The area, in sq. units, enclosed by the lines $x = 2$, $y = |x - 2| + 4$, the $X-$axis and the $Y-$axis is equal to

We need to find the area enclosed by:

Line x = 2 (a vertical line)

Curve y = |x - 2| + 4 (an absolute value function)

X-axis (y = 0)

Y-axis (x = 0)

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The key to solving this problem is understanding $y = |x - 2| + 4$.

The absolute value function |x - 2| creates a "V" shape with its vertex at x = 2.

Let's break it into cases:

When $x \geq 2$: $|x - 2| = x - 2$, so $y = (x - 2) + 4 = x + 2$

When $x < 2$: $|x - 2| = -(x - 2) = -x + 2$, so $y = (-x + 2) + 4 = -x + 6$

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To find the enclosed area, we need to identify where these boundaries intersect:

Point A: Y-axis meets X-axis → (0, 0)

Point B: X-axis meets x = 2 → (2, 0)

Point C: Line x = 2 meets y = |x - 2| + 4

At x = 2: $y = |2 - 2| + 4 = 0 + 4 = 4$ → (2, 4)

Point D: Y-axis meets y = |x - 2| + 4

At x = 0: $y = |0 - 2| + 4 = 2 + 4 = 6$ → (0, 6)

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The four corner points are: (0, 0), (2, 0), (2, 4), and (0, 6).

This forms a trapezium (trapezoid) with:

Bottom base: from (0, 0) to (2, 0) with length = 2

Left side: from (0, 0) to (0, 6) with height = 6

Right side: from (2, 0) to (2, 4) with height = 4

Top side: the slanted line from (0, 6) to (2, 4)

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Formula for trapezium area: Area = $\tfrac{1}{2} \times$ (sum of parallel sides) $\times$ height

The parallel sides are the vertical segments:

Left side = 6 units

Right side = 4 units

Horizontal distance (height) = 2 units

Area = $\tfrac{1}{2} \times (6 + 4) \times 2 = \tfrac{1}{2} \times 10 \times 2 = 10$ square units

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When dealing with absolute value functions, we always break them into cases based on where the expression inside the absolute value changes sign. This helps us understand the shape of the curve and find intersection points accurately.

Answer: 10 square units