Question

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

Answer 1

Let $I={\int }_{-1}^1(x+1)dx$
$\Rightarrow \int (x+1)dx=\frac{x^2}{2}+x=F\left(x\right)$
By second fundamental theorem of calculus, we obtain
$I=F\left(1\right)-F(-1)$
$=(\frac{1}{2}+1)-(\frac{1}{2}-1)=\frac{1}{2}+1-\frac{1}{2}+1=2$