Question

The value of $\underset{-5}{\overset{5}{\int }}f\left(x\right)dx=k,$ where $f\left(x\right)=min(\{x+1\},\{x-1\}),\forall x\in R$ and $\{\cdot \}$ denotes fractional part of $x$. Then the value of $k$ is equal to

Answer 1

We know,
$\{x+1\}=\{x-1\}=\left\{x\right\}$
Thus,
$f\left(x\right)=min(\{x+1\},\{x-1\})=\left\{x\right\}$
$\Rightarrow I=\underset{5}{\overset{-5}{\int }}f\left(x\right)dx\Rightarrow I=\underset{-5}{\overset{5}{\int }}\left\{x\right\}dx\Rightarrow I=[5-(-5\left)\right]\underset{0}{\overset{1}{\int }}\left\{x\right\}dx[\because \left\{x\right\}\text{is periodic with period}1]\therefore I=10\underset{0}{\overset{1}{\int }}xdx=5$