# How do you evaluate the definite integral int (x+4)dx from [-1,3]?

$\frac{39}{4}$

#### Explanation:

$\setminus {\int}_{-} {1}^{3} \left(x + 4\right) \setminus \mathrm{dx}$

$= {\left({x}^{2} / 2 + 4 x\right)}_{-} {1}^{3}$

$= {3}^{2} / 2 + 4 \left(3\right) - {1}^{2} / 2 - 4 \left(1\right)$

$= \frac{39}{4}$