Calculus
posted by Lindsay .
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
(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 
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
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