Question

The average value of the function F (T) =(t-4)^2 on [0,9] is

Answer 1

Answer:

7

Step-by-step explanation:

Average value of a function F(t) on interval [a,b]

is $\frac{1}{b-a} \int_a^b F(x)dx$

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

$\frac{1}{9-0} \frac{(x-4)^3}{3}|_0^9$

$\frac{1}{9-0} [\frac{(9-4)^3}{3}-\frac{(0-4)^3}{3}]$

$\frac{1}{9} [\frac{5^3}{3}-\frac{(-4)^3}{3}]$

$\frac{1}{9} \cdot \frac{125+64}{3}$

$\frac{1}{9} \cdot \frac{189}{3}$

7