Question

\[\int\limits_0^{15} \left[ x^2 \right] dx\]

Answer 1

We have,

\[I = \int\limits_0^{1 . 5} \left[ x^2 \right] dx\]

\[ = \int\limits_0^1 \left[ x^2 \right] dx + \int\limits_1^\sqrt{2} \left[ x^2 \right] dx + \int\limits_\sqrt{2}^{1 . 5} \left[ x^2 \right] dx\]

\[ = \int\limits_0^1 \left( 0 \right) dx + \int\limits_1^\sqrt{2} \left( 1 \right) dx + \int\limits_\sqrt{2}^{1 . 5} \left( 2 \right) dx ..............\left(\because \left[ x^2 \right] = \begin{cases}0 &where,& 0 < x < 1 \\ 1 &where,& 1 < x < \sqrt{2}\\2 &where,& \sqrt{2} < x < 1.5 \end{cases}\right)\]

\[ = 0 + \left[ x \right]_1^\sqrt{2} + \left[ 2x \right]_\sqrt{2}^{1 . 5} \]

\[ = \left[ x \right]_1^\sqrt{2} + 2 \left[ x \right]_\sqrt{2}^{1 . 5} \]

\[ = \left( \sqrt{2} - 1 \right) + 2\left( 1 . 5 - \sqrt{2} \right)\]

\[ = \sqrt{2} - 1 + 3 - 2\sqrt{2}\]

\[ = 2 - \sqrt{2}\]