-1

I know that in greatest integer function, we have to divide this into limits where the value of function changes. But in case of fraction what happens is just confusing. Can anyone help? Like if the limit is from $- 3/5$ to $3/5$ to integrate [ 2x-1] where [. ] is gretest integere function

Ankita A
  • 1
  • You have me confused! Where does the "greatest integer function" come into this? Do you mean that the function to be integrated is the "greatest integer less than or equal" to 2x- 1? If that is the case then look at that "2". For x from -1 to -1/2 (so from -3/5 to -1/2) 2x- 1 goes from -3 to -2 so the "greatest integer" is -3. For x from -1/2 to 0, 2x- 1 goes from -2 to -1, so the "greatest integer" is -2. For x from 0 to 1/2 (so from 0 to 3/5), 2x- 1 goes from -1 to 0 so the "greatest integer is -1. – user247327 Feb 24 '19 at 20:18

1 Answers1

0

I think you mean the following $$\int_{-0.6}^{0.6} \lfloor2x-1\rfloor$$ $$=\frac{1}{2}\int_{-2.2}^{0.2} \lfloor u\rfloor$$ $$=\frac{1}{2}((-3)(0.2)+(-2)(1)+(-1)(1)+(0)(0.2))$$ $$=-1.8$$ Because for $-2.2\le u\lt-2$ the value of $\lfloor u\rfloor=-3$; $-2\le u\lt-1$ the value of $\lfloor u\rfloor=-2$; $-1\le u\lt0$ the value of $\lfloor u\rfloor=-1$; $0\le u\lt0.2$ the value of $\lfloor u\rfloor=0$.

Peter Foreman
  • 19,947