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Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Very short answer type question 43

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Answer: 0

Hint: you must know the rule of integration

Given: \int_{1}^{2} \log _{e}[x] d x

Solution:  \int_{1}^{2} \log _{e}[x] d x

Using rule of greatest integer

\begin{aligned} & \int_{0}^{1} 0 d x+\int_{1}^{2} 1 d x+\int_{2}^{3} 2 d x \\\\ I=& \int_{1}^{2} \log [x] d x=\int_{0}^{1} \log (0) d x+\int_{1}^{2} \log 1 d x \\\\ =& 0 \end{aligned}



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