One of the uses of the definite integral is that it can help us find the average value of a function on an interval [a, b] . The formula for the average value on an interval [a, b] is as follows:
|f avg = f (x)dx|
To see why this is the case, consider this form of the equation:
|f avg(b - a) = f (x)dx|
The left side of the equation is the area of rectangle with base of (b - a) and height of f avg . The right side of the equation is the area under the curve of f over the interval with length (b - a) . These areas are depicted below:
The equation for the average value is a statement of the intuitive fact that if we construct a rectangle with the height f avg and width (b - a) , its area should be the same as the area under curve from a to b .
The mean value theorem for integrals states the following: if f is a continuous function on [a, b] , there exists at least one c on [a, b] such that
|f (c) = f (x)dx|
In other words, the MVT for integrals states that every continuous function attains its average value at least once on an interval.
Consider the function
|g(x) = f (t)dt|
for the graph of f drawn below:
In words, what does the function g(x) represent on the graph above?
The function g(x) represents the area under the graph of f between a and x . Thus, g(a) = 0 , since the area between a and a is zero. According to the second fundamental theorem of calculus,
|f (t)dt = f (x)|
This tells us two things. First, an area function of f is always an antiderivative of f . Second, differentiation and integration are inverse operations, because integrating f and then differentiating it yields f again.