Question

The value of the integral $\underset{-1}{\overset{1}{\int }}\left|1-x\right|dx$ is ________________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{-1}{\overset{1}{\int }}\left|1-x\right|dx \\ \mathrm{We}\mathrm{know}, \\ \left|1-x\right|=\left\{\begin{array}{l}\left(1-x\right),\mathrm{if}\left(1-x\right)\ge 0\\ -\left(1-x\right),\mathrm{if}\left(1-x\right)<0\end{array}\right. \\ =\left\{\begin{array}{l}\left(1-x\right),\mathrm{if}x\le 1\\ -\left(1-x\right),\mathrm{if}x>1\end{array}\right. \\ \mathrm{Thus}, \\ I=\underset{-1}{\overset{1}{\int }}\left(1-x\right)dx \\ ={\left[x-\frac{x^2}{2}\right]}_{-1}^1 \\ =\left[1-\frac{{\left(1\right)}^2}{2}\right]-\left[-1-\frac{{\left(-1\right)}^2}{2}\right] \\ =\left[1-\frac{1}{2}\right]-\left[-1-\frac{1}{2}\right] \\ =\left[\frac{2-1}{2}\right]-\left[\frac{-2-1}{2}\right] \\ =\frac{1}{2}-\left[-\frac{3}{2}\right] \\ =\frac{1}{2}+\frac{3}{2} \\ =\frac{4}{2} \\ =2 \end{gathered}$$​

​Hence, the value of the integral $\underset{-1}{\overset{1}{\int }}\left|1-x\right|dx$ is 2.