butterflydreamer
  • butterflydreamer
Question ---> http://prntscr.com/7tc26v I just need a hint on how to approach Part (ii) :) Thank you !
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
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 \(f(\dfrac{1}{n})\).
MrNood
  • MrNood
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....
ganeshie8
  • ganeshie8
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 \(f(\dfrac{\color{red}{r}}{n})\).

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butterflydreamer
  • butterflydreamer
\[\therefore A_n =\frac{ 1 }{ n^4 } \sum_{r = 1}^{n-1}r^3 + \frac{ 1 }{ n^2 }\sum_{r=1}^{n-1}r\]oh.. so it will be something like this...? \[A_1 = \frac{ 1 }{ n } ( \frac{ 1 }{ n^3 } + \frac{ 1 }{ n }) = \frac{ 1 }{ n^4 } + \frac{ 1 }{ n^2 }\] \[A_2 = \frac{ 1 }{ n } (\frac{ 8 }{ n^3 } + \frac{ 2 }{ n }) = \frac{ 8 }{ n^4 } + \frac{ 2 }{ n^2 }\] \[A_3 = \frac{ 1 }{ n } ( \frac{ 27 }{ n^3} + \frac{ 3 }{ n }) = \frac{ 27 }{ n^4 } +\frac{ 3 }{ n^2 }\] . . . \[A _{n-1} = \frac{ 1 }{ n } \left[ \frac{ (n-1)^3 }{ n^3 }+\frac{ n-1 }{ n } \right]= \frac{ (n-1)^3 }{n^4 }+\frac{ n-1 }{ n^2 }\] \[A_1 + A_2 + ...+ A _{n-1} = \frac{ 1 }{ n^4 } + \frac{ 8 }{ n^4 } + \frac{ 27 }{ n^4 }+...+\frac{ (n-1)^3 }{ n^4}+ \frac{ 1 }{ n^2 }+\frac{ 2 }{ n^2 }+\frac{ 3 }{ n^2}+...+ \frac{ n-1 }{ n^2}\] \[A_n = \frac{ 1 }{ n^4 } \left[ 1 + 8 + 27 + ... + (n-1)^3\right]+ \frac{ 1 }{ n^2 }\left[ 1 + 2+...+(n-1) \right]\]
butterflydreamer
  • butterflydreamer
oops. I made a typo LOL. The first line " therefore .." is meant to be at the end xD
MrNood
  • MrNood
yes - but you do not help by putting the numbers in ]
butterflydreamer
  • butterflydreamer
putting the numbers in ?
MrNood
  • MrNood
\[area _{n} = \frac{ 1 }{ n }( \frac{ r ^{3} }{ n ^{3} } -\frac{ r }{ n })\]
butterflydreamer
  • butterflydreamer
hmm.............
MrNood
  • MrNood
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
butterflydreamer
  • butterflydreamer
ahh okay. So what i did was unnecessary?
MrNood
  • MrNood
^^ typo in formula should be + not -
MrNood
  • MrNood
just distribute the expression I wrote above...
butterflydreamer
  • butterflydreamer
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?
MrNood
  • MrNood
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....
MrNood
  • MrNood
another typo in formula it should be area r not area n
butterflydreamer
  • butterflydreamer
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!
butterflydreamer
  • butterflydreamer
oh typo *\[A_r = \frac{ 1 }{ n } (\frac{ r^3 }{ n^3 }+\frac{ r }{ n })\]
phi
  • phi
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
phi
  • phi
The first summation is given by Part 1 the second is well known
MrNood
  • MrNood
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.
butterflydreamer
  • butterflydreamer
okay.
MrNood
  • MrNood
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?
butterflydreamer
  • butterflydreamer
\[so. A_r = \frac{ 1 }{ n } (\frac{ r^3 }{ n^3 } + \frac{ r }{ n }) = \frac{ r^3 }{ n^4 } + \frac{ r }{ n^2 }\]
MrNood
  • MrNood
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..
butterflydreamer
  • butterflydreamer
okayyy

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