∫(0-> 1-x) 12x^3.y^2 dy
= 4x^3. [y^3]|(0->1-x)
=4x^3. ( 1-x)^3
∫(0->1) [∫(0-> 1-x) 12x^3.y^2 dy ] dx
=∫(0->1) [ 4x^3. ( 1-x)^3 ] dx
=4∫(0->1) (x^3 -3x^4 +3x^5 -x^6) dx
=4[ (1/4)x^4 -(3/5)x^5 + (1/2)x^6 - (1/7)x^7 ]|(0->1)
=4 ( 1/4 -3/5 +1/2 -1/7)
= 4 [(70-168+140-40)/280]
=1/35
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