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August 1, 2014

August 1, 2014

Posted by **rajiv n.** on Wednesday, May 18, 2011 at 11:02pm.

- calculus -
**Reiny**, Thursday, May 19, 2011 at 7:21ammake a sketch to see that in your region the straight line lies above the parabola.

Find the intersection:

x^2= x/2 + 2

2x^2 - x - 4 = 0

x = appr 1.686 or x = -1.183

so we want:

Area = [integral] (x/2 + 2 - x^2) dx from 0 to 1.686

= (1/4)x^2 + 2x - (1/3)x^3 from 0 to 1.686

= ...

I will let you do the arithmetic.

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