Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (a)/Solution 1

For negative values of $x,\,1-|x|=1+x$ and for positive values $1-|x|=1-x$ (as $|x|={\begin{cases}x,&x\geq 0\\-x,&x<0\end{cases}}$). We can split the integral so that one part has only the negative values of $x$ and the other the positive values:

$\int _{-1}^{2}(1-|x|)dx=\int _{-1}^{0}(1-|x|)dx+\int _{0}^{2}(1-|x|)dx=\int _{-1}^{0}(1+x)dx+\int _{0}^{2}(1-x)dx$

$\int _{-1}^{0}(1+x)dx=(x+{\frac {x^{2}}{2}}){\bigg |}_{-1}^{0}={\frac {1}{2}}\,$ and $\int _{0}^{2}(1-x)dx=(x-{\frac {x^{2}}{2}}){\bigg |}_{0}^{2}=0$

Answer: $\color {blue}{\frac {1}{2}}$