What is the integer that is greater than $\sqrt{11}$ but less than $\sqrt{19}$ ?
Since $3^{2}=9<11<16=4^{2}$, it must be that $3<\sqrt{11}<4$.

Similarly, $4^{2}=16<19<25=5^{2}$, so $4<\sqrt{19}<5$.

Therefore, $\sqrt{11}<4<\sqrt{19}$.

ANSWER: 4