IIT JEE , AIEEE Forum - Mathematics , Trignometry

mukulss

how can we say that the average value of cos²

is 1/2.

i mean what is the meaning of average value here.

is it defined or calculated

tango_goat

AVERAGE VALUE OF ANY FUNCTION IS GIVEN BY
AVG=INT FROM 0 TO X OF F[X]dX/X.....WHERE DENOMINATOR PART IS NOT INCLUDED IN THE INTEGRAL PART

krishna.gopal

av(cos²

)=(1/2

)_{[ 0]}

^{[2 ]} cos²

d

Use cos²

=(cos(2

)+1)/2

If you evaluate this integral you will find that its value is 1/2.

manoj_babu

cos^2x<=1=>INT[cos^2x]=1,0(1 for 0 & 180)

=>avg value=1 or 0

avinash.sharma

Hello manoj_babu,

krishna.gopal has given the correct guidance. Please read the following definitions to make it more clear :

Average of a function

(1) If f(x) is continuous and it's integration is possible on [a, b], then the average value y of y = f(x) for x in the interval [a, b] is

={1/(b-a)} _{[a ]}^{[ b]} f(x) dx

(2) Average Value Theorem: If f(x) is continuous and it's integration is possible on [a, b], then there exists a point c in the interval [a, b] such that

_{[a ]}^{[ b]} f(x) dx = f(c) (b-a)

Now come to the question : As Cos function is continuous and it's integration is possible in the interval 0 to 2p

_{[ 0]}^[2 ^p^] Cos²q dq = _{[ 0]}^[2pi^] (1+Cos2q)/2 dq = (1/2)[_{[ 0]}^[2pi^] dq + _{[ 0]}^[2pi^] Cos2q dq = (1/2)[ q + (Sin2q)/2] 0 to 2p

= (1/2)[2p-0+(sin4p)/2-(Sin 0 )/2]= (1/2) [2p] = p as sin4p = 0 =Sin 0

now by definition of average divide it by (b-a) = ( 2p - 0) = 2p

and you will get 1/2.

So the average value of a function is well defined and can be calculated for a particular function.

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