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