Question

$\underset{-1}{\overset{1}{\int }}\left\{\left[x\right]+\left|x\right|\right\}dx$ has the value is

• A. 0
• B. 1/2
• C. 1
• D. 2

Answer 1

Answer: (A)

0

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