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

The region is bounded by the line $x + y = 2$, the y-axis ($x = 0$), the vertical line $x = 3$, and the x-axis.

From $x + y = 2$:

$y = 2 - x$

---

To find where the line crosses the x-axis, set $y = 0$:

$0 = 2 - x$

$x = 2$

The line crosses the x-axis at $(2, 0)$.

---

The region splits into two parts at $x = 2$:

From $x = 0$ to $x = 2$: The line is above the x-axis

• At $x = 0$: $y = 2$
• At $x = 2$: $y = 0$

From $x = 2$ to $x = 3$: The line is below the x-axis

• At $x = 3$: $y = -1$

---

For the region from $x = 0$ to $x = 2$ (triangle above x-axis):

Base $= 2$ units

Height $= 2$ units

Area $= \dfrac{1}{2} \times 2 \times 2$

Area $= 2$

---

For the region from $x = 2$ to $x = 3$ (triangle below x-axis):

Base $= 1$ unit

Height $= 1$ unit

Area $= \dfrac{1}{2} \times 1 \times 1$

Area $= \dfrac{1}{2}$

---

Total Area $= 2 + \dfrac{1}{2}$

Total Area $= \dfrac{4}{2} + \dfrac{1}{2}$

Total Area $= \dfrac{5}{2}$

Therefore, the area of the region is $\dfrac{5}{2}$ square units.