# Find the average value of the function f on the given interval. f(x) = x, [0, 16].

**Solution:**

Given: Function f(x) = x and the interval [0, 16]

The average of a function can be defined by the mean value theorem.

The mean value theorem states that y = (x) be a continuous function on the closed interval [a,b].

The mean value theorem for integrals states that there exists a point c in that interval such that

f(c) = 1/(b - a) \(\int_{a}^{b} f(x)dx\)

Here the lower limit is 0 and the upper limit is 16.

i.e., a = 0 and b = 16

f(c) = 1/(16 - 0) \(\int_{0}^{16} f(x)dx\)

f(c) = 1/16 \(\frac{x^2}{2}]^{16}_{0}\)

f(c) = 1/16(16^{2}/2 - 0^{2})

f(c) = 16/2 = 8

f(c) = 8

The average value of f(x) is 8

## Find the average value of the function f on the given interval. f(x) = x, [0, 16].

**Summary:**

The average value of the function f on the given interval. f(x) = x, [0, 16] is 8.

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