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Evaluate 0n[[x]]dx\int_{0}^{n} [[x ]] d x, where n is a positive integer.

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0n[x]dx=01[x]dx+12[x]dx+...+n1n[x]dx\int_{0}^{n}[x]dx=\int_{0}^{1}[x]dx+\int_{1}^{2}[x]dx+...+\int_{n-1}^{n}[x]dx

acf(x)dx=abf(x)dx+bcf(x)dx\int_{a}^{c}f(x)dx=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx

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