# average value

• Dec 26th 2007, 09:29 AM
DINOCALC09
average value
The average value of the function f(x) = (x-1)^2 on the interval from x=1 to x=5 is:

a) -16/3
b) 16/3
c) 64/3
d) 66/3
e) 256/3
• Dec 26th 2007, 09:34 AM
colby2152
Quote:

Originally Posted by DINOCALC09
The average value of the function f(x) = (x-1)^2 on the interval from x=1 to x=5 is:

a) -16/3
b) 16/3
c) 64/3
d) 66/3
e) 256/3

Integrate the function and then divide by four.

$\displaystyle \int_1^5 (x-1)^2 dx = \frac{(x-1)^3}{3}|_1^5$

$\displaystyle \int_1^5 (x-1)^2 dx = \frac{4^3}{3}$

$\displaystyle \int_1^5 (x-1)^2 dx=64/3$

Divide that total area by four to get the average value of 16/3.
• Dec 26th 2007, 09:41 AM
Isomorphism
Quote:

Originally Posted by DINOCALC09
The average value of the function f(x) = (x-1)^2 on the interval from x=1 to x=5 is:

a) -16/3
b) 16/3
c) 64/3
d) 66/3
e) 256/3

I think
(b) $\displaystyle \frac{16}{3}$

$\displaystyle \frac{\int^4_0{t^2}\ dt }{4} =\frac14 \frac{4^3}{3} = \frac{16}{3}$
• Dec 26th 2007, 12:36 PM
Jhevon
why did they divide by 4? and how do they know what to integrate?

Definition: Let $\displaystyle f(x)$ be an integrable function on the closed interval $\displaystyle [a,b]$. Then the average value of $\displaystyle f(x)$ on $\displaystyle [a,b]$ is given by:

$\displaystyle \mbox{Average Value} = \frac 1{b - a} \int_a^b f(x)~dx$