# What is the area of the region enclosed between the curves y=x^2-2x+2 and -x^2+6 ?

## Expert Answers We have to find the area enclosed between y=x^2 - 2x + 2 and y = -x^2 + 6.

First lets find the points of intersection

x^2 - 2x + 2 = -x^2 + 6

=> 2x^2 - 2x - 4 = 0

=> x^2 - x - 2 = 0

=> x^2 - 2x + x - 2 = 0

=> x(x - 2) +1(x - 2) = 0

=> (x + 1)(x - 2) = 0

x = -1 and x = 2

Integrate the two expressions x^2 - 2x + 2 and -x^2 + 6 between the limits x = -1 and x = 2.

Int[ x^2 - 2x + 2 ] = x^3/3 - x^2 + 2x + C

Within the limits: 2^3/3 - 2^2 + 2*2 + C - (-1)^3/3 + (-1)^2 - 2*(-1) + C

=> 8/3 - 4 + 4 + C + 1/3 + 1 + 2 + C

=> 6

Int [ -x^2 + 6 ]  = -x^3/3 + 6x + C

Within the limits x = -1 and x = 2

=> -2^3/3 + 6*2 + C  + (-1)^3/3 + 6*(-1) + C

=> -8/3 + 12 - 1/3 - 6

=> 3

But in this interval the two curves are curved in different directions as one curve has the term x^2 and the other has -x^2. So we have to add the areas.

The enclosed area is 9.

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