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

Mathematics
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this is going to take like forever, but it is doable how much time you got?
iv got tons of time ,
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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okay i know how to do this when their is only one boundary line but not when there is two
first off they intersect at two place \((0,0)\) and \((1,1)\) so really you are approximating \[\int_0^1(x-x^3)dx\]
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
then comes the annoying computation part
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
oops picture had a typo, idea still the same http://www.wolframalpha.com/input/?i=y%3Dx^3%2Cy%3Dx+domain+0..1
.125 , .375 , .625 and .875 ?
midpoints are \[.125,.375,.625,.875\]
yay, got em' right
okay so know what ?
now *
plug and chug ick
\[(.125-.125^3+.375-.375^3+.625-.625^3+.875-.875^3)\times .25\]
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
actually since i wrote it all down, we can use wolfram to get it wolfram reads latex
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
okay im getting .2578
seems right :D , thanks for you help

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