# Process for finding average value of a function

The **average value of a function **is defined by the mean of values for a function taken over a given interval. In other words, it is the ratio of several values of the function to the number of instances. This process is same as that of the definite integration. It means that **finding average value of a function** is same as finding the integral over a closed period.

## Average value of a function formula in terms of integrals<

The **average value of a function formula **involving definite integrals is given below. Consider that g(x) is a continuous function for the interval [a, b]. Then the average value of g(x) from [a, b] is given by f_{avg} = (b – a)^{-1} ∫_{a}^{b} g(x) dx. This equation can be used to find average value for any given functions on bound limits. There also exist a mean value theorem which says that we can find a point 'z' in the given interval [a, b] so that ∫_{a}^{b} g(x) dx = (b – a) g(c). This equation is of great use for solving the examples.

## Solve the following average value of a function examples

Now let us solve **average value of a function examples **to learn the concept of average along with that of integration.

(1) Determine the average value for g(x) = x + 3 for the closed period [1,3]:

From the formula, f_{avg} = (b – a)^{-1} ∫_{a}^{b} g(x) dx. Here g(x) = x + 3, a = 1 and b = 3. So let us substitute all this values in the formula. f_{avg} = (3 – 1)^{-1} ∫_{1}^{3} (x + 3) dx = (1/2) [x^{2}/2 + 3x]_{1}^{3} = 0.5 [4.5 + 9 – 0.5 – 3]. Thus f_{avg} = 0.5 (4 + 6) = 0.5 x 10 = 5.

(2) Find the average value for g(x) = x for the closed period [1,2]:

Here g(x) = x, a = 1 and b =2. So the equation goes like f_{avg} = (b – a)^{-1} ∫_{a}^{b} g(x) dx = (2-1) ∫_{1}^{2} x dx. Therefore f_{avg} = [x^{2}/2]_{1}^{2} = 4/2 - ½ = 3/2.

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