2 x^3 \, dx \]

\[ = \left[ 2x^2 \right]_0^2 - \left[ \frac{x^4}{4} \right]_0^2 = 2(2^2) - \frac{2^4}{4} = 2(4) - \frac{16}{4} = 8 - 4 = 4 \]

Final Answer: (C) 4