The area bounded by the curves $$y = |x-1| + |x-2|$$ and $$y = 3$$ is equal to

To find the area bounded by $$y = |x - 1| + |x - 2|$$ and $$y = 3$$, we first analyze the shape of the absolute value function.

1. Identify the Curve

The function $$y = |x - 1| + |x - 2|$$ creates a "bucket" shape:

For $$1 \le x \le 2$$: $$y = (x - 1) - (x - 2) = 1$$ (The flat bottom).For $$x > 2$$: $$y = 2x - 3$$.For $$x < 1$$: $$y = -2x + 3$$.

2. Find Intersection Points

Set the side equations equal to $$y = 3$$:$$2x - 3 = 3 \implies 2x = 6 \implies \mathbf{x = 3}$$.

$$-2x + 3 = 3 \implies -2x = 0 \implies \mathbf{x = 0}$$.

The region is a trapezoid with vertices at $$(0, 3)$$, $$(1, 1)$$, $$(2, 1)$$, and $$(3, 3)$$.

3. Calculate Area

The area is the region between the line $$y = 3$$ and the curve:

$$\text{Area} = \int_{0}^{3} (3 - y) \, dx$$

Alternatively, use the trapezoid formula where height $$h = (3 - 1) = 2$$, top base $$a = (2 - 1) = 1$$, and bottom base $$b = (3 - 0) = 3$$:$$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$$

$$\text{Area} = \frac{1}{2} \times (1 + 3) \times 2$$

$$\text{Area} = 4 \text{ square units}$$

Final Result:The area is 4.