- anonymous

find mean value?
2cos2x+sinx=0
a= -pi\2, b=pi\2

- chestercat

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- anonymous

plz reply karo plzzzzzzzzzz

- myininaya

{f(b)-f(a)}/{b-a}={f(pi/2)-f(-pi/2)}/{pi}={[2cos(pi)+sin(pi/2)]-[2cos(-pi)+sin(-pi/2)]}/{pi}
={2(-1)+1-2(-1)-(-1)}/pi=(-2+1+2+1)/pi=2/pi
but f'(x)=4sin(2x)+cosx
f'(c)=4sin(2c)+cos(c)
set f'(c)=2/pi and solve for c

- anonymous

i got a value of c

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## More answers

- myininaya

ok cool!

- anonymous

i got !
4sin2c+(1\90)=cosc
now how can be find value of c

- myininaya

4sin(2c)+cos(c)=2/pi

- anonymous

plz reply fast myninaya............

- myininaya

-0.04535615498 is what I got using a calculator
so the problem says to use the mean value thm?

- anonymous

how can you find the value of c . give me the method?

- dumbcow

\[Average = \frac{1}{b-a}\int\limits_{a}^{b}2\cos 2x +\sin x dx\]

- anonymous

dumcow i don't understand this method . give me another method?

- dumbcow

are you looking for the mean value of the function from a to b
or does it say to use mean value theorem to find point that equals avg rate of change

- anonymous

but my teacher says that u can use mean value theorm

- myininaya

http://www.eldamar.org.uk/maths/calculus/node17.html

- myininaya

m(b-a)=int(f(x),a..b)
m is the mean value

- myininaya

cow has already said this though

- myininaya

do you know how to integrate?

- anonymous

no i don't know?

- anonymous

Okay, By the mean value theorem. The mean value is [f(b)-f(a)]/[b-a].
\[f(b)=f(\pi/2)=2\cos \pi +\sin(\pi/2)=-2+1=-1\]
\[f(a)=f(-{\pi \over 2})=2\cos (-\pi)+\sin (-\pi/2)=-2-1=-3\]

- anonymous

Therefore,
\[{f(b)-f(a) \over b-a}={-1-(-3) \over {\pi \over 2}-{-\pi \over 2}}={2 \over \pi}\]

- anonymous

Does that help?

- anonymous

its right but how we can find the value of c?

- anonymous

This value, we just found (2/pi) is equal to f'(c). You can use this relation to find c.

- anonymous

f`c= -4sin2c+cosc

- anonymous

-4sin2c+cosc=pi\2
so how we can find C?

- anonymous

Solve the equation, you may get more than one value for c. Take only the value that is in the given interval (-pi/2,pi/2).

- anonymous

how can it solve the equa? i don't understand this equation to find the value for C

- myininaya

i would graph it and approximate the solution

- myininaya

you could also use newton's method

- anonymous

You never solved quadratic equation?! Hmm I think myininaya got a point. It's difficult to solve it using identities. Probably graphing is a good method.

- anonymous

Just gimme a minute.

- anonymous

BRB

- anonymous

BRB what?

- myininaya

means be right back

- anonymous

ok

- anonymous

no body can solve this question?

- anonymous

oops myininaya has already solved the problem. Sorry I didn't see that.

- myininaya

lol

- anonymous

you are give my proper metjhod to find the valuc of C?

- anonymous

as myininaya, we can estimate the value by graphing.

- anonymous

c will be around 1.66

- anonymous

you said that C1.66 . how you can find tell me?

- anonymous

Wait. this value of c is out side our interval. The value of c, that's in the interval is around 0.045

- myininaya

look for x intercepts of f'(c)=2/pi

- anonymous

\[f'(c)={f(b)-f(a) \over b-a}\] this is the formula.

- anonymous

ok i know thic formula i completed..
-4sin2c+cosx=2\pi
after what can i do i don't understand?

- myininaya

##### 1 Attachment

- myininaya

this is what I got when I used newton's method to find c

- anonymous

f(x)=2cos2x+sinx
andf`(x)=-4sin2x+cosx

- myininaya

oops i forgot about the 2/pi
you try
i have to go

- anonymous

ok thanks!

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