Mathematics - Area Under Curves Question with Solution | TestHub

$A={\int }_{-1}^0\{(3+x)-(-x+1)\}dx+{\int }_0^1\{(3-x)-(-x+1)\}dx+{\int }_1^2\{(3-x)-(-x-1)\}dx$ $={\int }_{-1}^0(2+2x)dx+{\int }_0^{1}2dx+{\int }_1^2(4-2x)dx$ $={\left[2x-x^2\right]}_{-1}^0+[2x{]}_0^1+{\left[4x-x^2\right]}_1^2$ $=0-(-2+1)+(2-0)+(8-4)-(4-1)$ $=1+2+4-3=4$ sq. units