If f(x) = |x| and g(x) = x^{2}-2, here is how we would find the area enclosed by the region.

First, set up the integral:

Area=∫ f(x)-g(x) dx

Then, substitute with our functions:

Area=∫ |x|-(x^{2}-2) dx

Then, find the bounds. When does |x| = x^{2}-2? This happens when x=-2 or x= 2. These will be our lower and upper bounds, respectively. Add these into our expression.

Area=∫_{-2}^{2} |x|-(x^{2}-2) dx (The -2 and 2 are mean to be at the top and bottom of the ∫ sign.)

Recognize that we will need to write |x| as a piecewise function and evaluate this integral in two parts: from -2 to 0, and from 0 to 2.

Area = ∫_{-2}^{0} |x|-(x^{2}-2) dx + ∫_{0}^{2} |x|-(x^{2}-2) dx

Rewrite |x| in each expression. (When x is negative, |x| = -x. When x is positive, |x| = x.) Then simplify.

Area = ∫_{-2}^{0} -x-(x^{2}-2) dx + ∫_{0}^{2} x-(x^{2}-2) dx

Area = ∫_{-2}^{0} -x-x^{2}+2 dx + ∫_{0}^{2} x-x^{2}+2 dx

Find the antiderivative using the reverse power rule:

Area = [(-x^{2}/2)-(x^{3}/3)+2x]|_{-2}^{0} + [(x^{2}/2)-(x^{3}/3)+2x]|_{0}^{2 }(The "|_{-2}^{0} at the end means evaluated from -2 to 0)

Next, evaluate the antiderivative by plugging numbers in:

Area = [(-0^{2}/2)-(0^{3}/3)+2*0] - [(-(-2)^{2}/2)-((-2)^{3}/3)+2*(-2)]^{ }+ [(2^{2}/2)-(2^{3}/3)+2*2]^{ }-[(0^{2}/2)-(0^{3}/3)+2*0]

Area = [0] - [(-(-2)^{2}/2)-((-2)^{3}/3)+2*(-2)]^{ }+ [(2^{2}/2)-(2^{3}/3)+2*2]^{ }-[0]

Area = [0] - [(-4/2)-(-8/3)-4]^{ }+ [(4/2)-(8/3)+4]^{ }-[0]

Area = - [-2-(-8/3)-4]^{ }+ [2-(8/3)+4]^{ }

After some algebra, the result simplifies to the answer of 20/3.