Posted by Lindsay on Friday, August 12, 2011 at 4:56am.
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
Related Questions
Calculus:) - Find the area of the region enclosed by y−2x^2≥...
Calculus:) - Find the area of the region enclosed by y−2x^2≥...
Calculus 2 - Find c>0 such that the area of the region enclosed by the ...
math - Maximize P = 2x − 3y subject to x + 2y ≤ 14 6x + y &...
math - Maximize and minimize p = x + 2y subject to x + y ≥ 8 x + y &...
math - Maximize z = 16x + 8y subject to: 2x + y ≤ 30 x + 2y &#...
MATH - 1. Maximize z = 16x + 8y subject to: 2x + y ≤ 30 x + 2y &#...
linear programming - Maximize z = 16x + 8y subject to: 2x + y ≤ 30 x...
calculus - find the value of m so that the line y = mx divides the region ...
calculus HELP - Find the values of a and b that make f continuous everywhere. f(...
For Further Reading
