So I'm taking a calculus course and I admit I don't quite understand this topic in my course.
Given that the integral of the function $f(x,y)$ over the area $[0,2] \times [0,2]$ equals $4$ , how can we compute the value of the integral $$\iint_R(f(x,y) + 2) \,dx\,dy\,\,?$$
So $$\iint_R(f(x,y) + 2) \,dx\,dy = \iint_R f(x,y) \,dx\,dy + 2\iint_R \,dx\,dy$$
Then $4 + 2\iint_R \,dx\,dy$? But I'm not sure how to proceed from here.
I'm guessing the bound $R$ is equivalent to writing $\int_0^2\int_0^2(f(x,y) + 2) \,dx\,dy$? I would really appreciate some help as I'm super confused (as mentioned, new topic for me). Thanks in advance!
If $f=4$ over $[0,2] \times [0,2]$, then $$\int_0^2\int_0^2(f(x,y) + 2) dxdy= \int_0^2\int_0^2f(x,y) dxdy+8=4\cdot 4+8=24$$
Addition.
Of course, if $\int_0^2\int_0^2f(x,y) dxdy = 4$, then answer will be $12$: $$\int\limits_0^2\int_\limits0^2(f(x,y) + 2) dxdy= \int_\limits0^2\int_\limits0^2f(x,y) dxdy+2\int_\limits0^2\int_\limits0^2dxdy =\\ = 4+2\cdot \int\limits_{0}^{2}dx \cdot \int\limits_{0}^{2}dy =12$$