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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