09Iaverageval


 * Average Value of a Function **

For any function, y = f(x), the ** average value ** of the function over the interval [a, b] is given by:

math . \qquad y_{av} = \dfrac{1}{b-a} \displaystyle{\int_a^b} f(x) \, dx \qquad. math

Geometrically, the average value of the function is the height, h, such that the __rectangle__ h by (b – a) has the same __area__ as the area under the graph y = f(x) in the interval [a, b]


 * Example 1 **

Find the average value of f(x) = 2x 2 for the interval [1, 4]. Hence find the value of x that corresponds to the average value.


 * Solution:**

math . \qquad y_{av} = \dfrac{1}{4-1} \displaystyle{\int_1^4} 2x^2 \, dx \qquad. math

math . \qquad y_{av} = \dfrac{1}{3} \left[ \dfrac{2x^3}{3} \right]_1^4 \qquad \qquad. math

math . \qquad y_{av} = \dfrac{1}{3} \times \left( \dfrac{128}{3} - \dfrac{2}{3} \right) \qquad. math

math . \qquad y_{av} = 14 \qquad. math

Thus the average value of f(x) over [1, 4] is 14.

To find the value of x that corresponds to the average value, solve f(x) = 14

math \\ . \qquad 2x^2 = 14 \qquad. \\ . \\ . \qquad x^2 = 7 \qquad. \\ . \\ . \qquad x = \pm \sqrt{7} \qquad. math

but the domain is [1, 4] so reject negative value

math . \qquad \textit {Hence } x = \sqrt{7} \qquad. math

ie at this value of x, the y-value is equal to the average value of the function over the interval [1, 4].