Question: 0 1 [ 1 2 ( x + 6 y ) d x ] d y = 0 1 ( 3 2 + 6 y

01[12(x+6y)dx]dy=01(32+6y)dy
=[32y+()y2]|01|
=
0 1 [ 1 2 ( x + 6 y ) d x ] d y = 0 1 ( 3 2 + 6 y

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