Question

The value of the integral ${\int }_{-1}^3(|x|+|x-1|)dx$ is

Answer 1

$I={\int }_{-1}^3(|x|+|x-1|)dx$

$={\int }_{-1}^3\underset{I_1}{\underset{\downarrow }{|x|}}dx+{\int }_{-1}^3\underset{I_2}{\underset{\downarrow }{|x-1|}}dx$ …………$\left(1\right)$

$I_1={\int }_{-1}^3|x|dx$ we know $|x|=\{\begin{array}{ccc}x& for& x\ge 0\\ -x& for& x<0\end{array}$

$={\int }_{-1}^0-xdx+{\int }_0^{3}xdx$

$=-{\left[\frac{x^2}{2}\right]}_{-1}^0+{\left[\frac{x^2}{2}\right]}_0^3$

$=-[\frac{0}{2}-\frac{1}{2}]+[\frac{9}{2}-0]$

$\Rightarrow \frac{1}{2}+\frac{9}{2}=\frac{10}{2}=5$

$I_2={\int }_{-1}^3|x-1|dx$ We know $|x-1|=\{\begin{array}{ccc}(x-1)& for& x\ge 1\\ -(x-1)& for& x<1\end{array}$

$={\int }_{-1}^1-(x-1)dx+{\int }_1^3(x-1)dx$

$=-{\int }_{-1}^{1}xdx+{\int }_{-1}^{1}1dx+{\int }_{-1}^{3}xdx-{\int }_1^{3}dx$

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

$=-[\frac{1}{2}-\frac{1}{2}]+[1+1]+[\frac{9}{2}-\frac{1}{2}]-[3-1]$

$=0+2+4-2=4$

Putting the $I_1$ & $I_2$ value in equation $\left(1\right)$

$I=5+4=9$ Answer.