# Find the average value of the function on the given interval. f(x)=9cosx ,[0,pi/2]

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### 2 Answers

The average of a series is the sum of all the terms divided by the number of terms. Here we have a continuous function f(x) = 9*cos x.

The sum of the function for all the values of x from 0 to pi/2 is the definite integral of f(x) from x = 0 to x = pi/2. The average of these values would be the sum divide by (pi/2 - 0)

Int[ f(x) dx], x = 0 to x= pi/2

=> Int[ 9*cos x dx], x = 0 to x= pi/2

=> 9*sin x, x = 0 to x= pi/2

=> 9[ sin pi/2 - sin 0]

=> 9

The average is 9/(pi/2 - 0)

=> 9*2/pi

=> 18/pi

**The required average value is 18/pi**

To determine the average value of the given function, we'll have to compute the definite integral of the function over the interval [0 , pi/2].

f avg. = Int f(x)dx/(b-a), where the interval is [a,b]

To calculate the definite integral, we'll apply Leibniz-Newton formula, over the interval [a,b]:

Int f(x) = F(b) - F(a)

f avg. = Int 9cosx dx/(pi/2 - 0)

f avg. = 9*Int cosx dx/(pi/2)

f avg. = 9*(sinpi/2- sin0)/(pi/2)

f avg. = 9*(1-0)/(pi/2)

f avg. = 18/pi

**The requested average value of the function is f avg. = 18/pi.**