find ave. value of f(x) = x/(x+3) over interval (-a,a)
- Thread starter kika09
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It appears there is a bit of a communication problem. If there is some deficiency in a previous response, feel free to ask questions. That's what we do, here.
Posting the same problem multiple times, while showing no work or effort of any kind, definitely is not the way to go.
Do you have a process, procedure, or method for computing average values? If so, what is it? If so, why can you not apply it to this problem? If not, why are you being presented this problem?
If you get questions, rather than answers, try answering the questions. We'll all be happier, including you.
Posting the same problem multiple times, while showing no work or effort of any kind, definitely is not the way to go.
Do you have a process, procedure, or method for computing average values? If so, what is it? If so, why can you not apply it to this problem? If not, why are you being presented this problem?
If you get questions, rather than answers, try answering the questions. We'll all be happier, including you.
pka
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- Jan 29, 2005
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The usual calculus definition of average on an interval is \(\displaystyle ave_{f,[a,b]} = \frac{{\int\limits_a^b f }}{{b - a}}\).
\(\displaystyle \L \begin{array}{rcl}
\int\limits_{ - a}^a {\frac{x}{{x + 3}}dx} & = & \left. {x - 3\ln \left( {\left| {x + 3} \right|} \right)} \right|_{ - a}^a \\
& = & \left( {a - 3\ln \left( {\left| {a + 3} \right|} \right)} \right) - \left( { - a - 3\ln \left( {\left| { - a + 3} \right|} \right)} \right) \\
& = & 2a + \ln \left( {\left| {\frac{{ - a + 3}}{{a + 3}}} \right|^3 } \right) \\
\end{array}\)
\(\displaystyle \L \begin{array}{rcl}
\int\limits_{ - a}^a {\frac{x}{{x + 3}}dx} & = & \left. {x - 3\ln \left( {\left| {x + 3} \right|} \right)} \right|_{ - a}^a \\
& = & \left( {a - 3\ln \left( {\left| {a + 3} \right|} \right)} \right) - \left( { - a - 3\ln \left( {\left| { - a + 3} \right|} \right)} \right) \\
& = & 2a + \ln \left( {\left| {\frac{{ - a + 3}}{{a + 3}}} \right|^3 } \right) \\
\end{array}\)
