Let f(x)= |x| and g(x)= x^2-2

Question

Find the area of the region enclosed by f(x) and g(x)

Answer:

Answer 1

If f(x) = |x| and g(x) = x²-2, 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|-(x²-2) dx

Then, find the bounds. When does |x| = x²-2? This happens when x=-2 or x= 2. These will be our lower and upper bounds, respectively. Add these into our expression.

Area=∫_-2² |x|-(x²-2) 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 = ∫_-2⁰ |x|-(x²-2) dx + ∫₀² |x|-(x²-2) dx

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

Area = ∫_-2⁰ -x-(x²-2) dx + ∫₀² x-(x²-2) dx

Area = ∫_-2⁰ -x-x²+2 dx + ∫₀² x-x²+2 dx

Find the antiderivative using the reverse power rule:

Area = [(-x²/2)-(x³/3)+2x]|_-2⁰ + [(x²/2)-(x³/3)+2x]|₀² (The "|_-2⁰ at the end means evaluated from -2 to 0)

Next, evaluate the antiderivative by plugging numbers in:

Area = [(-0²/2)-(0³/3)+2*0] - [(-(-2)²/2)-((-2)³/3)+2*(-2)] + [(2²/2)-(2³/3)+2*2] -[(0²/2)-(0³/3)+2*0]

Area = [0] - [(-(-2)²/2)-((-2)³/3)+2*(-2)] + [(2²/2)-(2³/3)+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.

Answer 2

f(x) = |x|

g(x)= x^2 -2

Area between them = 6 2/3

It helps to graph them. g(x) is an upward opening parabola. f(x) is two half lines, y=x and y=-x,

minimum g(x) is (0,-2) minimum f(x) is the origin (0,0)

Calculate the area between them from x=0 to x=2, which is 3 1/3

then multiply by 2 to get 6 2/3 = total area between them

break up the area between them from x=0 to 2 into 2 areas, below the x axis and above the x axis

area below is the integral of of x^2-2 evaluated from x=0 to x= sqr2. that area is 4sqr2/3

area above is 2 - the integral of x^2-2 evaluated from x=sqr2 to x=2. that area is (10-4sqr2)/3

add them

10/3 -4sqr2/3 +4sqr2/3 = 10/3

multiply it by 2

total area = 2(10/3)

= 20/3

= 6 2/3