Evaluate the following integrals: ^ 1 _02^ x-[x] dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits^{1}_02^{\text{x}-[\text{x}]}\text{dx}$

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Answer

We have,
$\text{I}=\int\limits^{1}_02^{\text{x}-[\text{x}]}\text{dx}$
$=\int\limits^{1}_02^{\text{x}-0}\text{ dx}$ $\big(\because[\text{x}]=0,\text{ where}, 0<\text{x}<1\big)$
$=\int\limits^{1}_02^{\text{x}}\text{ dx}$
$=\Big[\frac{2^{\text{x}}}{\log_\text{e}2}\Big]^1_0$
$=\frac{2^1}{\log_\text{e}2}-\frac{2^0}{\log_\text{e}2}$
$=\frac{2}{\log_\text{e}2}-\frac{1}{\log_\text{e}2}$
$=\frac{1}{\log_\text{e}2}$

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