Question:
$\int_{-1}^{1}(x+1) d x$
Solution:
Let $I=\int_{-1}^{1}(x+1) d x$
$\int(x+1) d x=\frac{x^{2}}{2}+x=\mathrm{F}(x)$
By second fundamental theorem of calculus, we obtain
$I=\mathrm{F}(1)-\mathrm{F}(-1)$
$=\left(\frac{1}{2}+1\right)-\left(\frac{1}{2}-1\right)$
$=\frac{1}{2}+1-\frac{1}{2}+1$
$=2$
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