Given, the equation of curves are:

  y = √(5 - x² ) .................1

and y = |x - 1| .................2

Curve 1 is the part of circle x² + y² = 5 which is above the x-axis. 

From equation 1 and 2, we get

      √(5 - x² ) = |x - 1|

=> 5 - x²  = (x - 1)²

=> 5 - x²  = x²  + 1 - 2x

=> x²  + 1 - 2x + x²  - 5 = 0 

=> 2x² - 2x - 4 = 0

=> x² - x - 2 = 0

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

=> x = 2, -1

Now, required area = ²∫_-1 √(5 - x² ) dx - f|x - 1| dx

Now ²∫_-1 √(5 - x² ) dx = [x√(5 - x² )/2 + (5/2)*sin^-1 (x/√5) _-1]²

=>  ²∫_-1 √(5 - x² ) dx = 1 + (5/2)*sin^-1 (2/√5) - {-1 + (5/2)*sin^-1 (-1/√5)}

=>  ²∫_-1 √(5 - x² ) dx = 2 + (5/2)*[sin^-1 (2/√5) + sin^-1 (1/√5)}

=>  ²∫_-1 √(5 - x² ) dx = 2 + (5/2)*[sin^-1 {(2/√5)*{√(1 - 1/5)} + (1/√5)*{√(1 - 4/5)}}]

=>  ²∫_-1 √(5 - x² ) dx = 2 + (5/2)*sin^-1 {(2/√5)*(2/√5) + (1/√5)*(1/√5)}

=>  ²∫_-1 √(5 - x² ) dx = 2 + (5/2)*sin^-1 (1)

=>  ²∫_-1 √(5 - x² ) dx = 2 + (5/2)*(π/2)

=>  ²∫_-1 √(5 - x² ) dx = 2 + 5π/4

Again,

²∫_-1 |x - 1| dx = ¹∫_-1 |x - 1| dx + ²∫₁ |x - 1| dx 

=> ²∫_-1 |x - 1| dx = - ¹∫_-1 (x - 1) dx + ²∫₁ (x - 1) dx 

=> ²∫_-1 |x - 1| dx = [x - x² /2 _-1]¹ + [x² /2 - x ₁]² 

=> ²∫_-1 |x - 1| dx = {1 - (1/2) - (-1 - 1/2)} + {4/2 - 2 - (1/2 - 1) }

=> ²∫_-1 |x - 1| dx = 1/2 + 3/2 + 1/2

=> ²∫_-1 |x - 1| dx = 5/2

So, the required area =  2 + 5π/4 - 5/2 = 5π/4 - 1/2 square units