Find the average value fave of the function ff on

Khadija Wells 2021-09-28 Answered
Find the average value fave of the function ff on the given interval.
\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}},\ \ \ {x}\in{\left[-{3},{3}\right]}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

Clara Reese
Answered 2021-09-29 Author has 6869 answers
The average value of the function f on the interval [a,b] is equal to:
\(\displaystyle{f}_{{{a}{v}{g}}}={\frac{{{1}}}{{{b}-{a}}}}{\int_{{{a}}}^{{{b}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\)
The given function is
\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}},\ \ \ {x}\in{\left[-{3},{3}\right]}\)
Thus, the average value of f is given by
\(\displaystyle{f}_{{{a}{v}{g}}}={\frac{{{1}}}{{{3}-{\left(-{3}\right)}}}}{\int_{{-{3}}}^{{{3}}}}{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{6}}}}{\int_{{-{3}}}^{{{3}}}}{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}\)
Now first calculate the indefinite integral \(\displaystyle\int{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle{\int_{{-{3}}}^{{{3}}}}{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}=\int{\frac{{{1}}}{{{3}{u}^{{{2}}}}}}{d}{u}\) [Apply u - substitution: \(\displaystyle{u}={x}^{{{3}}}+{30}\)]
\(\displaystyle={\frac{{{1}}}{{{3}}}}\cdot{\frac{{{u}^{{-{2}+{1}}}}}{{-{2}+{1}}}}\)
\(\displaystyle=-{\frac{{{1}}}{{{3}{u}}}}\)
\(\displaystyle=-{\frac{{{1}}}{{{3}{\left({x}^{{{3}}}+{30}\right)}}}}\) [Substitute back: \(\displaystyle{u}={x}^{{{3}}}+{30}\)]
Therefore we get
\(\displaystyle{f}_{{{a}{v}{g}}}={\frac{{{1}}}{{{6}}}}{\int_{{-{3}}}^{{{3}}}}{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{6}}}}{\left({\left[-{\frac{{{1}}}{{{3}{\left({x}^{{{3}}}+{30}\right)}}}}\right]}_{{{x}={3}}}-{\left[-{\frac{{{1}}}{{{3}{\left({x}^{{{3}}}+{30}\right)}}}}\right]}_{{{x}={3}}}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{6}}}}{\left[-{\frac{{{1}}}{{{171}}}}-{\left(-{\frac{{{1}}}{{{9}}}}\right)}\right]}\)
\(\displaystyle={\frac{{{1}}}{{{6}}}}\cdot{\frac{{{2}}}{{{19}}}}\)
\(\displaystyle={\frac{{{1}}}{{{57}}}}\)
Result:
\(\displaystyle{f}_{{{a}{v}{g}}}={\frac{{{1}}}{{{6}}}}{\int_{{-{3}}}^{{{3}}}}{\frac{{{x}^{{{2}}}}}{{{\left({x}^{{{3}}}+{30}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{57}}}}\)
Have a similar question?
Ask An Expert
4
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-08-04
A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a standard deviation of 3900 kilometers.
Assume the distribution of measurements to be approximately normal.
a) Construct a \(\displaystyle{99}\%\) confidence interval for the average number of kilometers an automobile is driven annually in Virginia.
b) What can we assert with \(\displaystyle{99}\%\) confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year?
asked 2021-08-08
Assume that the true value, that is the population mean, the average height of french women in france is 167.5 cm. Students in a class of 160 students each took a random sample of french women of size 100, measured the height of the women and calculated a 95% confidence interval (confidence interval) for the average height.How many students can be expected to calculate a confidence interval that does not include the number 167.5?
15 8 5 20 1
asked 2021-08-10
The high price of medicines is a source of major expense for those seniors in the United States who have to pay for these medicines themselves. A random sample of 2000 seniors who pay for their medicines showed that they spent an average of \(\displaystyle\${4600}\) last year on medicines with a standard deviation of \(\displaystyle\${800}\). a) Make a \(\displaystyle{98}\%\) confidence interval for the corresponding population mean. b) Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss allpossible alternatives. Which alternative is the best?
asked 2021-08-09
Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. To accomplish this, the records of 200 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.3 seats. \(\displaystyle{s}_{{{x}}}=?,\ {n}=?,\ {n}-{1}=?\). Construct a \(\displaystyle{92}\%\) confidence interval for the population average number of unoccupied seats per flight. (State the confidence interval. (Round your answers to two decimal places.)
asked 2021-07-30
A paper reported a \(\displaystyle{\left({1}-\alpha\right)}\) confidence interval for the proportion of voters is (0.561,0.599) based on a sample of 2,056 people. However, the paper omitted the value of \(\displaystyle\alpha\). If you want to test the hypothesis that the proportion of voters is greater than \(\displaystyle{65}\%\) at \(\displaystyle{1}\%\) significance, find \(\displaystyle{z}_{{{c}{a}{l}{c}}}\) value for this problem? Please report your answer to 2 decimal places.
asked 2021-08-02
The city wishes to estimate the average commute distance for all city employees. They collect a random sample of 97 employees and find a sample mean of \(\displaystyle\overline{{{x}}}={7.1}\overline{{{x}}}={7.1}\) miles. They also assume a population standard deviation of \(\displaystyle\sigma={1.5}\sigma={1.5}\) miles. Calculate the z interval to estimate, \(\displaystyle\mu\mu\), the average commute distance for all city employees with \(\displaystyle{95}\%\) confidence.
Round your answers to 2 decimal places.
Find the Critical Value z*:
Point Estimate:
Margin of Error:
Write the interval as (lower bound, upper bound)
...