Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 18

First, you need to find the point of intersection between the curves y = sqrt(x-1) and y = x - 1 , by solving the equation:
sqrt(x-1) = x-1 => x - 1 = (x-1)^2 => (x-1)^2 - (x-1) = 0
Factoring out (x-1) yields:
(x-1)(x-1-1) = 0 => (x-1)(x-2) = 0 => x - 1 = 0 or x - 2 = 0
Hence, x = 1 and x = 2 and these values are the endpoints of the definite integral you need to evaluate to find the area enclosed by the given curves.
You must check what curve is greater than the other on interval [1,2] and you may notice that y = x - 1 is greater that y = sqrt(x-1) on interval [1,2].
x - 1 > sqrt(x - 1)
You may evaluate the area such that:
int_1^2 |((x - 1) - sqrt(x - 1))|dx = int_1^2 xdx - int_1^2 dx - int_1^2 sqrt(x - 1) dx
int_1^2 |((x - 1) - sqrt(x - 1))|dx = |(x^2)/2|_1^2 - x|_1^2 - (2/3)(x-1)^(3/2)|_1^2|
int_1^2 |((x - 1) - sqrt(x - 1))|dx = |4/2 - 1/2 - 2 + 1 - (2/3) + 0|
int_1^2 |((x - 1) - sqrt(x - 1))|dx = |3/2 - 1 - 2/3|
int_1^2 |((x - 1) - sqrt(x - 1))|dx = |(9 - 6 - 4)/6|
int_1^2 |((x - 1) - sqrt(x - 1))|dx = |-1/6| = 1/6
Hence, evaluating the area enclosed by the curves yields int_1^2 |((x - 1) - sqrt(x - 1))|dx = |-1/6| = 1/6.

The area evaluated above is the area of the region between the red line and orange curve, for x in [1,2].