$\int_{-2}^{1}[x+1] d x=$ (Where $[x]$ is greatest integer function not greater than $x$ )

Question

Aicharya · Accepted Answer

$\begin{aligned} \int_{-2}^{1}[x+1] d x &=\int_{-2}^{-1}([x]+1) d x+\int_{-1}^{0}([x]+1) d y+\int_{0}^{1}([x]+1) d x \ &=\int_{-2}^{-1}(-2+1) d x+\int_{-1}^{0}(-1+1) d x+\int_{0}^{1}(0+1) d x \ &=-[x]_{-2}^{-1}+0+[x]_{0}^{1}=-(-1+2)+0+(1-0) \ &=0 \end{aligned}$

$\int_{-2}^{1}[x+1] d x=$ (Where $[x]$ is greatest integer function not greater than $x$ )

Solution