Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 26

You need to find the intersection points beteen the curves, hence, since |x| = y, you need to discuss two cases, y = -x and y = x.
For y = -x, you need to find the intersection between curves by solving the equation, such that:
-x = x^2 - 2 => x^2 + x - 2 = 0
x_(1,2) = (-1+-sqrt(1 + 8))/2 => x_(1,2) = (-1+-3)/2
x_1 = 1; x_2 = -2
Hence, you need to check what curve is greater on interval [-2,1] and you may notice that -x > x^2 - 2 on [-2,0] and -x < x^2-2 on [0,1], hence, you may evaluate the area. such that:
int_(-2)^1 | -x - x^2 + 2| dx = int_(-2)^0 ( -x - x^2 + 2) dx + int_0^1 (x^2 + x - 2) dx
int_(-2)^1 | -x - x^2 + 2| dx = (- x^2/2 - x^3/3 + 2x)|_(-2)^0 + (x^3/3 + x^2/2 - 2x)|_0^1
int_(-2)^1 | -x - x^2 + 2| dx = ( (-2)^2)/2 + ((-2)^3)/3 - 2*(-2) + 1/3 + 1/2 - 2
int_(-2)^1 | -x - x^2 + 2| dx = 2 - 8/3 + 4 - 2 + 1/3 + 1/2
int_(-2)^1 | -x - x^2 + 2| dx = 4 - 7/3 + 1/2
int_(-2)^1 | -x - x^2 + 2| dx = (24 - 14 + 3)/6
int_(-2)^1 | -x - x^2 + 2| dx = 13/6
Hence, evaluating the area enclosed by the curves y = -x and y = x^2 - 2 yields int_(-2)^1 | -x - x^2 + 2| dx = 13/6.

The region whose area is A = 13/6 is enclosed by the red line and the red curve.