Answer
a) 1
b) c = 2, c = 4
Work Step by Step
Given: $f(x) = (x-3)^2 $ on [2,5]
a) Substitute:
$=\frac{1}{5-2} \int_{2}^{5} ((x-3)^2 )dx$ Expand the function
$= \frac{1}{3} \int_{2}^{5} (x^2 -6x +9)dx$
Find the antiderivative
$= \frac{1}{3} [(\frac{1}{3}x^3 - \frac{6}{2}x^2 +9x)]_{2}^{5}$
$=\frac{1}{3} ((\frac{125}{3} - 75 +45) - (\frac{8}{3} - 12 + 18))$ Combine terms
$= \frac{1}{3}(\frac{117}{3} - 36) = \frac{117}{9} - 12 = \frac{117}{9} - \frac{108}{9} = \frac{9}{9} = 1$
b) Find $c$ where $f_{avg} = f(c)$
Mean Value Theorem for Integrals:
$\frac{1}{b-a} \int_{a}^{b} f(x)dx = f(c)(b-a)$
$f(c) = (c-3)^2 = c^2 -6c +9$
Substitute:
$\int_{2}^{5} (x^2 -6x+9)dx = (c^2 -6c +9)(5-2)$
$[\frac{1}{3}x^3 - \frac{6}{2}x^2 +9x]_{2}^{5} = 3c^2 -18c +27$
$(\frac{117}{3} - \frac{108}{3}) = 3c^2 -18c +27$
$3 = 3c^2 -18c +27$
$0 = 3c^2 - 18c + 24$ Divide both sides by 3
$0 = c^2 - 6c + 8$ Factor out
$0 = (c-4)(c-2)$
Therefore $c = 2$ and $c=4$ because they both lie within the interval [2,5]
