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Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Very short answer type question 39

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Answer: \frac{1}{2}

Hint: you must know the rule of integration

Given: \int_{0}^{1}\{x\} d x

Solution:  \int_{0}^{1}\{x\} d x

\begin{aligned} &=\int_{0}^{1}(x-[x]) d x \\\\ &=\int_{0}^{1} x d x-\int_{0}^{1}[x] d x \end{aligned}

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



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