Answer:
7
Step-by-step explanation:
Average value of a function F(t) on interval [a,b]
is [tex] \frac{1}{b-a} \int_a^b F(x)dx[/tex]
So let's plug it in!
a=0
b=9
F(x)=(x-4)^2
To integrate (x-4)^2 just use power rule for integration.
So this what we get
[tex] \frac{1}{9-0} \frac{(x-4)^3}{3}|_0^9 [/tex]
[tex] \frac{1}{9-0} [\frac{(9-4)^3}{3}-\frac{(0-4)^3}{3}] [/tex]
[tex] \frac{1}{9} [\frac{5^3}{3}-\frac{(-4)^3}{3}] [/tex]
[tex] \frac{1}{9} \cdot \frac{125+64}{3} [/tex]
[tex] \frac{1}{9} \cdot \frac{189}{3} [/tex]
7
