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Find the exact area enclosed between the curve y=sqrt(4-x^2) and the line x-y+2=0

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You first need to find the interception points between the curves, hence, you should solve the system of equations `y=x+2`  and `y=sqrt(4-x^2)`  such that:

`x+2 = sqrt(4-x^2)`

You need to raise to square to remove the square root such that:

`(x+2)^2 = 4 - x^2`

You need to expand the binomial such that:

`x^2 + 4x + 4 = 4 - x^2`

You need to move all terms to the left side such that:

`x^2 + 4x + 4 - 4 + x^2 = 0`

`2x^2 + 4x = 0`

Factoring out 2x yields:

`2x(x + 2) = 0 =gt x = 0 and x+2 = 0 =gt x = -2`

You need to verify what is the upper limit curve and what is the lower limit curve, between -2 and 0 values, hence, you should select the value -1 to evaluate both functions such that:

`y = -1+2 = 1` 

`sqrt(4 - 1) = sqrt 3`

Notice that sqrt>1, hence the...

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