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anonymous
 one year ago
Use the midpoint rule with n = 4 to approximate the area of the region bounded by y = x^3 and y = x.
anonymous
 one year ago
Use the midpoint rule with n = 4 to approximate the area of the region bounded by y = x^3 and y = x.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is going to take like forever, but it is doable how much time you got?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0iv got tons of time ,

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok it is not that it is that hard to understand, it is just that it is a bunch of computation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay i know how to do this when their is only one boundary line but not when there is two

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0first off they intersect at two place \((0,0)\) and \((1,1)\) so really you are approximating \[\int_0^1(xx^3)dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0here 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
 one year ago
Best ResponseYou've already chosen the best response.0then comes the annoying computation part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0interval 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
 one year ago
Best ResponseYou've already chosen the best response.0oops picture had a typo, idea still the same http://www.wolframalpha.com/input/?i=y%3Dx^3%2Cy%3Dx+domain+0..1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0.125 , .375 , .625 and .875 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0midpoints are \[.125,.375,.625,.875\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(.125.125^3+.375.375^3+.625.625^3+.875.875^3)\times .25\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0probably 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
 one year ago
Best ResponseYou've already chosen the best response.0actually since i wrote it all down, we can use wolfram to get it wolfram reads latex

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0copy 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
 one year ago
Best ResponseYou've already chosen the best response.0okay im getting .2578

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0seems right :D , thanks for you help
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