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

Question

Emily Ammerman · Accepted Answer

The average value is #=1#

The average value of a function #f(x)# over an interval #[a,b]# is

#barx=1/(b-a)int_a^bf(x)dx#

Here,

#f(x)=(x-3)^2=x^2-6x+9#

and #[a,b]=[2,5]#

Therefore,

#barx=1/(5-2)int_2^5(x-3)^2dx#

#=1/3int_2^5(x^2-6x+9)dx#

#=1/3[x^3/3-6x^2/2+9x]_2^5#

#=1/3((125/3-75+45)-(8/3-12+18))#

#=1/3(117/3-36)#

#=1/3*3#

#=1#

Christopher Agresta · Answer

undefined To find the average value of $ f(x) = (x - 3)^2 $ as $ x $ varies between $ [2, 5] $, you use the following formula:

$$ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx $$

Substitute $ a = 2 $ and $ b = 5 $ into the formula and find the integral:

$$ \text{Average value} = \frac{1}{5 - 2} \int_{2}^{5} (x - 3)^2 \, dx $$

Now, integrate $ (x - 3)^2 $ with respect to $ x $:

$$ \int_{2}^{5} (x - 3)^2 \, dx = \left[ \frac{(x - 3)^3}{3} \right]_{2}^{5} $$

$$ = \left[ \frac{(5 - 3)^3}{3} - \frac{(2 - 3)^3}{3} \right] $$

$$ = \left[ \frac{2^3}{3} - \frac{-1^3}{3} \right] $$

$$ = \left[ \frac{8}{3} + \frac{1}{3} \right] $$

$$ = \frac{9}{3} $$

$$ = 3 $$

Thus, the average value of $ f(x) = (x - 3)^2 $ as $ x $ varies between $ [2, 5] $ is $ 3 $.