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butterflydreamer

  • one year ago

Question ---> http://prntscr.com/7tc26v I just need a hint on how to approach Part (ii) :) Thank you !

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  1. ganeshie8
    • one year ago
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    Simply multiply the width by corresponding height to get the area of \(r\)th rectangle. Let \(f(x) = x^3+x\) and notice below : Width of each rectangle is given by \(\dfrac{1-0}{n} = \dfrac{1}{n}\) Height of \(r\)th rectangle is given by \(f(\dfrac{1}{n})\).

  2. MrNood
    • one year ago
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    if you label the rectangles with r=1 r=2 .... r =n-1 (instead of the a1 a2 you have drawn on the diagram) you can see that the x value of the left side of th erectangle is r/n for each rectangle SO using the equation for y you can work out the HEIGHT of the rectangle Height = (r/n)^3 + r/n The WIDTH of each rectangle is 1/n SO Multiply the height by the width to get the area of each rectangle You will see the form of the answer begin to appear when you write this expression. The total are is simply the sum from r = 1 to r = n-1 of all those areas, which leads tot the answer shown....

  3. ganeshie8
    • one year ago
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    typo fixed : Let \(f(x) = x^3+x\) and notice below : Width of each rectangle is given by \(\dfrac{1-0}{n} = \dfrac{1}{n}\) Height of \(r\)th rectangle is given by \(f(\dfrac{\color{red}{r}}{n})\).

  4. butterflydreamer
    • one year ago
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    \[\therefore A_n =\frac{ 1 }{ n^4 } \sum_{r = 1}^{n-1}r^3 + \frac{ 1 }{ n^2 }\sum_{r=1}^{n-1}r\]oh.. so it will be something like this...? \[A_1 = \frac{ 1 }{ n } ( \frac{ 1 }{ n^3 } + \frac{ 1 }{ n }) = \frac{ 1 }{ n^4 } + \frac{ 1 }{ n^2 }\] \[A_2 = \frac{ 1 }{ n } (\frac{ 8 }{ n^3 } + \frac{ 2 }{ n }) = \frac{ 8 }{ n^4 } + \frac{ 2 }{ n^2 }\] \[A_3 = \frac{ 1 }{ n } ( \frac{ 27 }{ n^3} + \frac{ 3 }{ n }) = \frac{ 27 }{ n^4 } +\frac{ 3 }{ n^2 }\] . . . \[A _{n-1} = \frac{ 1 }{ n } \left[ \frac{ (n-1)^3 }{ n^3 }+\frac{ n-1 }{ n } \right]= \frac{ (n-1)^3 }{n^4 }+\frac{ n-1 }{ n^2 }\] \[A_1 + A_2 + ...+ A _{n-1} = \frac{ 1 }{ n^4 } + \frac{ 8 }{ n^4 } + \frac{ 27 }{ n^4 }+...+\frac{ (n-1)^3 }{ n^4}+ \frac{ 1 }{ n^2 }+\frac{ 2 }{ n^2 }+\frac{ 3 }{ n^2}+...+ \frac{ n-1 }{ n^2}\] \[A_n = \frac{ 1 }{ n^4 } \left[ 1 + 8 + 27 + ... + (n-1)^3\right]+ \frac{ 1 }{ n^2 }\left[ 1 + 2+...+(n-1) \right]\]

  5. butterflydreamer
    • one year ago
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    oops. I made a typo LOL. The first line " therefore .." is meant to be at the end xD

  6. MrNood
    • one year ago
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    yes - but you do not help by putting the numbers in ]

  7. butterflydreamer
    • one year ago
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    putting the numbers in ?

  8. MrNood
    • one year ago
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    \[area _{n} = \frac{ 1 }{ n }( \frac{ r ^{3} }{ n ^{3} } -\frac{ r }{ n })\]

  9. butterflydreamer
    • one year ago
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    hmm.............

  10. MrNood
    • one year ago
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    you have put the numbers 1,2, 3 etc for r but oyu just need to leave it in terms of r and n expand the above - you will see the answer form appear

  11. butterflydreamer
    • one year ago
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    ahh okay. So what i did was unnecessary?

  12. MrNood
    • one year ago
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    ^^ typo in formula should be + not -

  13. MrNood
    • one year ago
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    just distribute the expression I wrote above...

  14. butterflydreamer
    • one year ago
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    xD i know i know. I'm just saying.. like using the method you demonstrated, i'd be able to show the expression faster right?

  15. MrNood
    • one year ago
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    it's not just faster - the numbers do not appear in the answer - so you are not going to need them to derive the answer after you have distributed - then you need to write the 'summation' equation remember n is a constant - so it can be taken outside the summation....

  16. MrNood
    • one year ago
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    another typo in formula it should be area r not area n

  17. butterflydreamer
    • one year ago
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    so we're using: \[A_r = \frac{ 1 }{ n } ( \frac{ 1 }{ n^3 } + \frac{ 1 }{ n }) \] ? And i just plug in r = 1, 2 , 3 ..... (leaving it in terms of r and n )and eventually i'll reach the summation equation after expanding and everything :) Okay thank you!

  18. butterflydreamer
    • one year ago
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    oh typo *\[A_r = \frac{ 1 }{ n } (\frac{ r^3 }{ n^3 }+\frac{ r }{ n })\]

  19. phi
    • one year ago
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    once you have \[ A_n =\frac{ 1 }{ n^4 } \sum_{r = 1}^{n-1}r^3 + \frac{ 1 }{ n^2 }\sum_{r=1}^{n-1}r \] your are done. the next step would be to use "closed forms" to replace the summations

  20. phi
    • one year ago
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    The first summation is given by Part 1 the second is well known

  21. MrNood
    • one year ago
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    NO multiply out the formual as I wrote it (with corrections) You NEVER need to put 1,2, etc into it - it does not ask you to work out abn answer - it asks oyu to prove the expression given at the bottom of the section. The expression above when multiplied gives you an expression for the area of ONE rectangle you then need to put this int a 'summation' form to show that the expression is true.

  22. butterflydreamer
    • one year ago
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    okay.

  23. MrNood
    • one year ago
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    I re-iterate: You are not asked to work out a value for the area or summation Since you do not know what n is you CANNOT Just derive the summation formula using the method above do you want to write your expansion of the expression here?

  24. butterflydreamer
    • one year ago
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    \[so. A_r = \frac{ 1 }{ n } (\frac{ r^3 }{ n^3 } + \frac{ r }{ n }) = \frac{ r^3 }{ n^4 } + \frac{ r }{ n^2 }\]

  25. MrNood
    • one year ago
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    OK good now - that is the area of 1 rectangle so the area of all the rectangles is the summation of that expression from r=1 to r-n-1 if you write that out in the notation used in the answer you will see that you have the right answer remember 2 things in simplifying: sum (A+B) = sum (A) + sum(B) and sum (A/n) = 1/n sum(A) if n is a constant use these and you will derive the answer given..

  26. butterflydreamer
    • one year ago
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    okayyy

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