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

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- Dec 26th 2007, 09:29 AMDINOCALC09average 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 AMcolby2152
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 AMIsomorphism
- Dec 26th 2007, 12:36 PMJhevon
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:

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