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Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 42 maths textbook solution

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Answer: \frac{1}{\log _{e} 2}

Hint: you must know the rule of integration

Given: I=\int_{0}^{1} 2^{x-[x]} d x


We know the values of greatest integer function on 0<x<1, \quad[x]=0

\begin{aligned} &I=\int_{0}^{1} 2^{x-0} d x \\\\ &=\int_{0}^{1} 2^{x} d x \\\\ &=\left[\frac{2^{x}}{\log 2}\right]_{0}^{1} \end{aligned}

\begin{aligned} &=\left[\frac{2^{1}}{\log 2}-\frac{2^{0}}{\log 2}\right] \\\\ &=\frac{(2-1)}{\log 2} \\\\ &=\frac{1}{\log 2} \end{aligned}

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