How do you find the average value of f(x)=-x^3+7x^2-11x+3 as x varies between [1,5]?

Nov 11, 2016

$- 2$

Explanation:

Average

$= \frac{\int f \left(x\right) \mathrm{dx}}{5 - 1}$, between x = 1 and x = 5

$\left(\frac{1}{4}\right) \left[- {x}^{4} / 4 + 7 {x}^{3} / 3 - 11 {x}^{2} / 3 + 3 x\right] ,$ between the limits

$= \left(\frac{1}{4}\right) \left[- \left({5}^{4} - 1\right) + 7 \left({5}^{3} - 1\right) - 11 \left({5}^{2} - 1\right) + 3 \left(5 - 1\right)\right]$

$= \left(\frac{1}{4}\right) \left[- 624 + 868 - 264 + 12\right]$

$= \left(\frac{1}{4}\right) \left(- 8\right)$

$= - 2$