Find the area of the region enclosed by
y−2x^2 ≥ 0 and |x|+|y| ≤ 1
the graph of |x|+|y| ≤ 1 is a square with vertices
(1,0), (0,1), (-1,0) and (0,-1)
y ≥ 2x^2 is the region above y = 2x^2
so we need their intersection:
The line of the square in the first quadrant is
y = -x + 1
then 2x^2 = -x + 1
2x^2 + x - 1 = 0
(2x-1)(x+1) = 0
x = 1/2 or x = -1
from the diagram we want x=1/2, then y = 1/2)
From symmetry we can take the region form x = 0 to x = 1/2, then double that answer.
Area = ∫(-x + 1 - 2x^2) dx from 0 to 1/2
= -x^2/2 + x - 2x^3/3 | from 0 to 1/2
= -1/8 + 1/2 - 1/12 - 0
= 7/24
oops, forgot to double our answer ...
since this was only the area in the first quadrant, the area in II would be the same, so the
total area = 14/24 or
7/12 square units
To find the area of the region enclosed by the given inequalities, we can break it down into smaller parts and then calculate the total area.
Step 1: Graph the inequalities
The first inequality, y − 2x^2 ≥ 0, represents a parabolic curve. To graph it, first, set it equal to zero: y − 2x^2 = 0. This equation represents the vertex of the parabola. Since the coefficient of x^2 is negative, the parabola opens downwards. You can plot several points around the vertex to sketch the graph.
The second inequality, |x| + |y| ≤ 1, represents a square. Since the absolute value function always returns a positive value, we can rewrite the inequality as x + y ≤ 1 and -x + y ≤ 1.
Now, graph the square by plotting the line x + y = 1 and -x + y = 1. Shade the area that satisfies both inequalities.
Step 2: Determine the enclosed regions
Looking at the graph, we can see that the region enclosed by the inequalities consists of two parts: the area inside the parabola and the area inside the square.
Step 3: Calculate the area of individual regions
To calculate the area enclosed by the parabolic curve, we need to integrate the equation y = 2x^2 over the appropriate interval. The interval can be determined by finding the x-coordinates where the curve intersects the square.
To calculate the area enclosed by the square, we can find the area of the square directly.
Step 4: Add the areas together
Finally, add the areas of the individual regions to find the total area enclosed by the given inequalities.