{"id":384,"date":"2025-02-13T19:44:30","date_gmt":"2025-02-13T19:44:30","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/the-definite-integral-learn-it-4\/"},"modified":"2025-02-13T19:44:30","modified_gmt":"2025-02-13T19:44:30","slug":"the-definite-integral-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/the-definite-integral-learn-it-4\/","title":{"raw":"The Definite Integral: Learn It 4","rendered":"The Definite Integral: Learn It 4"},"content":{"raw":"\n

Average Value of a Function<\/h2>\n
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We often need to find the average of a set of numbers, such as an average test grade.<\/p>\n

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Suppose you received the following test scores in your algebra class: [latex]89, 90, 56, 78, 100[\/latex], and [latex]69[\/latex]. Your semester grade is your average of test scores and you want to know what grade to expect.<\/p>\n

We can find the average by adding all the scores and dividing by the number of scores. In this case, there are six test scores. Thus,<\/p>\n

[latex]\\dfrac{89+90+56+78+100+69}{6}=\\dfrac{482}{6}\\approx 80.33[\/latex]<\/div>\n

Therefore, your average test grade is approximately [latex]80.33[\/latex].<\/p>\n<\/section>\n

Suppose, however, that we have a function [latex]v(t)[\/latex] that gives us the speed of an object at any time [latex]t[\/latex], and we want to find the object\u2019s average speed. The function [latex]v(t)[\/latex] takes on an infinite number of values, so we can\u2019t use the process just described. Fortunately, we can use a definite integral to find the average value of a function<\/strong> such as this.<\/p>\n

Let [latex]f(x)[\/latex] be continuous over the interval [latex][a,b][\/latex] and let [latex][a,b][\/latex] be divided into [latex]n[\/latex] subintervals of width [latex]\\Delta x=\\frac{(b-a)}{n}[\/latex]. Choose a representative [latex]x_i^*[\/latex] in each subinterval and calculate [latex]f(x_i^*)[\/latex] for [latex]i=1,2, \\cdots , n[\/latex].<\/p>\n

In other words, consider each [latex]f(x_i^*)[\/latex] as a sampling of the function over each subinterval. The average value of the function may then be approximated as,<\/p>\n

[latex]\\dfrac{f(x_1^*)+f(x_2^*)+ \\cdots +f(x_n^*)}{n}[\/latex],<\/div>\n

which is basically the same expression used to calculate the average of discrete values.<\/p>\n

But we know [latex]\\Delta x=\\frac{b-a}{n}[\/latex], so [latex]n=\\frac{b-a}{\\Delta x}[\/latex], and we get,<\/p>\n

[latex]\\dfrac{f(x_1^*)+f(x_2^*)+ \\cdots +f(x_n^*)}{n}=\\dfrac{f(x_1^*)+f(x_2^*)+ \\cdots +f(x_n^*)}{\\dfrac{(b-a)}{\\Delta x}}[\/latex].<\/div>\n

Following through with the algebra, the numerator is a sum that is represented as [latex]\\displaystyle\\sum_{i=1}^{n} f(x_i^*)[\/latex], and we are dividing by a fraction. <\/p>\n

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To divide by a fraction, invert the denominator and multiply.<\/p>\n<\/section>\n

Thus, an approximate value for the average value of the function is given by,<\/p>\n

[latex]\\begin{array}{ll}\\frac{\\displaystyle\\sum_{i=1}^{n} f(x_i^*)}{\\dfrac{\\left(b-a\\right)}{\\Delta x}} & =\\left(\\dfrac{\\Delta x}{b-a}\\right)\\displaystyle\\sum_{i=1}^{n} f(x_i^*) \\\\ & =\\left(\\dfrac{1}{b-a}\\right)\\displaystyle\\sum_{i=1}^{n} f(x_i^*) \\Delta x \\end{array}[\/latex]<\/div>\n

This is a Riemann sum. To get the exact<\/em> average value, take the limit as [latex]n[\/latex] goes to infinity. Thus, the average value of a function is given by,<\/p>\n

[latex]\\dfrac{1}{b-a}\\underset{n\\to \\infty }{\\lim}\\displaystyle\\sum_{i=1}^{n} f(x_i) \\Delta x=\\dfrac{1}{b-a} \\displaystyle\\int_a^b f(x) dx[\/latex].<\/div>\n
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average value of a function<\/h3>\n

Let [latex]f(x)[\/latex] be continuous over the interval [latex][a,b][\/latex]. Then, the average value of the function [latex]f(x)[\/latex] (or [latex]f_{\\text{ave}}[\/latex]) on [latex][a,b][\/latex] is given by,<\/p>\n

[latex]f_{\\text{ave}}=\\dfrac{1}{b-a} \\displaystyle\\int_a^b f(x) dx[\/latex].<\/div>\n<\/section>\n
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Find the average value of [latex]f(x)=x+1[\/latex] over the interval [latex][0,5][\/latex].<\/p>\n

[reveal-answer q=\"fs-id1170572443612\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1170572443612\"]<\/p>\n

First, graph the function on the stated interval, as shown in Figure 11.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Figure 11. The graph shows the area under the function [latex]f(x)=x+1[\/latex] over [latex][0,5][\/latex].[\/caption]\n\n

The region is a trapezoid lying on its side, so we can use the area formula for a trapezoid [latex]A=\\frac{1}{2}h(a+b)[\/latex], where [latex]h[\/latex] represents height, and [latex]a[\/latex] and [latex]b[\/latex] represent the two parallel sides. Then,<\/p>\n

[latex]\\begin{array}{ll} \\displaystyle\\int_0^5 x+1 dx & =\\frac{1}{2}h(a+b) \\\\ & =\\frac{1}{2} \\cdot 5 \\cdot (1+6) \\\\ & =\\frac{35}{2} \\end{array}[\/latex]<\/div>\n

Thus the average value of the function is<\/p>\n

[latex]\\frac{1}{5-0} \\displaystyle\\int_0^5 x+1 dx = \\frac{1}{5} \\cdot \\frac{35}{2}=\\frac{7}{2}.[\/latex]<\/div>\n

Watch the following video to see the worked solution to Example: Finding the Average Value of a Linear Function.<\/p>\n