Posted by Lindsay on .
Find the area of the region enclosed by
y−2x^2 ≥ 0 and x+y ≤ 1

Calculus 
Reiny,
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
(2x1)(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 
continuation  Calculus 
Reiny,
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