Hello,
plz try to solve that question.
Thanks.
Do the inside integral, aka the y integral, first:
Int[(4x + 2)dy, x^2, 2x]
= [(4x + 2)y, x^2, 2x] <-- x is a constant for the y integral
= (4x + 2)*2x - (4x + 2)*x^2
= 8x^2 + 4x - 4x^3 - 2x^2
= -4x^3 + 6x^2 + 4x
Now do the outside integral, aka the x integral:
Int[(-4x^3 + 6x^2 + 4x)dx, 0, 2]
= [-4*(1/4)x^4 + 6*(1/3)x^3 + 4*(1/2)x^2, 0, 2]
= -2^4 + 2*2^3 + 2*2^2
= -16 + 16 + 8
= 8
-Dan