The area (in square units) of the region enclosed between the lines $x + y = 2, x = 0, x = 3$ and x-axis is equal to

$\frac{5}{2}$

The correct answer is Option (3) → $\frac{5}{2}$

Correct interpretation:

The region enclosed by the lines is NOT just the triangular region from $x=0$ to $x=2$.

Since the line $x = 3$ is also a boundary, the full enclosed region is a trapezium formed by:

• $x = 0$ (left vertical line)

• $x = 3$ (right vertical line)

• $y = 0$ (x-axis)

• $x + y = 2$ (slanted line)

Between $x = 2$ and $x = 3$, the line $x + y = 2$ gives negative $y$, so the region is only between $x = 0$ and $x = 2$ — but the boundary $x=3$ makes the enclosed figure include the triangle between $x = 2$ and $x = 3$ cut by the x-axis.

Thus the region consists of:

1. Triangle between $x = 0$ and $x = 2$ under $y = 2 - x$

2. Vertical strip from $x = 2$ to $x = 3$ with height $0$ → adds zero area.

So enclosed area = area of triangle with base 2 and height 2:

$\text{Area} = \frac{1}{2} \cdot 2 \cdot 2 = \frac{5}{2}$

The correct area is $\frac{5}{2}$ square units.