anonymous
  • anonymous
Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x^3 and y = x.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
this is going to take like forever, but it is doable how much time you got?
anonymous
  • anonymous
iv got tons of time ,
anonymous
  • anonymous
ok it is not that it is that hard to understand, it is just that it is a bunch of computation

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anonymous
  • anonymous
okay i know how to do this when their is only one boundary line but not when there is two
anonymous
  • anonymous
first off they intersect at two place \((0,0)\) and \((1,1)\) so really you are approximating \[\int_0^1(x-x^3)dx\]
anonymous
  • anonymous
here is a nice picture to show what it looks like http://www.wolframalpha.com/input/?i=y%3Dx^2%2Cy%3Dx+domain+0..1
anonymous
  • anonymous
then comes the annoying computation part
anonymous
  • anonymous
interval has length 1, divide in to four parts, length of each is \(0.25\) endpoints are \[0,.25,.5,.75,1\] then you need the midpoints
anonymous
  • anonymous
oops picture had a typo, idea still the same http://www.wolframalpha.com/input/?i=y%3Dx^3%2Cy%3Dx+domain+0..1
anonymous
  • anonymous
.125 , .375 , .625 and .875 ?
anonymous
  • anonymous
midpoints are \[.125,.375,.625,.875\]
anonymous
  • anonymous
yay, got em' right
anonymous
  • anonymous
okay so know what ?
anonymous
  • anonymous
now *
anonymous
  • anonymous
plug and chug ick
anonymous
  • anonymous
\[(.125-.125^3+.375-.375^3+.625-.625^3+.875-.875^3)\times .25\]
anonymous
  • anonymous
probably a more artful way to arrange this but that literally what you need to compute i did factor out the \(.25\) and put it at the end, since that is the length of each interval
anonymous
  • anonymous
actually since i wrote it all down, we can use wolfram to get it wolfram reads latex
anonymous
  • anonymous
copy and paste this might work check for typos http://www.wolframalpha.com/input/?i=%28.125-.125^3%2B.375-.375^3%2B.625-.625^3%2B.875-.875^3%29\times+.25
anonymous
  • anonymous
okay im getting .2578
anonymous
  • anonymous
seems right :D , thanks for you help

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