Evaluate the integral Z ZR(x+y)^2dxdy over the triangle R with vertices (0, 0), (2, 2)and (0, 1

The integral you're asking about is a double integral over a triangular region. Here's how you can solve it:

Step 1: Identify the limits of integration The triangular region R is defined by the vertices (0,0), (2,2), and (0,1). This means that x ranges from 0 to 2, and for each x, y ranges from x/2 to x.

Step 2: Set up the double integral The double integral is ∫ from 0 to 2 ∫ from x/2 to x (x+y)^2 dy dx.

Step 3: Evaluate the inner integral First, we integrate with respect to y. The antiderivative of (x+y)^2 with respect to y is (x+y)^3/3. Evaluating this from y=x/2 to y=x gives [x^3/3 + x^2x/3 - (x^3/12 + x^2x/4)] = x^3/4 + x^3/12 = x^3/3.

Step 4: Evaluate the outer integral Now we integrate x^3/3 with respect to x from 0 to 2. The antiderivative of x^3/3 is x^4/12. Evaluating this from x=0 to x=2 gives 2^4/12 - 0 = 16/12 = 4/3.

So, the value of the double integral is 4/3.