Integration and floor function

$\large I = \displaystyle \int_0^{64} \color{#D61F06}{\lfloor x \rfloor }\text{dx}$ $\large I = \displaystyle \int_0^{64} \color{#3D99F6}{\lfloor 64 - x \rfloor }\text{dx}$ $\large 2I = \displaystyle \int_0^{64} \left( \color{#D61F06}{\lfloor x \rfloor} + \color{#3D99F6}{\lfloor 64 - x \rfloor }\right)\text{dx}$ $\large 2I = \displaystyle \int_0^{64} \left(\color{#D61F06}{\lfloor x \rfloor} + \color{#3D99F6}{63 - \lfloor x \rfloor }\right) \text{dx}$ $\large 2I = 64 \times 63$ $\large \boxed{I = 2016 }$

I hope my handwriting isn't too much of a burden, perhaps I'll rewrite this solution in LaTeX tomorrow after school :)

It is clear that $\int _{ n-1 }^{ n }{ \left\lfloor x \right\rfloor } dx=n-1$ and then, by induction, it is easy to prove that $\int _{0}^{ n }{ \left\lfloor x \right\rfloor } dx=\frac{(n-1)n}{2}$ .

The integrand is a step function consisting of bars with unit width and heights

$0,1,2\dots,62,63$

Interpreting the definite integral as the 'area under the curve' gives the answer

$0+1+2+\dots+62+63=\frac{63\times64}{2}=2016$