UPSC NDA I 2025 Mathematics Paper

Given,
The function is f(x)=⌊x2⌋.
We are tasked with finding the value of ∫√22f(x)‌dx.
We can break the integral into two parts as follows:
∫√22f(x)‌dx=∫√2√32‌dx+∫√323‌dx
For the range √‌2≤x≤√‌3,⌊x2⌋=2. So the first part of the integral is:
∫√2√32dx=2×(√‌3−√‌2)
For the range √‌3≤x≤2,⌊x2⌋=3. So the second part of the integral is:
∫√323‌dx=3×(2−√‌3)
Now, we calculate the values:
‌2×(√‌3−√‌2)=2√‌3−2√‌2
‌3×(2−√‌3)=6−3√‌3
Combining the two parts:
2√‌3−2√‌2+6−3√‌3=6−2√‌2−√‌3