Question

Question #e30bc

Answer 1

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

Explanation:

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`