Average Value Theorem

Like we have mean value theorem in differential calculus, there is a theorem in integral calculus. While the mean value theorem relates the mean rate of change to instantaneous rate of the function, the average value theorem relates the average value of the function to the average all the instantaneous values.

As you are aware, the integration of a function within an interval is area under the graph of the function within the same interval. Therefore, the average area is same as the exact area of the graph of a linear function defined in the same interval.


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Description of Average Value Theorem

average value theorem

The above diagram shows the function y = f(x). Consider a thin vertical strip of height y and width dx. The area of the strip is given by dA = ydx.

The area A under the curve in the interval [a, b] is the integration of all such thin strips. In other words,

A = $\int_{a}^{b}dA$ = $\int_{a}^{b}ydx$

Now the average value theorem says if a function is continuous and differentiable in an interval, then there exists a point c, such that

(b – a)f(c) = $\int_{a}^{b}ydx$

Referring to the same diagram, as per average value theorem the area aABb under the curve is same as the area aA’B’b.
Example Problem on Average Value Theorem

The area under a curve f(x) = x2  between the interval [-4, 4] is stretched into a rectangle of the same horizontal width. What is the length of the rectangle that has been formed.


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The area of under the curve f(x) = x2  between [-4, 4] is,

$\int_{-4}^{4}x^2dx$ = [ 43/3] - [ (-4)3/3] = 128/3 units

As per average value theorem, this area is same as the area with a base of 8 units and a length of l (say).

Therefore, 8l = 128/3

or, l ˜ 5.33 units.