Thursday

July 31, 2014

July 31, 2014

Posted by **AIRA** on Thursday, April 7, 2011 at 2:55am.

- MATH -
**Reiny**, Thursday, April 7, 2011 at 8:18amMake 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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