Find the average value of the function on the given interval:
f(x)=cos(16x), [0,pi/2]
The answer I got is 1.
That cannot be the correct value because the cos 16x function undergoes 4 complete oscillations from 0 to pi/2. The average value must be zero.
The indefinite integral is (1/16)sin(16x).
Evaluate it at x = pi/2 and 0 and take the difference.
0 - 0 = 0
Divide by the interval (pi/2) and you still get zero for the average value.
I got that the first time, but I though something was wrong.
This is a problem that could have been answered by inspection. You have an integer number of oscillations about a mean value of zero.
To find the average value of a function on a given interval, you need to calculate the definite integral of the function over that interval and then divide it by the width of the interval.
In this case, the function is f(x) = cos(16x) and the interval is [0, π/2].
The definite integral of the function over the interval [0, π/2] can be calculated as follows:
∫[0,π/2] cos(16x) dx
To evaluate this integral, you can use a u-substitution. Let u = 16x, then du = 16 dx, and dx = du/16. The integral becomes:
∫[0,π/2] (1/16) cos(u) du
Now, evaluate the integral:
(1/16) ∫[0,π/2] cos(u) du
Applying the fundamental property of integrals for the cosine function, we get:
(1/16) [sin(u)] [0,π/2]
Now substitute back in the original variable: u = 16x:
(1/16) [sin(16x)] [0,π/2] = (1/16) [sin(16(π/2)) - sin(16(0))]
Since sin(0) = 0, the integral simplifies to:
(1/16) [sin(8π) - 0] = (1/16) [0 - 0] = 0
The definite integral of the function f(x) = cos(16x) over the interval [0, π/2] is 0.
To find the average value of the function, divide the definite integral by the width of the interval, which is π/2 - 0 = π/2:
Average = (definite integral) / (width of interval) = 0 / (π/2) = 0
Therefore, the average value of the function f(x) = cos(16x) on the interval [0, π/2] is 0, not 1.