Question

Find the average value of the function

Answer 1

Interval is [.5 , 2]

Function is

\[f(x) = \frac{ x^2 + 1 }{ x^2 }\]

Answer 2

The answer when done on a calculator is .325 but i do not know what I keep doing wrong and keep getting 2

Answer 3

\[1/(b-a) \int\limits_{a}^{b}f(x)\]

Answer 4

so b-a=2-0.5=1.5
\[\int\limits_{0.5}^{2}1+1/x ^{2}\]

Answer 5

I know that so if I rewrite the function and the dividing number in front I get

\[\frac{ 2 }{ 3 }\int\limits_{.5}^{2}1 + x^{-2}\]

Answer 6

(x-1/x) then 2-1/2=1.5 and 1/2-2=-1.5 1.5-(-1.5)=3

Answer 7

integral equals to 3 so 2/3.3=2

Answer 8

Yea buutttt when done on a calculator the answer turns out ot be .325

Answer 9

impossible you can check it by putting values instead of x for example for 2 it is 5/4=1.25
for 1/2 it is 5 so the average must be between these two

Answer 10

ok I see thank you

Answer 11

nop

Answer 12

I've gone through the necessary calculations and find that @kagıtucak arrived at the correct answer for the average value of this function on the interval [.5,2]; it is 2.

Let's see what we can all agree upon:

The area under the curve of the given function between .5 and 2 is 3.
The length of this interval is 2-0.5 is 3/2.

Dividing that area by that length produces the average value of the function on that interval: 3/(3/2) = 2.

Answer 13

@hilbertboy96 : Try calculating the area under the graph of the function on the interval [.5,2]. Then verify that the interval length is 3/2. Finally, divide the area by the length to get the average value of the function over that interval.

I'm curious what procedure you're using to calculate that area. Of course there are multiple ways of doing this, but I've chosen to use the fnInt( function on my TI-83 Plus calculator.