Let the area of the region {(x,y):0≤x≤3,0≤y≤ min{x2+2,2x+2}} be A. Then 12 A is equal to ______

To solve this problem, we need to find the area under the curves y = x^2 + 2 and y = 2x + 2 in the interval [0, 3] and then multiply it by 12.

Step 1: Find the intersection points of the two curves. Set x^2 + 2 = 2x + 2 and solve for x. The solution is x = 0 and x = 2.

Step 2: Calculate the area under each curve in the interval [0, 2].

For y = x^2 + 2, the integral from 0 to 2 is ∫(x^2 + 2) dx from 0 to 2 = [x^3/3 + 2x] from 0 to 2 = 8/3 + 4 = 20/3.

For y = 2x + 2, the integral from 0 to 2 is ∫(2x + 2) dx from 0 to 2 = [x^2 + 2x] from 0 to 2 = 4 + 4 = 8.

Step 3: Subtract the smaller area from the larger area to get the area of the region. A = 20/3 - 8 = 4/3.

Step 4: Multiply the area by 12. 12A = 12 * 4/3 = 16.

So, 12A is equal to 16.