Question

Let $\left[t\right]$ denote the greatest integer $\le t$. Then the value of $8\cdot \underset{-\frac{1}{2}}{\overset{1}{\int }}\left(\right[2x]+|x|)dx$ is

Answer 1

$\underset{-\frac{1}{2}}{\overset{1}{\int }}\left(\right[2x]+|x|)dx=\underset{-\frac{1}{2}}{\overset{1}{\int }}\left[2x\right]dx+\underset{-\frac{1}{2}}{\overset{1}{\int }}|x|dx=\underset{-\frac{1}{2}}{\overset{1}{\int }}\left[2x\right]dx+\underset{-\frac{1}{2}}{\overset{0}{\int }}-xdx+\underset{0}{\overset{1}{\int }}xdx=\underset{-\frac{1}{2}}{\overset{0}{\int }}-1dx+\underset{0}{\overset{\frac{1}{2}}{\int }}0dx+\underset{\frac{1}{2}}{\overset{1}{\int }}1dx+\underset{-\frac{1}{2}}{\overset{0}{\int }}-xdx+\underset{0}{\overset{1}{\int }}xdx=(-\frac{1}{2}+0+\frac{1}{2})+{\left[\frac{-x^2}{2}\right]}_{-\frac{1}{2}}^0+{\left[\frac{x^2}{2}\right|}_0^1=0+\frac{1}{8}+\frac{1}{2}=\frac{5}{8}\therefore 8\cdot \underset{-\frac{1}{2}}{\overset{1}{\int }}\left(\right[2x]+|x|)dx=8\times \frac{5}{8}=5$