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on the interval <math>[a,b]</math>. | on the interval <math>[a,b]</math>. | ||
| − | I will give Josh Hunsberger's proof here. | + | I will give Josh Hunsberger's proof here. --[[User:Bell|Bell]] 15:04, 29 September 2008 (UTC) |
Proof. Notice that | Proof. Notice that | ||
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<math>V(c)= \int_a^b \pi (f(x)-c)^2\ dx = Ac^2-Bc+D</math> | <math>V(c)= \int_a^b \pi (f(x)-c)^2\ dx = Ac^2-Bc+D</math> | ||
| − | where <math>A | + | where the constants <math>A</math>, <math>B</math>, and <math>D</math> are given by |
| − | + | ||
| − | + | <math>A=\pi(b-a)</math>, | |
| − | + | ||
| − | + | <math>B=2\pi\int_a^b f(x)\ dx</math>, and | |
| − | + | ||
| − | parabola. | + | <math>D=\pi \int_a^b f(x)^2\ dx</math>. |
| + | |||
| + | The graph of <math>V(c)</math> is an upward opening | ||
| + | parabola. Such a parabola has an absolute minimum at the value <math>c_0</math> such that | ||
| + | <math>V'(c_0)=0</math>. Since <math>V'(c)=2Ac-B</math>, this value of <math>c_0</math> is | ||
| + | |||
| + | <math>c_0 = \frac{B}{2A} = \frac{1}{b-a}\int_a^b f(x)\ dx,</math> | ||
| + | |||
| + | the average value of <math>f(x)</math> on the interval <math>[a,b]</math>. To complete the argument, | ||
| + | we must show that this value of <math>c</math> falls in the interval between | ||
| + | <math>m</math> and <math>M</math>. But this is easy because | ||
| + | |||
| + | <math>m \le f(x) \le M</math> | ||
| + | |||
| + | on <math>[a,b]</math> implies that | ||
| + | |||
| + | <math> \int_a^b m\ dx \le \int_a^b f(x)\ dx \le \int_a^b M\ dx,</math> | ||
| + | |||
| + | and so | ||
| + | |||
| + | <math> m(b-a) \le \int_a^b f(x)\ dx \le M(b-a),</math> | ||
| + | |||
| + | and dividing by <math>(b-a)</math> yields | ||
| + | |||
| + | <math>m\le c_0\le M.</math> | ||
| + | |||
| + | The proof is complete. | ||
Latest revision as of 11:16, 29 September 2008
Suppose that $ f(x) $ is a continuous function on the interval $ [a,b] $. Let $ V(c) $ denote the volume of the solid obtained by revolving the area between the graph of $ y=f(x) $ and the line $ y=c $ about the line $ y=c $.
As $ c $ ranges from $ m=\{\text{min }f(x):a\le x\le b\} $ and $ M=\{\text{Max }f(x):a\le x\le b\} $, prove that the minmum value of $ V(c) $ is attained at $ c $ equal to the average value of $ f(x) $ on the interval $ [a,b] $.
I will give Josh Hunsberger's proof here. --Bell 15:04, 29 September 2008 (UTC)
Proof. Notice that
$ V(c)= \int_a^b \pi (f(x)-c)^2\ dx = Ac^2-Bc+D $
where the constants $ A $, $ B $, and $ D $ are given by
$ A=\pi(b-a) $,
$ B=2\pi\int_a^b f(x)\ dx $, and
$ D=\pi \int_a^b f(x)^2\ dx $.
The graph of $ V(c) $ is an upward opening parabola. Such a parabola has an absolute minimum at the value $ c_0 $ such that $ V'(c_0)=0 $. Since $ V'(c)=2Ac-B $, this value of $ c_0 $ is
$ c_0 = \frac{B}{2A} = \frac{1}{b-a}\int_a^b f(x)\ dx, $
the average value of $ f(x) $ on the interval $ [a,b] $. To complete the argument, we must show that this value of $ c $ falls in the interval between $ m $ and $ M $. But this is easy because
$ m \le f(x) \le M $
on $ [a,b] $ implies that
$ \int_a^b m\ dx \le \int_a^b f(x)\ dx \le \int_a^b M\ dx, $
and so
$ m(b-a) \le \int_a^b f(x)\ dx \le M(b-a), $
and dividing by $ (b-a) $ yields
$ m\le c_0\le M. $
The proof is complete.
