Question

Average value on the given interval? Please help!

Answer 1

@amistre64 Can you please check my answer on this one?

Answer 2

f(x) = sin nx, 0≤x≤ pi/n where n is a positive integer.

Answer 3

I think the indefinite integral is (-1/pi)(cos(nx)

Answer 4

\[\frac{n}{\pi }\int_0^{\frac{\pi}{n}}\sin(nx)dx\] is a start

Answer 5

So with the boundaries the definite integral is 2/n?

Answer 6

Yeah I did that. I let u = nx so
integral sin(nx) dx = integral sin udn/n

Answer 7

= 1/n integral sinu du = -1/n cos u + c

Answer 8

= -1/n (cos(nx)) / pi/n

Answer 9

= (-1/pi)(cos(nx)

Answer 10

dividing by \(\frac{\pi}{n}\) is the same as multiplying by \(\frac{n}{\pi}\)

Answer 11

Oh so I'm wrong?

Answer 12

no i think you are right, i didn't get to your second line

Answer 13

Oh okay. :) So I think my indefinite integral is correct I just need verification of my definite integral solution.

Answer 14

think you have it
\[-\frac{1}{\pi}\cos(nx)\] evaluated at \(\frac{\pi}{n}\) and at \(0\)

Answer 15

Yes. Does that give me the average value?

Answer 16

yes, because it is
\[\frac{1}{b-a}\int _a^bf(x)dx\]

Answer 17

Yay! Thank so much! :)

Answer 18

i get \(\frac{2}{\pi}\) but you should check my arithmetic

Answer 19

Ok (-1/pi)(cos(nx) I put pi/n in first?

Answer 20

yes

Answer 21

You're probably right I'm really bad at math. lol

Answer 22

I'm not getting graded for this anyway I just wanted someone to help me through the practice problems. Thank you!

Answer 23

me too

Answer 24

but if you plug in \(\frac{\pi}{n}\) you get \(-\frac{1}{\pi}\cos(n\frac{\pi}{n})=-\frac{1}{\pi}\cos(\pi)=-\frac{1}{\pi}(-1)=\frac{1}{\pi}\)

Answer 25

Oh okay that makes sense. That's where I messed up. I really appreciate it.

Answer 26

yw