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