(1) I=∫x2(1+2x3)3dx By substitution method, let u=1+2x3Such that du = 6x2 dx I=∫x2(1+2x3)3dx =I=∫\x2u36\x2du I=41∫u3du = 24u4+CI=24(1+2x3)4+C(2) I =∫x e7x dxUsing integration by part u=x , dv=e7x dxdu=1 , v=7e7x ∫udv=uv − ∫v du I=7xe7x −71∫ e7x dxI=7xe7x −49e7x+C(3) y=1+x2Using method of diskV = ∫−22πy2dx = ∫−22π(1+x2)2dxV = ∫−22π(1+2x2+x4)dxV=15412π
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