Question

Evaluate the definite integrals.
${\int }_{-1}^1(x+1)dx.$

Answer 1

Let $I={\int }_{-1}^1(x+1)dx={[\frac{x^2}{2}+x]}_{-1}^1=[\frac{1^2}{2}+1]-[\frac{(-1{)}^2}{2}-1]$
$\Rightarrow I=\frac{3}{2}+\frac{1}{2}=\frac{4}{2}=2$