Between which two consecutive integers does \( \sqrt{2 \pi 1} \) lie?

To find between which two consecutive integers \( \sqrt{2 \pi 1} \) lies, we first need to calculate \( 2 \pi \). Using the approximation \( \pi \approx 3.14 \): \[ 2 \pi \approx 2 \times 3.14 \approx 6.28 \] Now we compute \( \sqrt{2 \pi} \): \[ \sqrt{2 \pi} \approx \sqrt{6.28} \] To estimate \( \sqrt{6.28} \), we can find the squares of some consecutive integers. The squares of 2 and 3 are: \[ 2^2 = 4 \quad \text{and} \quad 3^2 = 9 \] Since \( 6.28 \) lies between \( 4 \) and \( 9 \), we check \( \sqrt{6.28} \): \[ 2 < \sqrt{6.28} < 3 \] For a more refined estimate, we can calculate: \[ 2.5^2 = 6.25 \quad \text{and} \quad 2.6^2 = 6.76 \] Since \( 6.28 \) is between \( 6.25 \) and \( 6.76 \), it follows that: \[ 2.5 < \sqrt{6.28} < 2.6 \] Thus, the integer parts are 2 and 3. Therefore, \( \sqrt{2 \pi} \) lies between the consecutive integers \( 2 \) and \( 3 \). In conclusion, \( \sqrt{2 \pi 1} \) lies between the integers \( 2 \) and \( 3 \).