20 Evaluate the following iterated integral 0 to 1 x2 to 1x2

Solution

(0 to 1) (x/2 to 1-x/2) (0 to 2-x-2y) 1 dz dy dx
integrate 1 with dz then you will get z
(0 to 1) (x/2 to 1-x/2) [z] from (0 to 2-x-2y) dy dx
(0 to 1) (x/2 to 1-x/2) [2-x-2y]dy dx
integrate [2-x-2y] with dy you will get [2y-xy-y²]
(0 to 1) [2y-xy-y²]from (x/2 to 1-x/2) dx
(0 to 1) [2(1-x/2 -x/2 ) -x(1-x/2 -x/2 ) - (1-x/2)²+(x/2)²] dx
(0 to 1) [ 2- 2x -x +x² -1+x] dx
(0 to 1) (x²-2x+1)dx
(0 to 1)(x-1)² dx
(x-1)³/3 from 0 to 1
= -1/3