# Evaluating Integral Functions

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I'm taking a DE calculus class and I'm having problems figuring out the logic in solving some of the problems.

The given integral is improper because both the interval of integration is unbounded and the integrand is unbounded near zero. Investigate its convergence by expressing it a sum of two intergrands-one from 0 to 1 and one from 1 to infinity. Evaluate the integrals if they converge.

1/(x^2/3 + x^4/3) dx

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##### Solution Summary

The expert evaluates integral functions. The convergence expressing of two integrands are provided.

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Solution. Since 1/(x^2/3+x^4/3)=3/(x^2+x^4)=3{1/x^2-1/(x^2+1)}

We know that if the integrand f(x)=1/(x^2+1), then integrate x from 0 to infinity, ...

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- BSc , Wuhan Univ. China
- MA, Shandong Univ.

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- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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