Let ff be the function defined by f, of, x, equals, cosine, left bracket, 2, pi, x, right bracket, .f(x)=cos(2πx). What is the average value of ff on the interval open square bracket, start fraction, 5, divided by, 12, end fraction, comma, start fraction, 11, divided by, 12, end fraction, close square bracket[ 125​ , 1211​ ] written in simplest form?

The average value of a function f(x) on the interval [a, b] is given by the formula:

1/(b - a) ∫ from a to b f(x) dx

Here, we are asked to find the average value of the function f(x) = cos(2πx) on the interval [5/12, 11/12].

Substituting the given values into the formula, we get:

1/(11/12 - 5/12) ∫ from 5/12 to 11/12 cos(2πx) dx

Simplify the fraction in front of the integral:

1/(6/12) ∫ from 5/12 to 11/12 cos(2πx) dx = 2 ∫ from 5/12 to 11/12 cos(2πx) dx

Now, we need to compute the integral. The antiderivative of cos(2πx) is (1/(2π))sin(2πx). So, we have:

2 * [(1/(2π))sin(2πx)] from 5/12 to 11/12

Evaluate the antiderivative at the upper and lower limits of the integral:

2 * [(1/(2π))sin(2π*(11/12)) - (1/(2π))sin(2π*(5/12))]

Simplify the expression:

2 * [(1/(2π))(sin(2π11/12) - sin(2π5/12))]

Since sin(2π11/12) = sin(2π5/12) = 0, the expression simplifies to 0.

So, the average value of the function f(x) = cos(2πx) on the interval [5/12, 11/12] is 0.