Question

Solve
${\int }_{-1}^1(x+1)dx$

Answer 1

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

$={\int }_{-1}^0(x+1)dx+{\int }_0^1(x+1)dx$

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

$=\left[\right(\frac{0}{2}+0)-(\frac{1}{2}-1\left)\right]+\left[\right(\frac{1}{2}+1)-(\frac{0}{2}+0\left)\right]$

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

$=\frac{1}{2}+\frac{3}{2}=\frac{4}{2}=2$