anonymous
  • anonymous
Use the right endpoint and 6 rectangles to find the approximation of the area of the region between the x axis and between the interval [2, 5] f(x) = 2x^2 - x -1. I'll post my work so far done.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\sum_{i = 1}^{6} f(x)\Delta x\] Delta x is the width of each rectangle which is: \[\frac{ 5 - 2 }{ 6 } = \frac{ 1 }{ 2 }\] The counter for the right endpoint is: \[i (\frac{ 1 }{ 2 }) = \frac{ i }{ 2 }\] \[\sum_{i = 1}^{6} (2x^2 - x - 1)(\frac{ 1 }{ 2 })\] since 1/2 is constant it can be in front \[(\frac{ 1 }{ 2 })\sum_{i = 1}^{6} (2x^2 - x - 1)\] \[(\frac{ 1 }{ 2 }) \sum_{i =1}^{6} 2x^2 - \sum_{i =1}^{6} x - \sum_{i =1}^{6} 1\] \[(\frac{ 1 }{ 2 }) \sum_{i =1}^{6} 2(\frac{ i }{ 2 })^2 - \sum_{i =1}^{6} (\frac{ i }{ 2 }) - \sum_{i =1}^{6} 1\] Is what I did until now correct?
jim_thompson5910
  • jim_thompson5910
Something seems off. Let me think
anonymous
  • anonymous
@jim_thompson5910 I'm supposed to use sigma (summation) notation not the definite integral method I think I should have a parenthesis or bracket around all the sigma notations

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jim_thompson5910
  • jim_thompson5910
ok I see what went wrong you start at a = 2 each term xi is defined as xi = a + i*(delta x) so plug in a = 2 and delta x = 0.5 to get xi = a + i*(delta x) xi = 2 + i*(0.5) xi = 2 + 0.5i so you should have \[\Large \sum_{i = 1}^{6} f(x_i)\Delta x\] \[\Large \sum_{i = 1}^{6} f(2+0.5i)0.5\] \[\Large 0.5\sum_{i = 1}^{6}[2x_i^2 - x_i - 1]\] \[\Large 0.5\sum_{i = 1}^{6}[ 2(2+0.5i)^2-(2+0.5i)-1]\] hopefully that makes sense
anonymous
  • anonymous
@jim_thompson5910 Thank you. Looking it over. Didn't digest all of it yet.
anonymous
  • anonymous
I think I get it now. Will crunch the numbers.
jim_thompson5910
  • jim_thompson5910
tell me what you get
anonymous
  • anonymous
I didn't realize that xi is: a + i(delta x). The problems we did in class a equaled 0. So now I realize xi for this example is: 2 + i(delta x) so I need to replace each x with 2 + (i/2). Please let me know if I'm still missing something
jim_thompson5910
  • jim_thompson5910
well in this case, a = 2 and b = 5 so maybe what your teacher did was shift all of f(x) 2 units to the left, so that a = 0 is the start of your interval
anonymous
  • anonymous
Correct. So I was thrown off with this one.

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