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# Integration Question

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For every one-dimensional set C for which the integral exists, let Q(C) = &#8747;c f(x) dx , where f(x) = 6x(1 - x) , 0 < x < 1, zero elsewhere, otherwise let Q(C) be undefined. If C1 = { x : ¼ < x < ¾ } , C2 = {1/2}, C3 = {x: 0 < x < 10}. Find Q(C1), Q(C2) and Q(C3).

Without doing any work I would think the C2 = 0. So I am concentrating on the other 2 problems. And I would also think that once I properly integrate the problem I would just plug in the numbers and get my solutions for C1 and C3.

So the problem as I see is this
&#8747; 6x(1-x) dx
= 6 &#8747;x(1 - x)dx
I would think that I would use Integration by parts.
u = x dv/dx = 1 -x
du/dx = 1 v = (x2 - 2x) / 2 ???????

Assuming the above are correct and I am not sure that they are what are the next step to solving this problem. I know from the answers in the back of the book that they should be Q(C1) = 11/16, Q(C2) = 0, Q(C3) = 1.

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A probability question is thoroughly explored. Integration questions for probability and calculus are provided.

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This is a probability question. However, my real problem is with the integration of the problem.
For every one-dimensional set C for which the integral exists, let Q(C) = ∫c f(x) dx , where f(x) = 6x(1 - x) , 0 < x < 1, zero elsewhere, otherwise let Q(C) be undefined. If C1 = { ...

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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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