Blog: `h(u) = (3 - 2u)^-1 , [-1, 1]` Find the average value of the function on the given interval.

Tuesday, August 9, 2016

$\displaystyle{h}{\left({u}\right)}={{\left({3}-{2}{u}\right)}}^{{-{{1}}}},{\left[-{1},{1}\right]}$ Find the average value of the function on the given interval.

The average value is the integral of a function over an interval divided by the length of an interval. The length here is 2, let's found the integral:

$\displaystyle{\int_{{-{1}}}^{{1}}}\frac{{1}}{{{3}-{2}{u}}}{d}{u}=-\frac{{1}}{{2}}{\ln{{\left({3}-{2}{u}\right)}}}{{\mid}_{{{u}=-{1}}}^{{1}}}=-\frac{{1}}{{2}}{\left({\ln{{1}}}-{\ln{{5}}}\right)}=\frac{{\ln{{5}}}}{{2}}.$

So the average is $\displaystyle\frac{{\ln{{5}}}}{{4}}\approx{0.40}.$

Note that the given interval [-1, 1] doesn't contain the point 1.5 where $\displaystyle{3}-{2}{u}={0}.$

The average value is the integral of a function over an interval divided by the length of an interval. The length here is 2, let's found the integral:

So the average is $\displaystyle\frac{{\ln{{5}}}}{{4}}\approx{0.40}.$

Note that the given interval [-1, 1] doesn't contain the point 1.5 where $\displaystyle{3}-{2}{u}={0}.$

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Tuesday, August 9, 2016

$\displaystyle{h}{\left({u}\right)}={{\left({3}-{2}{u}\right)}}^{{-{{1}}}},{\left[-{1},{1}\right]}$ Find the average value of the function on the given interval.

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What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Tuesday, August 9, 2016

Find the average value of the function on the given interval.

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What is the Exposition, Rising Action, Climax, and Falling Action of &quot;One Thousand Dollars&quot;?

$\displaystyle{h}{\left({u}\right)}={{\left({3}-{2}{u}\right)}}^{{-{{1}}}},{\left[-{1},{1}\right]}$ Find the average value of the function on the given interval.

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?