$\int_{-2}^{2.24}[x] d x=$ where $[x]$ is the greatest integer function

Question

Aicharya · Accepted Answer

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

$\int_{-2}^{2.24}[x] d x=$ where $[x]$ is the greatest integer function

Solution