Find the total area enclosed by the curves
y= absolute value of x; y=x^2–2
Dr WLS already gave you quite a hint in your typo below.
OK, it looks like the question you submitted just before this was incorrectly stated, as I expected.
See my response to your previous question, where one of the functions was x^2 +2. Integrate x - x^2 +2 from 0 to 2, and double the result. This will take advantage of symmetry of the enclosed zone about the y axis.
how come they will integrate from 0 to 2 and not -1 to 2? that is what i got when i made the two equations equal to find where they intersect
They intersect at -2 and +2. The areas on opposite sides of the y axis are equal, so you can double the area of one side, from x = 0 to 2.
To find the total area enclosed by the curves, you need to calculate the area between the two curves. This can be done by finding the points of intersection of the curves and then integrating the difference between the two curves over the interval of their intersection.
Step 1: Find the points of intersection.
To find the points of intersection, set the two equations equal to each other and solve for x:
|x| = x^2 - 2
This equation has two possible solutions, one when x is positive and one when x is negative.
For x ≥ 0:
x = x^2 - 2
Rearranging the equation:
x^2 - x - 2 = 0
Factoring the quadratic equation:
(x - 2)(x + 1) = 0
So, x = 2 or x = -1.
For x < 0:
-x = x^2 - 2
Rearranging the equation:
x^2 + x - 2 = 0
Factoring the quadratic equation:
(x - 1)(x + 2) = 0
So, x = 1 or x = -2.
Therefore, the points of intersection are (-2, 2), (-1, 1), (1, 1), and (2, 2).
Step 2: Calculate the area between the curves.
To calculate the area between the curves, you need to integrate the difference between the curves over the intervals formed by the points of intersection.
For the interval -2 ≤ x ≤ -1:
∫(x^2 - 2) - |x| dx
For the interval -1 ≤ x ≤ 1:
∫|x| - (x^2 - 2) dx
For the interval 1 ≤ x ≤ 2:
∫(x^2 - 2) - |x| dx
Evaluate each integral separately, and then add up the results to find the total area enclosed by the curves.