# How do you find the average value of the function for f(x)=1+x, -1<=x<=1?

Feb 11, 2018

$1$

#### Explanation:

$\text{average of f(x) in interval [a,b] is : }$
$\left(\frac{1}{b - a}\right) {\int}_{a}^{b} f \left(x\right)$
$\text{So here we have}$
$a = - 1$
$b = 1$
$f \left(x\right) = 1 + x$
$\implies a v g = \left(\frac{1}{2}\right) {\int}_{- 1}^{1} \left(1 + x\right) \left(\mathrm{dx}\right)$
$= \left(\frac{1}{2}\right) {\left[x + {x}^{2} / 2 + C\right]}_{- 1}^{1}$
$= \left(\frac{1}{2}\right) \left[1 + {1}^{2} / 2 + \cancel{C} - \left(- 1\right) - {\left(- 1\right)}^{2} / 2 - \cancel{C}\right]$
$= \left(\frac{1}{2}\right) \left[1 + \cancel{\frac{1}{2}} + 1 - \cancel{\frac{1}{2}}\right]$
$= \left(\frac{1}{2}\right) \left[1 + 1\right]$
$= 1$
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