Question

Integral Question:

Find the average value of the function f on the given interval

f(x) = (1/x^2) [1,5]

Answer 1

are you just asking to integrate that function? I'm not sure what you mean by average value

Answer 2

Average value is when you take the term and multiply it by 1/(b-a) or just f(c) = b-a

Answer 3

Ah one of the calc1 theorems, sorry I can't remember those off the top of my head.

Answer 4

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

Answer 5

So far I got:

The integral for 5 to 1 (x^-2) = x^-1/1

Answer 6

That is the average value of a function, using calculus

Answer 7

Correct Andras. Did I integrate well so far? I don't want to get confused with the signs and negative exponents.

Answer 8

This integral is a polynomial. So I really hope that you can do the integral

Answer 9

You are missing a minus sign!

Answer 10

Oh yeah lol my teacher keeps reinforcing that you can't take the area of a polynomial.

I got -x^-1 = -1/5 - 1 = -1/5 -5/5 = -6/5

Answer 11

Then would I just multiply it by 1/(b-a), which is 1/4 ?

-6/5 multiplied by 1/4

= -6/20 = -3/10

That's what I got but I wanted to confirm it.

Answer 12

Sorry had to leave for a sec, Im back now.

Answer 13

f(x) = (1/x^2)
Think!!!!
Can this: -3/10 be the average value of the function between 1 and 5??

Answer 14

Oh no it can't. It's not in the domain. Where did I go wrong?

Answer 15

I got -x^-1 for the integral. You are correct
But when you put the values in make sure you are treating all the - signs correctly. Will you give it another go?

Answer 16

Yeah sure just half a minute.

Answer 17

It should be -1/5 +1 right?
Which is -1/5 + 5/5 = 4/5

Answer 18

Yep

Answer 19

Oh wow oh wow.

4/5(1/4) = 4/20 = 1/5

Is that in the domain?

Answer 20

Yeah that is it. Not the domain! The question should be: Is that in the range?

Answer 21

Gotcha :)
Thank you so much!