Question

If ${\int }_{-a}^a(|x|+|x-2|)dx=22,(a>2)$ and $\left[x\right]$ denotes the greatest integer $\le x$, then ${\int }_a^{-a}(x+[x\left]\right)dx$ is equal to

• A. 3.00
• B. 3.0
• C. 3

Answer 1

Answer: (A), (B), (C)

${\int }_{-a}^0(-2x+2)dx+{\int }_0^2(x+2-x)dx+{\int }_2^a\left(2x-2\right)dx=22$
$\Rightarrow (x^2-2x){|}_0^{-a}+\left(2x\right){|}_0^2+(x^2-2x){|}_2^a=22$
$\Rightarrow a^2+2a+4+a^2-2a-(4-4)=22$
$\Rightarrow 2a^2=18\Rightarrow a=3$
Now ${\int }_3^{-3}(x+[x\left]\right)dx=-\left({\int }_{-3}^3\right(x+\left[x\right]\left)dx\right)=-\left({\int }_{-3}^3\right[x\left]dx\right),(\because {\int }_{-3}^{3}xdx=0)$
$=-(-3-2-1+0+1+2)=3$