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anonymous

  • one year ago

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x^3 and y = x.

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  1. anonymous
    • one year ago
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    this is going to take like forever, but it is doable how much time you got?

  2. anonymous
    • one year ago
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    iv got tons of time ,

  3. anonymous
    • one year ago
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    ok it is not that it is that hard to understand, it is just that it is a bunch of computation

  4. anonymous
    • one year ago
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    okay i know how to do this when their is only one boundary line but not when there is two

  5. anonymous
    • one year ago
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    first off they intersect at two place \((0,0)\) and \((1,1)\) so really you are approximating \[\int_0^1(x-x^3)dx\]

  6. anonymous
    • one year ago
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    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

  7. anonymous
    • one year ago
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    then comes the annoying computation part

  8. anonymous
    • one year ago
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    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

  9. anonymous
    • one year ago
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    oops picture had a typo, idea still the same http://www.wolframalpha.com/input/?i=y%3Dx^3%2Cy%3Dx+domain+0..1

  10. anonymous
    • one year ago
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    .125 , .375 , .625 and .875 ?

  11. anonymous
    • one year ago
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    midpoints are \[.125,.375,.625,.875\]

  12. anonymous
    • one year ago
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    yay, got em' right

  13. anonymous
    • one year ago
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    okay so know what ?

  14. anonymous
    • one year ago
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    now *

  15. anonymous
    • one year ago
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    plug and chug ick

  16. anonymous
    • one year ago
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    \[(.125-.125^3+.375-.375^3+.625-.625^3+.875-.875^3)\times .25\]

  17. anonymous
    • one year ago
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    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

  18. anonymous
    • one year ago
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    actually since i wrote it all down, we can use wolfram to get it wolfram reads latex

  19. anonymous
    • one year ago
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    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

  20. anonymous
    • one year ago
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    okay im getting .2578

  21. anonymous
    • one year ago
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    seems right :D , thanks for you help

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