Question
Evaluate: $\int_0^3[x] d x$

Answer


$\begin{array}{l}\int_0^3[x] d x=\int_0^1[x] d x+\int_1^2[x] d x+\int_2^3[x] d x \\ =\int_0^1 0 d x+\int_1^2 1 d x+\int_2^3 2 d x \\ =0+[x]_1^2+[2 x]_2^3=1+6-4=3\end{array}$

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