Compute lower and upper sum for Riemann integral

  • Thread starter rayman123
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  • #1
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Homework Statement


let [tex]f(x)=x^2[/tex] Calculate upper sum and lower sum on the interval [tex][-2,2][/tex] when n=2


The Attempt at a Solution


since n=2 I divide the interval into
[tex][-2,0]\cup[0,2][/tex]

then on the interval [tex][-2,0][/tex] the function [tex]f(x)=x^2[/tex] has the highest valute at [tex]x=-2, f(-2)=4=M_{0}[/tex] and the lowest value is at [tex] x=0, f(0)=0=m_{0}[/tex]

on the interval the situation is the same [tex]x=0, f(0)=0=m_{1}
[/tex](again the lowest value) ,and at [tex] x=2, f(2)=4=M_{1}[/tex](the highest value)

thus upper sum will be
[tex]S_{n}=M_{0}\cdot \Delta x+M_{1}\cdot \Delta x[/tex] where [tex]\Delta x=2[/tex]
[tex]S_{n}=4\cdot 2+4\cdot 2=16[/tex]
lower sum
[tex]s_{n}=0\cdot 2+0\cdot 2=0[/tex]
and here I am a bit confused cause in my homework it says'' if you calculated correctly then the difference between lower sum and upper sum should be 16'' well here it would not work....where do I make mistake?

any help appreciated
 

Answers and Replies

  • #2
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[tex]4\cdot 2+4\cdot 2=8[/tex]

This does not look correct...
 
  • #3
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This does not look correct...
it should be 16, it was a typo but this still does not show me the error
 
  • #4
22,129
3,298
Isn't the difference between the lower sum and the upper sum 16 now??
 
  • #5
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the difference is [tex] s_{n}-S_{n}=-16[/tex]
 
  • #6
22,129
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Yeah, of course since [itex]S_n[/itex] is always larger than [itex]s_n[/itex].

But with difference, they don't literally mean [itex]s_n-S_n[/itex] here. Rather, they mean something like [itex]|s_n-S_n|[/itex].
 
  • #7
152
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thank you :) now I see
 

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