Question

The value of ${\int }_{-5}^{5}f\left(x\right)dx$; where $f\left(x\right)=min\left(\right\{x+1\},\{x-1\left\}\right),\forall xϵR$ and $\{.\}$ denotes fractional part of $x$ is equal to

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

The correct option is C 5

${\int }_{-5}^{5}f\left(x\right)dx$$={\int }_{-5}^5(x-1-[x-1])dx$

$={\int }_{-5}^{-4}(x+5)dx+{\int }_{-4}^{-3}(x+4)dx+{\int }_{-3}^{-2}(x+3)dx+{\int }_{-2}^{-1}(x+1)dx$

$+{\int }_{-1}^0\left(x\right)dx+{\int }_0^1(x-1)dx+{\int }_1^2(x-2)dx+{\int }_2^3(x-3)dx$

$+{\int }_3^4(x-4)dx+{\int }_4^5(x-5)dx=5$