\[ y^2 \leq 4x, \quad x < 4 \]
\[ \frac{xy(x-1)(x-2)}{(x-3)(x-4)} > 0 \]
Case - I: \( y > 0 \)
\[ \frac{x(x-1)(x-2)}{(x-3)(x-4)} > 0, \quad x \in (0,1) \cup (2,3) \]
Case - II: \( y < 0 \)
\[ \frac{x(x-1)(x-2)}{(x-3)(x-4)} < 0, \quad x \in (1,2) \cup (3,4) \]
Area:
\[ \text{Area} = 2 \int_{0}^{4} \sqrt{x} \, dx \]
\[ = 2 \cdot \frac{2}{3} \left[ x^{3/2} \right]_{0}^{4} = \frac{32}{3} \]
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).