Question

The average (mean) value of tan x on the interval from x = 0 to x = pi/3 is

Answer 1

I thought the mean value theorem was f'(c) = f(b)-f(a)/b-a
I plugged in the values and I got
\[f'(c) = \sqrt{3}/(\pi/3)\]

Answer 2

But it wasn't one of the answer choices. Was I supposed to do something else?

Answer 3

what are you answer choices ?

Answer 4

\(3 \sqrt 3/ \pi\)
?

Answer 5

a. ln(1/2)
b. (3/pi)ln2
c. ln2
d. rad(3) / 2
e. 9/pi

Answer 6

why are there so many ln 2 s
is the function 'tan x' only ?

Answer 7

Yes

Answer 8

i kinda hope it was tan^2 x

Answer 9

I don't know. I wrote it exactly how my book had it. And I don't think the book would make too many errors

Answer 10

lets see what parth has to say :)

Answer 11

Pardon me, but isn't the mean value of a function given by\[\dfrac{\int_{a}^{b} f(x) dx}{b-a}\]Then,\[\frac{-\ln |\cos \pi/3| + \ln |\cos 0|}{\pi /3}\]\[= \frac{-\ln |1/2|}{\pi /3}\]\[= 3\ln (2)/\pi \]

Answer 12

Ya now I'm pretty sure thats what I should've done

Answer 13

Because AFAIK, the right side of the mean value theorem is not actually the mean value of the function.

Answer 14

I just saw the word mean and applied the mean value theorem. My bad >.<

Answer 15

you're mean
apply MVT

Answer 16

i was thinking something else
\(\LARGE \dfrac{f(b)-f(a)}{b-a}\)