`y = |x|, y = x^2 - 2` Sketch the region enclosed by the given curves and find its area.

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 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.