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In this lesson, weβll learn how to use the mean value theorem for integrals to find the average value of a function.
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At this stage, you should feel confident in finding definite integrals of a variety of functions, in particular polynomial functions.
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In this video, weβll look to take these ideas and extend them into finding the average value of a given function over some closed interval.
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We begin then by recalling the mean value theorem for integrals.
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This is that if π is a continuous function on some closed interval π to π, then there exists a number π in this interval such that the integral evaluated between π and π of π of π₯ with respect to π₯ is equal to π of π times π minus π.
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And in this theorem, the value of π of π is the average value of our function π on that closed interval π to π.
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But what does this actually mean?
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It essentially tells us that we guarantee that a continuous function will have at least one point, where the function itself equals the average value of the function over that closed interval.
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We can rearrange this equation to make π of π the subject.
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And we find the formula for the average value of our function.
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If π is integrable on some closed interval π to π, then the average value of the function over that closed interval is given by one over π minus π times the integral evaluated between π and π of π of π₯ with respect to π₯.
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Weβre now going to look at the application of this formula through a variety of examples.
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Determine the average value of π of π₯ equals three π₯ squared minus two π₯ on the closed interval negative three to five.
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Remember, the formula for the average value of the function π over some closed interval π to π is one over π minus π times the integral evaluated between π and π of π of π₯ with respect to π₯.
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In this case, we see that our π of π₯ is equal to three π₯ squared minus two π₯.
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Our closed interval is from negative three to five.
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So we let π be equal to negative three and π be equal to five.
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One over π minus π becomes one over five minus negative three.
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And this is multiplied by the integral evaluated between negative three and five of three π₯ squared minus two π₯ with respect to π₯.
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Five minus negative three is eight.
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So we need to evaluate this definite integral.
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Here, we recall that the indefinite integral of a general polynomial term ππ₯ to the power of π is ππ₯ to the power of π plus one over π plus one plus π, where π and π are constants and π is not equal to negative one.
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We also recall that we can integrate the sum of polynomial terms by integrating each term individually.
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And we see that the integral of three π₯ squared is three π₯ cubed over three.
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And the integral of negative two π₯ is negative two π₯ squared over two.
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And weβre going to evaluate this between negative three and five.
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Now this simplifies somewhat to π₯ cubed minus π₯ squared.
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So letβs substitute our limits in.
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Weβre looking to find an eighth of five cubed minus five squared minus negative three cubed minus negative three squared.
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Thatβs an eighth of 100 minus negative 36, which is equal to 17.
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So the average value of our function three π₯ squared minus two π₯ on the closed interval negative three to five is 17.
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Now that we have an idea of how the average value of a function formula works, weβll look at an example which involves a little more work on the integration site.
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Find the average value of π of π₯ equals π₯ squared over π₯ cubed minus five all squared on the closed interval negative one to one.
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Remember, the formula for the average value of a function over a closed interval π to π is given as one over π minus π times the integral of π of π₯ with respect to π₯ evaluated between π and π.
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In this case, we can see that our function is equal to π₯ squared over π₯ cubed minus five all squared.
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And we see our closed interval is from negative one to one.
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So we let π be equal to negative one and π be equal to one.
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And this means the average value of our function is given as one over one minus negative one multiplied by the integral of π₯ squared over π₯ cubed minus five all squared with respect to π₯ evaluated between negative one and one.
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One minus negative one is two.
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But how do we evaluate this definite integral?
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Well, we need to notice that the numerator is a scalar multiple of the derivative of part of the denominator.
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The derivative of π₯ cubed minus five is three π₯ squared.
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And this tells us we can use integration by substitution.
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We let π’ be equal to π₯ cubed minus five, and dπ’ by dπ₯ is therefore equal to three π₯ squared.
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We know that dπ’ by dπ₯ is not a fraction.
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But for the purposes of integration by substitution, we do treat it a little like where we say that this is equivalent to a third dπ’ equals π₯ squared dπ₯.
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Weβre now in a position to replace the various parts of our integral.
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We replace π₯ squared dπ₯ with a third dπ’.
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And we replace π₯ cubed minus five with π’.
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And we see that the average value of our function is equal to a half times the integral of a third times one over π’ squared with respect to π’.
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But what do we do about these limits?
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Well, we use our definition of π’.
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We said that π’ was equal to π₯ cubed minus five.
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So for our upper limit, when π₯ is equal to one, π’ is equal to one cubed minus five, which is negative four.
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And when π₯ is equal to negative one, π’ is equal to negative one cubed minus five, which is equal to negative six.
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We can take the constant a third outside of the integral sign and rewrite one over π’ squared as π’ to the negative two.
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And we know that the integral of π’ to the negative two is π’ to the negative one divided by negative one, or negative π’ to the negative one, which can then be written as negative one over π’.
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We substitute negative four and negative six.
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And we get a sixth of negative one over negative four minus negative one over negative six.
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Well, negative one over negative four is just a quarter.
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And negative one over negative six is a sixth.
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We subtract these fractions by creating a common denominator.
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And we see that the average value of our function is a sixth times a twelfth, which is one over 72.
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In our next example, weβll look how to use the inverse of the average value of a function formula to help us calculate missing values.
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The average value of π of π₯ equals negative six π₯ squared plus six π₯ minus one on the closed interval zero to π is zero.
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Find all possible values of π.
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Remember, the formula for the average value of a function π over a closed interval π to π is one over π minus π times the integral evaluated between π and π of π of π₯ with respect to π₯.
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We can see that π of π₯ is equal to negative six π₯ squared plus six π₯ minus one.
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And we see that π is equal to zero, and we donβt know the value of π.
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So we begin by substituting what we know about the average value of our function into the formula.
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We get one over π minus zero, which is of course one over π.
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And we multiply that by the definite integral of negative six π₯ squared plus six π₯ minus one evaluated between zero and π.
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We do, however, know that the average value of our function is equal to zero.
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So we can set this equal to zero.
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Letβs look to evaluate our integral.
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The integral of negative six π₯ squared is negative six π₯ cubed over three.
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The integral of six π₯ is six π₯ squared over two.
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And the integral of negative one is negative π₯.
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Letβs substitute π and zero into this expression.
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We see that zero is equal to one over π times negative two π cubed plus three π squared minus π.
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And then we divide through by π.
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And so we obtain that negative two π squared plus three π minus one is equal to zero.
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And we see we have a quadratic equation.
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So letβs multiply through by negative one.
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And then weβll factor the expression two π squared minus three π plus one.
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When we do, we see that two π minus one times π minus one must be equal to zero.
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And for this statement to be true, either two π minus one must be equal to zero or π minus one must be equal to zero.
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And we solve for π.
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And we see that π must be equal to one-half or one.
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Find all π such that π of π equals the average value of π of π₯ equals π₯ minus two all squared over the closed interval negative one to five.
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We begin by recalling the formula for the average value of a function over some closed interval π to π.
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Itβs one over π minus π times the integral of π of π₯ with respect to π₯ evaluated between π and π.
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We see that π of π₯ here is π₯ minus two all squared.
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And our interval is from negative one to five inclusive.
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So we say π is equal to negative one and π is equal to five.
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So the average value of this function over that closed interval is one over five minus negative one times the integral evaluated between negative one and five of π₯ minus two all squared with respect to π₯.
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And there are a couple of ways that we can evaluate this definite integral.
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We could look to distribute the parentheses.
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Alternatively, we spot that if we let π’ be equal to π₯ minus two, then we obtain dπ’ by dπ₯ to be equal to one, which in turn means we can say that dπ’ must be equal to dπ₯.
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And we can therefore use integration by substitution.
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We replace π₯ minus two with π’ and dπ₯ with dπ’.
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But we do still need to deal with the limits.
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So we use our definition of π’.
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And we see that when π₯ is equal to five, π’ is equal to five minus two, which is of course three.
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We also see that when π₯ is equal to negative one, π’ is negative one minus two, which is negative three.
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And the average value of our function is a sixth of the integral evaluated between negative three and three of π’ squared with respect to π’.
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And the integral of π’ squared is π’ cubed over three.
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And when we substitute our limits into the expression, we get a sixth of three cubed over three minus negative three cubed over three, which is simply three.
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Weβre not quite finished there.
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Weβre trying to find the values of π such that π of π equals the average value of π of π₯ over that closed interval, in other words, when is π of π equal to three.
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Well, π of π is π minus two all squared.
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So weβre looking to find when π minus two all squared is equal to three.
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So weβll solve this equation for π.
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We take the square root of both sides of our equation, remembering to find both a positive and negative square root of three.
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And then we add two to both sides.
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And we obtain π to be equal to two plus root three and two minus root three.
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In our final example, weβll consider the geometric interpretation of this formula with regards to graphs of functions.
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What is the average value of this function on the closed interval negative five to four?
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We begin by recalling the formula for the average value of a function in a closed interval π to π.
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Itβs one over π minus π times the integral of π of π₯ evaluated between π and π.
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In our case, the limits are four and negative five.
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So itβs one over four minus negative five times that definite integral.
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This graph, however, shows a piecewise function made up of a number of different functions.
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Instead of working out the function at each point, weβre going to recall the most basic definition of the integral of a function.
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It allows us to find the net area between the graph of the function and the π₯-axis.
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We can therefore find the area between the graph of this function and the π₯-axis by splitting into subintervals and remembering that when we evaluate the area underneath the π₯-axis, weβll end up with a negative value.
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Weβll begin by finding the area of this triangle.
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The formula for area of a triangle is half times base times height.
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So the area of this triangle is a half times one times four, which is two square units.
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Since this triangle sits below the π₯-axis, we give this a value of negative two in our integral.
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Our next triangle sits above the π₯-axis.
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Its area is a half times two times one, which is just one square unit.
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So we add one.
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The next shape weβll come across is a trapezium, though we couldβve split this into a triangle and a square.
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Its area is a half times four plus three times three, which is 10.5 square units.
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So we add 10.5.
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We then find a second trapezium, which has an area of a half times three plus two times one, which is 2.5 square units.
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And we have one final trapezium, which has an area of a half times four plus two times two, which is three square units.
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Or we find the sum of these values, and thatβs equal to the integral evaluated between negative five and four of π of π₯ with respect to π₯.
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We see that the average value of our function is therefore a ninth times 15, which is five-thirds.
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In this video, weβve learnt that we can find the average value of a function π over some closed interval π to π given that π is integrable by using the formula one over π minus π times the integral evaluated between π and π of π of π₯ with respect to π₯.
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And we saw that this process can be applied to more complicated functions such as those that require integration by substitution.
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And finally, we saw that when given the graph of a function, itβs sometimes easier to find the area between the graph and the π₯-axis than try to evaluate an integral.