9.3average Valueap Calculus

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What I want to do in this video is think about the idea of an average value of a function over some closed interval so what do I mean by that and how could we think about what average value of a function even means so let's say that's my y-axis and let's say that this is my this right over here is my x-axis and let me draw a function here so let's say the function looks something like that.

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Section 6-1 : Average Function Value

The first application of integrals that we’ll take a look at is the average value of a function. The following fact tells us how to compute this.

Average Function Value

  • The average of some finite set of values is a familiar concept. If, for example, the class scores on a quiz are 10, 9, 10, 8, 7, 5, 7, 6, 3, 2, 7, 8, then the average score is the sum of these numbers divided by the size of the class: (9.4.1) average score = 10 + 9 + 10 + 8 + 7 + 5 + 7 + 6 + 3 + 2 + 7 + 8 12 = 82 12 ≈ 6.83.
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The average value of a function (fleft( x right)) over the interval (left[ {a,b} right]) is given by,

[{f_{avg}} = frac{1}{{b - a}}int_{{,a}}^{{,b}}{{fleft( x right),dx}}]

To see a justification of this formula see the Proof of Various Integral Properties section of the Extras chapter.

Let’s work a couple of quick examples.

Example 1 Determine the average value of each of the following functions on the given interval.
  1. (fleft( t right) = {t^2} - 5t + 6cos left( {pi ,t} right)) on (left[ { - 1,frac{5}{2}} right])
  2. (Rleft( z right) = sin left( {2z} right){{bf{e}}^{1 - cos left( {2z} right)}}) on (left[ { - pi ,pi } right])
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9 3 Average Valueap Calculus 14th Edition

Hide All Solutionsa (fleft( t right) = {t^2} - 5t + 6cos left( {pi ,t} right)) on (left[ { - 1,frac{5}{2}} right]) Show Solution

There’s really not a whole lot to do in this problem other than just use the formula.

[begin{align*}{f_{avg}} & = frac{1}{{frac{5}{2} - left( { - 1} right)}}int_{{, - 1}}^{{,frac{5}{2}}}{{{t^2} - 5t + 6cos left( {pi ,t} right),dt}} & = frac{2}{7}left. {left( {frac{1}{3}{t^3} - frac{5}{2}{t^2} + frac{6}{pi }sin left( {pi t} right)} right)} right _{ - 1}^{frac{5}{2}} & = frac{{12}}{{7pi }} - frac{{13}}{6} & = - 1.620993end{align*}]

You caught the substitution needed for the third term right?

So, the average value of this function of the given interval is -1.620993.


Valueapb (Rleft( z right) = sin left( {2z} right){{bf{e}}^{1 - cos left( {2z} right)}}) on (left[ { - pi ,pi } right]) Show Solution

Again, not much to do here other than use the formula. Note that the integral will need the following substitution.

[u = 1 - cos left( {2z} right)]

Here is the average value of this function,

[begin{align*}{R_{avg}} & = frac{1}{{pi - left( { - pi } right)}}int_{{, - pi }}^{{,pi }}{{sin left( {2z} right){{bf{e}}^{1 - cos left( {2z} right)}},dz}} & = left. {frac{1}{{4pi }}{{bf{e}}^{1 - cos left( {2z} right)}}} right _{ - pi }^pi & = 0end{align*}]

So, in this case the average function value is zero. Do not get excited about getting zero here. It will happen on occasion. In fact, if you look at the graph of the function on this interval it’s not too hard to see that this is the correct answer.

There is also a theorem that is related to the average function value.

The Mean Value Theorem for Integrals

9 3 average valueap calculus calculator

If (fleft( x right)) is a continuous function on (left[ {a,b} right]) then there is a number (c) in (left[ {a,b} right]) such that,

[int_{{,a}}^{{,b}}{{fleft( x right),dx}} = fleft( c right)left( {b - a} right)]

Note that this is very similar to the Mean Value Theorem that we saw in the Derivatives Applications chapter. See the Proof of Various Integral Properties section of the Extras chapter for the proof.

Note that one way to think of this theorem is the following. First rewrite the result as,

[frac{1}{{b - a}}int_{{,a}}^{{,b}}{{fleft( x right),dx}} = fleft( c right)]

and from this we can see that this theorem is telling us that there is a number (a < c < b) such that ({f_{avg}} = fleft( c right)). Or, in other words, if (fleft( x right)) is a continuous function then somewhere in (left[ {a,b} right]) the function will take on its average value.

Let’s take a quick look at an example using this theorem.

Example 2

9 3 Average Valueap Calculus Calculator

Determine the number (c) that satisfies the Mean Value Theorem for Integrals for the function (fleft( x right) = {x^2} + 3x + 2) on the interval (left[ {1,4} right]). Show Solution

First let’s notice that the function is a polynomial and so is continuous on the given interval. This means that we can use the Mean Value Theorem. So, let’s do that.

[begin{align*}int_{{,1}}^{{,4}}{{{x^2} + 3x + 2,dx}} & = left( {{c^2} + 3c + 2} right)left( {4 - 1} right) & left. {left( {frac{1}{3}{x^3} + frac{3}{2}{x^2} + 2x} right)} right _1^4 = 3left( {{c^2} + 3c + 2} right) & frac{{99}}{2} = 3{c^2} + 9c + 6 & 0 = 3{c^2} + 9c - frac{{87}}{2}end{align*}]

This is a quadratic equation that we can solve. Using the quadratic formula we get the following two solutions,

[begin{align*}c & = frac{{ - 3 + sqrt {67} }}{2} = 2.593 c & = frac{{ - 3 - sqrt {67} }}{2} = - 5.593end{align*}]

9 3 Average Valueap Calculus Formula

Clearly the second number is not in the interval and so that isn’t the one that we’re after. The first however is in the interval and so that’s the number we want.

9 3 Average Valueap Calculus Equation

Note that it is possible for both numbers to be in the interval so don’t expect only one to be in the interval.