Mathematics - Definite Integration Question with Solution | TestHub

${\int }_{-1}^3\{|x-2|+[x]\}dx$ $={\int }_1^0\{|x-2|+[x]\}dx+{\int }_0^1\{|x-2|+[x]\}dx+{\int }_1^2\{|x-2|+[x]\}dx+{\int }_2^3\{|x-2|+[x]\}dx$ $={\int }_{-1}^0(2-x-1)dx+{\int }_0^1(2-x+0)dx+{\int }_1^2(2-x+1)dx+{\int }_2^3(x-2+2)dx$ $={\left[x-\frac{x^2}{2}\right]}_{-1}^0+{\left[2x-\frac{x^2}{2}\right]}_0^1+{\left[3x-\frac{x^2}{2}\right]}_1^2+{\left[\frac{x^2}{2}\right]}_2^3=-\left(-1-\frac{1}{2}\right)+\left(2-\frac{1}{2}\right)+(6-2)-\left(3-\frac{1}{2}\right)+\frac{9}{2}-2=7$