Complete Question is:

Using integration find the area of the following region {(x,y): |x +1| ≤ y ≤ √(5 - x² )}

Solution:

Given, y ≥ |x - 1| ................1

So, the upper part of the curve is V.

Again, given y ≤ √(5 - x² )

=> y² ≤ 5 - x²

=> y² + x² ≤ 5

=> x² + y² ≤ 5 .............2

Which is an equation of the circle.

Now from the figure, the dotted region of the required region.

To find the point of intersection,

Put y = x - 1 in equation 2, we get

     x² + (x - 1)² = 5

=> x² + x² + 1 - 2x = 5

=> 2x² + 1 - 2x - 5 = 0

=> 2x² - 2x - 4 = 0

=> x² - x - 2 = 0

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

=> x = -1, 2

Now, the required area  = _-1∫² [√(5 - x² ) - (x - 1)] dx

                                = [{(x/2)*√(5 - x² ) + (5/2)*sin^-1 (x/√5)} - x² /2 + x _-1]²

                                = [{(2/2)*√(5 - 2² ) + (5/2)*sin^-1 (2/√5)} - 2² /2 + 2] - [{(-1/2)*√(5 - (-1)² ) + (5/2)*sin^-1 (-1/√5)} - (-1)² /2 - 1]

                                = [1 + (5/2)*sin^-1 (2/√5)] - [-1 + (5/2)*sin^-1 (-1/√5) - 3/2]

                                = 1 + (5/2)*sin^-1 (2/√5) + 1 - (5/2)*sin^-1 (-1/√5) + 3/2

                                = 2 + 3/2 + (5/2)*sin^-1 (2/√5) - (5/2)*sin^-1 (-1/√5)

                                = 7/2 + (5/2)*sin^-1 (2/√5) - (5/2)*sin^-1 (-1/√5)