Question:
What is \(\int_0^{\sqrt{2}} [x^2]\,dx\) equal to (where \([.]\) is the greatest integer function)?
\(\int_0^{\sqrt{2}} [x^2]\,dx\) किसके बराबर है (जहाँ \([.]\) महत्तम पूर्णांक फलन है)?
Options:
- \(\sqrt{2}-1\)
- \(1-\sqrt{2}\)
- \(2(\sqrt{2}-1)\)
- \(\sqrt{3}-1\)
- \(\sqrt{2}-1\)
- \(1-\sqrt{2}\)
- \(2(\sqrt{2}-1)\)
- \(\sqrt{3}-1\)
Correct Answer: a)
Explanation:
For \(x \in [0, 1)\), \(x^2 \in [0, 1) \implies [x^2] = 0\).
For \(x \in [1, \sqrt{2}]\), \(x^2 \in [1, 2] \implies [x^2] = 1\).
Therefore:
\[\int_0^{\sqrt{2}} [x^2]\,dx = \int_0^1 0\,dx + \int_1^{\sqrt{2}} 1\,dx = [x]_1^{\sqrt{2}} = \sqrt{2} – 1\]
Hence, the correct option is (a).
जब \(x \in [0, 1)\), तब \(x^2 \in [0, 1) \implies [x^2] = 0\).
जब \(x \in [1, \sqrt{2}]\), तब \(x^2 \in [1, 2] \implies [x^2] = 1\)।
अतः समाकलन:
\[\int_0^{\sqrt{2}} [x^2]\,dx = \int_0^1 0\,dx + \int_1^{\sqrt{2}} 1\,dx = [x]_1^{\sqrt{2}} = \sqrt{2} – 1\]
अतः सही विकल्प (a) है।