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September 20, 2014

September 20, 2014

Posted by **Lindsay** on Friday, August 12, 2011 at 4:56am.

y−2x^2 ≥ 0 and |x|+|y| ≤ 1

- Calculus -
**Reiny**, Friday, August 12, 2011 at 7:38amthe 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

- continuation - Calculus -
**Reiny**, Friday, August 12, 2011 at 7:39amoops, 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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