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प्रश्न

\[\int\limits_0^1 2^{x - \left[ x \right]} dx\]
बेरीज

उत्तर

\[\text{We have}, \]
\[I = \int\limits_0^1 2^{x - \left[ x \right]} dx\]
\[ = \int\limits_0^1 2^{x - 0} dx ...............\left( \because \left[ x \right] = 0\text{ where, }0 < x < 1 \right)\]
\[ = \int\limits_0^1 2^x dx\]
\[ = \left[ \frac{2^x}{\log_e 2} \right]_0^1 \]
\[ = \frac{2^1}{\log_e 2} - \frac{2^0}{\log_e 2}\]
\[ = \frac{2}{\log_e 2} - \frac{1}{\log_e 2}\]
\[ = \frac{1}{\log_e 2}\]

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Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Very Short Answers [पृष्ठ ११६]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Very Short Answers | Q 42 | पृष्ठ ११६

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