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.
[tex]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}\\[/tex]
[tex]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[/tex]
Total area bounded = 4/3 + 49/6 = 57/6
