Question

\[\int\limits_0^2 x\left[ x \right] dx .\]

Answer 1

\[\text{We have}, \]
\[I = \int\limits_0^2 x\left[ x \right] dx\]
\[\text{We know that}, \]
\[x\left[ x \right] = \begin{cases}x \times 0&,& 0 < x < 1\\x \times 1&,& 1 < x < 2\end{cases}\]
\[i . e . , \]
\[x\left[ x \right] = \begin{cases}0&,& 0 < x < 1\\x&,& 1 < x < 2\end{cases}\]
\[ \therefore I = \int\limits_0^2 x\left[ x \right] dx\]
\[ = \int\limits_0^1 x\left[ x \right] dx + \int\limits_1^2 x\left[ x \right] dx\]
\[ = \int\limits_0^1 \left( 0 \right) dx + \int\limits_1^2 \left( x \right) dx\]
\[ = 0 + \left[ \frac{x^2}{2} \right]_1^2 \]
\[ = \frac{2^2}{2} - \frac{1^2}{2}\]
\[ = \frac{4}{2} - \frac{1}{2}\]
\[ = \frac{3}{2}\]