# Real Analysis : Mean Value Theorem

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Let f(x) be integrable on [a,b], and let g(x) be nondecreasing and continuously differentiable on [a,b]. Let {p be element of P} be a partition of [a,b], and define

U(f,g,p) = SIGMA (Mi(g(the ith term of x) - g(the (i-1)th term of x))) as i=1 to n

L(f,g,p) = SIGMA (Ni(g(the ith term of x)-g(the (i-1)th term of x))) as i=1 to n

Use mean value theorem to prove that (inf U(f,g,p), for p is element of P) = (sup L(f,g,p), for p is element of P) = ( INTEGRAL f(x)g'(x)dx, as x from a to b)

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The Mean Value Theorem is used to prove an an integral relation. The mean value theorem to prove a function is found.

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Let f(x) be integrable on [a,b], and let g(x) be nondecreasing and continuously differentiable on [a,b]. Let {p be element of P} be a partition of [a,b], and define

Use mean value theorem to prove that

Proof. Given a partition P: of [a,b]. Since g(x) is nondecreasing and continuously differentiable on ...

###### Education

- 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."
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