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Question

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

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

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Chapter 19: Definite Integrals - Very Short Answers [Page 116]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Very Short Answers | Q 42 | Page 116

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