To solve the inequality
\[ 5 - 2r + 3 \geq -11, \]
first, combine like terms on the left side.
\[ 5 + 3 = 8, \]
so we have:
\[ 8 - 2r \geq -11. \]
Next, subtract 8 from both sides:
\[ -2r \geq -11 - 8, \]
which simplifies to:
\[ -2r \geq -19. \]
Now, divide by -2. Remember that when you divide or multiply by a negative number, you must reverse the inequality sign:
\[ r \leq \frac{-19}{-2} \Rightarrow r \leq 9.5. \]
Since the problem asks for integers, we can conclude that:
\[ r \leq 9. \]
Since there is no lower boundary given in the original problem, let's assume \( r \) can take any integer value less than or equal to 9. Thus, we can express this as:
\[ r \in (-\infty, 9]. \]
Therefore, as a compound inequality for integers, the solution is:
\[ r \leq 9. \]
In terms of integers, the answer is:
\[ r \in {\ldots, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. \]