Question
Evaluate : $\int_0^3 x[x] \cdot d x$, where $[x]$ denote greatest integrate function not greater than $x$.

Answer

$
\begin{aligned}
\text { Let I } & =\int_0^3 x[x] \cdot d x \\
\qquad & =\int_0^1 x[x] \cdot d x+\int_1^2 x[x] \cdot d x+\int_2^3 x[x] \cdot d x \\
& =\int_0^1 x(0) \cdot d x+\int_1^2 x(1) \cdot d x+\int_2^3 x(2) \cdot d x \\
& =0+\left[\frac{x^2}{2}\right]_1^2+\left[x^2\right]_2^3 \\
& =0+\left(\frac{4}{2}-\frac{1}{2}\right)+(9-4) \\
& =\frac{3}{2}+5=\frac{13}{2} \\
\therefore \quad \int_0^3 x[x] \cdot d x & =\frac{13}{2}
\end{aligned}
$

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