Posted by AIRA on Thursday, April 7, 2011 at 2:55am.
Make a sketch and then solve for the intersection of the two curves
x^2 - 2x + 3 = -x^2 + 3
.....
x = 0 or x = 1
so
area = [integral] (-x^2 + 3 -(x^2 - 2x+3)) dx from 0 to 1
= integral (-2x^2 + 2x)dx from 0 to 1
= (-2/3)x^3 + x^2 | from 0 to 1
= -2/3 + 1 - 0
= 1/3
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