How in the second step the limit of x is 0 to 2 in the 3rd integal ???? please explain the limit of x should be 0 to 1 .

EEduncle (0,3) X+y=3 x+y=2 (0,2) X+y=1\ (0,1) X A Solution. The domain of integration R (.e. region ABDE) IS divided into three domains ARE BCEF and CDE and [x + y] = 0 in the region AB, X + Y =1 n region BCEF and (x + yl = 2 in the renion CDE ,0)(1,0) (2,0) (3,0) Chang new C Therefore, we have l +yldkdy-Jdxx+yldy +J, dxJ,, x+ yldy + J. dx x + yldy - e'o dy+ ax1dy- dx2 dy - 2-x-1+ xbx+2 [2 -2+ xdx (r. A -3dk+2 xdx