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) → $\frac{5}{2}$

The line is:

$y = 2 - x$

The region is bounded by:

• $x = 0, x = 3$
• x-axis ($y = 0$)
• the line $y = 2 - x$

Since the line crosses the x-axis at $x = 2$, we split the area:

From $x = 0$ to $x = 2$ (above x-axis)

$\text{Area}_1 = \int_{0}^{2} (2 - x) \, dx$

From $x = 2$ to $x = 3$ (below x-axis $\rightarrow$ take positive area)

$\text{Area}_2 = \int_{2}^{3} (x - 2) \, dx$

$\int\limits_{0}^{2} (2 - x) \, dx + \int\limits_{2}^{3} (x - 2) \, dx$

Now evaluating:

• First part = 2
• Second part = 1/2

Total area:

$2 + \frac{1}{2} = \frac{5}{2}$