(1 +(x-1)^2*y+4y^2) with first limit 0 to 2 and second limit 0 to 3

It seems like you're asking for a double integral of the function (1 +(x-1)^2*y+4y^2) over the region defined by 0 ≤ x ≤ 2 and 0 ≤ y ≤ 3. Here's how you can solve it:

Step 1: Integrate with respect to y first. The integral of the function with respect to y from 0 to 3 is:

∫ from 0 to 3 [1 +(x-1)^2*y+4y^2] dy

Step 2: Break this integral up into three separate integrals:

= ∫ from 0 to 3 dy + ∫ from 0 to 3 (x-1)^2*y dy + ∫ from 0 to 3 4y^2 dy

Step 3: Evaluate these integrals:

= [y] from 0 to 3 + [(x-1)^2 * y^2 / 2] from 0 to 3 + [4 * y^3 / 3] from 0 to 3

= 3 + (x-1)^2 * 9 / 2 + 4 * 27 / 3 - 0

= 3 + 4.5*(x-1)^2 + 36

Step 4: Now integrate this result with respect to x from 0 to 2:

∫ from 0 to 2 [3 + 4.5*(x-1)^2 + 36] dx

Step 5: Break this integral up into three separate integrals:

= ∫ from 0 to 2 3 dx + ∫ from 0 to 2 4.5*(x-1)^2 dx + ∫ from 0 to 2 36 dx

Step 6: Evaluate these integrals:

= [3x] from 0 to 2 + [4.5 * (x^3/3 - x^2 + x/3)] from 0 to 2 + [36x] from 0 to 2

= 6 + 4.5 * (8/3 - 4 + 2/3) + 72 - 0

= 6 + 4.5 * (4/3) + 72

= 6 + 6 + 72

= 84

So, the double integral of the function (1 +(x-1)^2*y+4y^2) over the region defined by 0 ≤ x ≤ 2 and 0 ≤ y ≤ 3 is 84.