The area of the region $\left\{(x,y): 0 \le x \le \frac{9}{4},\ 0 \le y \le 1,\ x \ge 3y,\ x + y \ge 2\right\}$ is

The area of the region {(x,y): 0 ≤ x ≤ 9/4, 0 ≤ y ≤ 1, x ≥ 3y, x + y ≥ 2} is $0 \le x \le \frac{9}{4},\quad 0 \le y \le 1$ $x \ge 3y \Rightarrow y \le \frac{x}{3};\qquad x + y \ge 2 \Rightarrow y \ge 2 - x$ $2 - x \le y \le \frac{x}{3},\qquad 0 \le y \le 1$ $\text{Find area of this region}$ $\frac{x}{3} = 2-x \Rightarrow x=\frac{3}{2},\quad y=\frac{1}{2}$ $A = \int_{3/2}^{2}\left(\frac{x}{3} - (2-x)\right)dx + \int_{2}^{9/4}\left(\frac{x}{3} - 0\right)dx$ $= \int_{3/2}^{2}\left(\frac{4x}{3}-2\right)dx + \int_{2}^{9/4}\frac{x}{3}\,dx$ $\int\left(\frac{4x}{3}-2\right)dx = \frac{2x^2}{3}-2x$ $\left[\frac{2x^2}{3}-2x\right]_{3/2}^{2} = -\frac{4}{3}-\left(-\frac{3}{2}\right) = \frac{1}{6}$ $\int_2^{9/4}\frac{x}{3}\,dx = \left[\frac{x^2}{6}\right]_2^{9/4} = \frac{1}{6}\left(\frac{81}{16}-4\right) = \frac{17}{96}$ $A = \frac{1}{6} + \frac{17}{96} = \frac{16+17}{96} = \frac{33}{96}$ $= \frac{11}{32} \Rightarrow \text{option (1)}$