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√ 2 ∫ 0 [ X 2 ] D X .

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प्रश्न

\[\int\limits_0^\sqrt{2} \left[ x^2 \right] dx .\]
बेरीज
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उत्तर

\[\text{We have}, \]
\[I = \int\limits_0^\sqrt{2} \left[ x^2 \right] dx\]
\[ = \int\limits_0^1 \left[ x^2 \right] dx + \int\limits_1^\sqrt{2} \left[ x^2 \right] dx\]
\[ = \int\limits_0^1 \left( 0 \right)dx + \int\limits_1^\sqrt{2} \left( 1 \right)dx .................\left( \because \left[ x^2 \right] = \begin{cases}0&& 0 < x < 1\\1&& 1 < x < \sqrt{2}\end{cases} \right)\]
\[ = 0 + \left[ x \right]_1^\sqrt{2} \]
\[ = \sqrt{2} - 1\]

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पाठ 19: Definite Integrals - Very Short Answers [पृष्ठ ११६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Very Short Answers | Q 44 | पृष्ठ ११६

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