$ \text { The limits of y are }y= x^2 \text{ and } y = 2-x \Longrightarrow x+y =2 \text { and x=0 to x=1} $

$ \text { The point of intersection of x+y = 2 and y = } x^2 $

$ x + x^2 = 2 \Longrightarrow x^2 +x-2 = 0 \Longrightarrow \text{x=1, x=-2} $

The region of integration is OABC. To change the order we consider a strip parallel to the x-axis, the region has to be divided into two parts OAC & ABC.In the region OAC, x varies from 0 to and y varies from 0 to 1. In region ACB, x varies from 0 to 2-y and y varies from 1 to 2.

$ I = \int_{x=0}^{1} \int_{y=0}^{\sqrt{y}} xy dy dx + \int_{x=1}^{2} \int_{y=0}^{2-y} xy dy dx $

$ = \int_{x=0}^{1} \left[ \frac{x^2}{2 } \right]_0^{\sqrt y} y dy + \int_{x=1}^{2} \left[ \frac{x^2}{2 } \right]_0^{2-y} y dy $

$ = \frac{1}{2} \int_{0}^{1} y^2 dy + \frac{1}{2} \int_{1}^{2} (2-y)^2 ydy $

$ = \frac{1}{2} \left[\frac{y^3}{3} \right ]_0^1 + \frac{1}{2} \int_1^2(4-4y+y^2)ydy $

$ = \frac{1}{6} + \frac{1}{2} \left[\frac{4y^2}{2} - \frac{4y^3}{3} +\frac{y^4}{4} \right ]_1^2 $

$ = \frac{1}{6} + \frac{1}{2} \left[\frac{4(4)}{2} - \frac{4(8)}{3} +\frac{16}{4} - \frac{4}{2} +\frac{4}{3} - \frac{1}{4} \right ] $

$ = \frac{1}{6} + \frac{1}{2} \frac{5}{12} $

$ = \frac{3}{8} $