Question
Evaluate the definite integral in Exercise:
$\int\limits_{-1}^{1}\text{(x}+1)\ \text{dx}$

Answer

$\text{Let}\ \text{I}=\int\limits_{-1}^{1}\text{(x+1)}\text{dx} $

$\int\text{(x}+1)\text{dx}=\frac{\text{x}^{2}}{2}+\text{x}$

By second theorem of calculus, we obtain

$\text{I}=\text{F}(1)-\text{F}(-1)$

$=\bigg(\frac{1}{2}+1\bigg)-\bigg(\frac{1}{2}-1\bigg)$

$=\frac{1}{2}+1-\frac{1}{2}+1$

$=2$

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