Find the area enclosed between the line `y=2` and the curve `y=x^2+1` .

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First we will find the points of intersection between the line `y=2` and `y= x^2 +1`

==> `x^2 +1 = 2`

==> `x^2 = 1`

==> `x = +-1`

Then, we need to find the area bounded by the line y=2 and the curve x^2+1 and between x=-1 and x= 1

`==gt Area = int_-1^1 (2-(x^2+1)) dx `

`==gt Area = int_-1^1 1-x^2 dx`

= x-x^3/3 Between x=-1 and x= 1

`==gt (1-1/3 - (-1+1/3) `

`==gt 2/3 + 1 - 1/3 = 4/3`

**Then, the area is 4/3 square units.**

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