This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
See diagram.   it is always good to draw a diagram and do the work with areas and volumes bounded by curves and planes.

The limits for the integration and the ends of the bounded area are obtained by intersection of axes, the curve and the straight line.
     y² = x  and x+y =2  =>   y² = x = (2-x)²   
       =>  x²-5x+4 = 0    =>  x = 1 and 4 
   hence points of intersection  are  (1,1,) and (4,-2)

Area bounded is divided in to two regions, one enclosed with in the parabola, from x = 0 to x = 1.  The other the area between the line x+y=2 and parabolic curve.

area\ A = 2 \int\limits^1_0 {y} \, dx =2 \int\limits^1_0 {\sqrt{x}} \, dx =2*[\frac{2}{3}x^{3/2}]_0^1\\\\=\frac{4}{3}*[1^{3/2}-0]=\frac{4}{3}\\

area\ B: \int\limits^4_1 {(y1-y2)} \, dx =\int\limits^4_1 {[(2-x)-(-\sqrt{x})]} \, dx\\\\ =\int\limits^4_1 {(2-x+\sqrt{x})} \, dx=[2x-\frac{1}{2}x^2+\frac{2}{3}x^{\frac{3}{2}}]_1^4\\\\=8-2+16/3-2+1/2-2/3\\\\=7/2+14/3=49/6

Total area bounded = 4/3 + 49/6 = 57/6