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f ( c ) = 1 b a a b f ( x ) d x ,

and the proof is complete.

Finding the average value of a function

Find the average value of the function f ( x ) = 8 2 x over the interval [ 0 , 4 ] and find c such that f ( c ) equals the average value of the function over [ 0 , 4 ] .

The formula states the mean value of f ( x ) is given by

1 4 0 0 4 ( 8 2 x ) d x .

We can see in [link] that the function represents a straight line and forms a right triangle bounded by the x - and y -axes. The area of the triangle is A = 1 2 ( base ) ( height ) . We have

A = 1 2 ( 4 ) ( 8 ) = 16 .

The average value is found by multiplying the area by 1 / ( 4 0 ) . Thus, the average value of the function is

1 4 ( 16 ) = 4 .

Set the average value equal to f ( c ) and solve for c .

8 2 c = 4 c = 2

At c = 2 , f ( 2 ) = 4 .

The graph of a decreasing line f(x) = 8 – 2x over [-1,4.5]. The line y=4 is drawn over [0,4], which intersects with the line at (2,4). A line is drawn down from (2,4) to the x axis and from (4,4) to the y axis. The area under y=4 is shaded.
By the Mean Value Theorem, the continuous function f ( x ) takes on its average value at c at least once over a closed interval.
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Find the average value of the function f ( x ) = x 2 over the interval [ 0 , 6 ] and find c such that f ( c ) equals the average value of the function over [ 0 , 6 ] .

Average value = 1.5 ; c = 3

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Finding the point where a function takes on its average value

Given 0 3 x 2 d x = 9 , find c such that f ( c ) equals the average value of f ( x ) = x 2 over [ 0 , 3 ] .

We are looking for the value of c such that

f ( c ) = 1 3 0 0 3 x 2 d x = 1 3 ( 9 ) = 3 .

Replacing f ( c ) with c 2 , we have

c 2 = 3 c = ± 3 .

Since 3 is outside the interval, take only the positive value. Thus, c = 3 ( [link] ).

A graph of the parabola f(x) = x^2 over [-2, 3]. The area under the curve and above the x axis is shaded, and the point (sqrt(3), 3) is marked.
Over the interval [ 0 , 3 ] , the function f ( x ) = x 2 takes on its average value at c = 3 .
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Given 0 3 ( 2 x 2 1 ) d x = 15 , find c such that f ( c ) equals the average value of f ( x ) = 2 x 2 1 over [ 0 , 3 ] .

c = 3

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Fundamental theorem of calculus part 1: integrals and antiderivatives

As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, Part 1 , is stated here. Part 1 establishes the relationship between differentiation and integration.

Fundamental theorem of calculus, part 1

If f ( x ) is continuous over an interval [ a , b ] , and the function F ( x ) is defined by

F ( x ) = a x f ( t ) d t ,

then F ( x ) = f ( x ) over [ a , b ] .

Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note that we have defined a function, F ( x ) , as the definite integral of another function, f ( t ) , from the point a to the point x . At first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks like it’s a function. The key here is to notice that for any particular value of x , the definite integral is a number. So the function F ( x ) returns a number (the value of the definite integral) for each value of x .

Second, it is worth commenting on some of the key implications of this theorem. There is a reason it is called the Fundamental Theorem of Calculus. Not only does it establish a relationship between integration and differentiation, but also it guarantees that any integrable function has an antiderivative. Specifically, it guarantees that any continuous function has an antiderivative.

Practice Key Terms 4

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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