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Introduction to Integrals

Average Value and Second Fundamental Theorem

Problems for "The Definite Integral"

Problems for "Average Value and Second Fundamental Theorem"

The Average Value of a Function

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:

Figure %: Two separate, equal areas

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

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.

The Second Fundamental Theorem of Calculus

Consider the function

g(x) = f (t)dt    

for the graph of f drawn below:

Figure %: Graph of f

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.

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