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Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Very short answer type question 44

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Answer: \sqrt{2}-1

Hint: The values of greatest integer function when 0<x<1,[x]=0 

        And when 1<x<2,[x]=1

Given: \int_{0}^{\sqrt{2}}\left[x^{2}\right] d x

Solution:  \int_{0}^{\sqrt{2}}\left[x^{2}\right] d x

=\int_{0}^{1}\left[x^{2}\right] d x+\int_{1}^{\sqrt{2}}\left[x^{2}\right] d x        [ using the value of greatest integer function ]

\begin{aligned} &=0+\int_{1}^{\sqrt{2}} 1 d x \\\\ &=[x]_{1}^{\sqrt{2}} \\\\ &=\sqrt{2}-1 \end{aligned}

Note: Final answer is not matching with the book.


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