Question

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

Answer 1

Step-1: Split the given integral using properties of modulus function

$\left|x\right|=x,x>0$

$\left|x\right|=-x,x<0$

$\left|x-2\right|=x-2,x>2$

$\left|x-2\right|=2-x,x<2$

Hence, the given integral can be split as

${\int }_{-a}^0\left(-x+2-x\right)dx+{\int }_0^2\left(x+2-x\right)dx+{\int }_2^a\left(x+x-2\right)dx=22$ ;[assuming $a>2$]

Step-2: Solve the integral to calculate the value of $a$

$\Rightarrow$${\int }_{-a}^0\left(-2x+2\right)dx+{\int }_0^2\left(2\right)dx+{\int }_2^a\left(2x-2\right)dx=22$

$\Rightarrow$ ${\left[-x^2+2x\right]}_{-a}^0+{\left[2x\right]}_0^2+{\left[x^2-2x\right]}_2^a=22$

$\Rightarrow$ $0+0+a^2+2a+4-0+a^2-2a-4+4=22$

$\Rightarrow$ $2a^2=18$

$\Rightarrow$ $a^2=9$

$\Rightarrow$ $a=3$

Step -3: Substitute the value of $a$ to solve the required integral

The given integral now becomes ${\int }_{-3}^3\left(x+\left[x\right]\right)dx$

$\Rightarrow$${\int }_{-3}^3\left(x+\left[x\right]\right)dx={\int }_{-3}^{3}xdx+{\int }_{-3}^3\left[x\right]dx$

$\Rightarrow$ $=0+{\int }_{-3}^{-2}\left[x\right]dx+{\int }_{-2}^{-1}\left[x\right]dx+{\int }_{-1}^0\left[x\right]dx+{\int }_0^1\left[x\right]dx+{\int }_1^2\left[x\right]dx+{\int }_2^3\left[x\right]dx$ $...\because \left[{\int }_{-a}^{a}f\right(x)dx=0$ for an odd function like $f\left(x\right)=x]$

$\Rightarrow$ $={\int }_{-3}^{-2}-3dx+{\int }_{-2}^{-1}-2dx+{\int }_{-1}^0-1dx+{\int }_0^{1}0dx+{\int }_1^{2}dx+{\int }_2^{3}2dx$ $...\left(\because \left[x\right]=-3\forall x\in [-3,-2),\left[x\right]=-2\forall x\in [-2,-1)\right)$ and so forth

$\Rightarrow$ $=-3{\left[x\right]}_{-3}^{-2}+-2{\left[x\right]}_{-2}^{-1}-{\left[x\right]}_{-2}^{-1}+{\left[x\right]}_1^2+{\left[x\right]}_2^3$

$\Rightarrow$ $=-3-2-1+1+2$

$\Rightarrow$${\int }_{-3}^3\left(x+\left[x\right]\right)dx=-3$

Hence, the value of the integral ${\int }_{-a}^a\left(x+\left[x\right]\right)dx$ is $-3$ for the given conditions.