Evaluate the following definite integral with respect to box function

Question

I would like to evaluate the following integral:

$$\int_0^3 x\, d[x]$$

where $[x]$ is the greatest integer $\leq x$

Answer 1

Using integration by parts, we have $$\int_{0^+}^{3^-} x d\lfloor x\rfloor = \left(x\lfloor x \rfloor \right)_{0^+}^{3^-} - \int_0^3 \lfloor x \rfloor dx = 6 - \left(0+1+2\right) = 3$$

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Another way is recognize that $$d\lfloor x \rfloor = \sum_{k=-\infty}^{\infty}\delta(x-k)dx$$ Hence, we have $$\int_{0^+}^{3^-} xd\lfloor x \rfloor = \int_{0^+}^{3^-} x\sum_{k=-\infty}^{\infty}\delta(x-k)dx = 0 + 1 + 2 = 3$$

In general, $$\int_{0^+}^{n^-} xd\lfloor x \rfloor = \dfrac{n(n-1)}2$$