The area of the region bounded by the line y = | x – 2 |, x = 1, x = 3 and x-axis is

The correct option is 1 sq. units.

To find the area of the region bounded by the curve $y=|x-2|$, the vertical lines $x=1$ and $x=3$, and the $x$-axis, we can set up a definite integral representing the area.

The absolute value function $y=|x-2|$ changes its definition at the point where the expression inside the absolute value is zero, which is at $x=2$:
$$|x-2|=\{\begin{array}{ll}-(x-2)=2-x,& \text{if }x<2\\ x-2,& \text{if }x\ge 2\end{array}$$

Since our interval of integration is from $x=1$ to $x=3$, we split the integral at the boundary point $x=2$:
$$\text{Area}={\int }_1^3|x-2|dx$$

Splitting this into two intervals:
$$\text{Area}={\int }_1^2(2-x)dx+{\int }_2^3(x-2)dx$$

Now, let's evaluate each integral separately.

First integral:
$${\int }_1^2(2-x)dx={[2x-\frac{x^2}{2}]}_1^2$$
Substituting the limits:
$$=(2(2)-\frac{2^2}{2})-(2(1)-\frac{1^2}{2})$$
$$=(4-2)-(2-\frac{1}{2})=2-\frac{3}{2}=\frac{1}{2}$$

Second integral:
$${\int }_2^3(x-2)dx={[\frac{x^2}{2}-2x]}_2^3$$
Substituting the limits:
$$=(\frac{3^2}{2}-2(3))-(\frac{2^2}{2}-2(2))$$
$$=(\frac{9}{2}-6)-(2-4)=-\frac{3}{2}-(-2)=-\frac{3}{2}+2=\frac{1}{2}$$

Adding the two values together:
$$\text{Total Area}=\frac{1}{2}+\frac{1}{2}=1\text{sq. units}$$

Thus, the area of the region is indeed 1 sq. units.