Question #e30bc

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Answer 1 · 2017-02-17T18:46:40

$int_0^x [t]dt = ( [x] (2x- [x]-1))/2$ for $x > 0$

Define the function as:

$[x] = n$ for $n<=x<(n+1)$

The function is continuous almost everywhere, that is in all $RR-ZZ$

Consider $x>0$ and $[x] = N$ , its integral can be calculated as:

$int_0^x [t]dt = int_0^1 0*dt + int_1^2 1*dt + ... + int_(N-1)^N (N-1)dt + int_N^x Ndt$

$int_0^x [t]dt = 1+2+...+(N-1)+N(x-N)$

$int_0^x [t]dt = (N(N-1))/2 +Nx-N^2$

$int_0^x [t]dt = (N^2-N+2Nx-2N^2)/2$

$int_0^x [t]dt = (N(2x-1)-N^2)/2$

$int_0^x [t]dt = (N(2x-N-1))/2$

Consider $x<0$ and $[x] = -N$ , its integral can be calculated as:

$int_0^x [t]dt = -int_x^(-N) (-N-1)*dt - int_(-N)^(-N+1) -N*dt + ... - int_(-1)^0 (-1)dt$

$int_0^x [t]dt = 1+2+...+N+(N+1)(x-N)$

$int_0^x [t]dt = (N(N+1))/2 +Nx-N^2+x-N$

$int_0^x [t]dt = (N^2+N+2Nx-2N^2+2x-2N)/2$

$int_0^x [t]dt = (N(2x-1)-N^2+2x)/2$

$int_0^x [t]dt = x+(N(2x-N-1))/2$