08 5avevalue

Average Value of a Function

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For any function, $y = f(x)$, the average value of the function over the interval $\big[a,\; b\big]$ is given by:

… … $y_{av} = \dfrac{1}{b-a} \displaystyle{\int_a^b} f(x) \, dx$

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Geometrically, the average value of the function is:

  • the height, y = h, such that the rectangle $h \times \big(b - a\big)$ has the same area
  • as the area under the graph $y = f(x)$ in the interval $\big[a,\; b\big]$
08.5aveval.gif

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CAUTION: Do not confuse the average value of a function with the average rate of change of a function.

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Example 1

Find the average value of $f(x) = 2x^2$ for the interval $\big[1,\; 4\big]$.
Hence find the value of x that corresponds to the average value.

Solution:

… … $y_{av} = \dfrac{1}{4-1} \displaystyle{\int_1^4} 2x^2 \, dx$
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… … $y_{av} = \dfrac{1}{3} \left[ \dfrac{2x^3}{3} \right]_1^4$
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… … $y_{av} = \dfrac{1}{3} \times \left( \dfrac{128}{3} - \dfrac{2}{3} \right)$
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… … $y_{av} = 14$
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… … Thus the average value of $f(x)$ over $\big[1,\; 4\big]$ is 14.

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… … To find the value of x that corresponds to the average value:
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… … solve $f(x) = 14$
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… … $2x^2 = 14$
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… … $x^2 = 7$
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… … $x = \pm \sqrt{7}$
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… … but the domain is $\big[1,\; 4\big]$ so reject negative value
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… … Hence $x = \sqrt{7}$

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Hence: At $x = \sqrt{7}$, the y-value $\Big( f\big( \sqrt{7} \big) = 14 \Big)$ is equal to the average value of the function over the interval $\big[1,\; 4\big]$.
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08.5aveval.gif
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