f X,Y​ (x,y)= 32​ x(2y+y 2 )

It seems like you're asking for the integration of the function f(x,y) = 32x(2y+y^2). However, you didn't specify the limits of the integration or whether it's a definite or indefinite integral.

If you're looking for the indefinite integral, here's how you can solve it:

First, recognize that the function can be separated into two parts: 32x2y and 32xy^2.

The integral of the function f(x,y) = 32x(2y+y^2) with respect to x and y is then:

∫∫f(x,y) dx dy = ∫(∫32x(2y+y^2) dx) dy

This can be separated and solved as:

= ∫(322y∫x dx + 32y^2∫x dx) dy = ∫(322y(1/2)x^2 + 32y^2(1/2)x^2) dy = ∫(32x^2y + 16x^2y^2) dy

Then, integrate with respect to y:

= 16x^2*∫y dy + 16x^2*∫y^2 dy = 16x^2*(1/2)y^2 + 16x^2*(1/3)y^3 = 8x^2y^2 + 16/3x^2*y^3 + C

So, the indefinite integral of the function f(x,y) = 32x(2y+y^2) is 8x^2y^2 + 16/3x^2*y^3 + C, where C is the constant of integration.