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

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

Answers (1)

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

Hint: you must know the rule of integration

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

Solution:  

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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