Question

Find the average value of f(x) = 48-12x^2 of the interval -5<=x<=5

Answer 1

This sounds like an integration question, but I don't think I've seen integration start iwht y or f(x) so I was thinking derivation? Just confused how to start this.

Answer 2

@ParthKohli

Answer 3

average value is
\[\frac{1}{10}\int_{-5}^5 (48-12x^2)dx\]

Answer 4

how did you figure that out?

Answer 5

that is the definition of average value
integral divided by the length of the path
but i am going to guess that it is zero in any case since this is a parabola with axis of symmetry as the \(y\) axis

Answer 6

scratch that remark it is wrong

Answer 7

http://archives.math.utk.edu/visual.calculus/5/average.1/index.html

From this it makes sense 1/(5--5)

Answer 8

it would be zero if the function was odd, sorry ignore that last remark

Answer 9

actually no -5-5 = 0

Answer 10

1/0 is undefined though?

Answer 11

it is \(\frac{1}{b-a}\) in your case \(\frac{1}{5-(-5)}=\frac{1}{10}\) but you should think of it simply as the length of the path

Answer 12

from minus 5 to 5 is ten

Answer 13

oh wait b is 5 lol I'm confusing myself.

Answer 14

that is why it is called "average value" it is analogous to the mean

Answer 15

if you have a bunch of numbers to find the average you add up and divide by how many you have
this is the continuous version of the same thing. adding, then take the limit is the integral, how many you have is the length of the interval
so it really is an average but with continuous data instead of discrete

Answer 16

yeah I would have thought for 2 values it would have been 2/10 but this formula is different.

Do I now do the integral, or am I done?