Question

@freckles

Answer 1

couold you explain this as simple as possible?? im so confused :/

Answer 2

@satellite73 could you try to explain?

Answer 3

i could try

Answer 4

:)

Answer 5

\\[y=x^3\] and \[y=x\] intersect at \((0,0)\) and \((1,1)\)

Answer 6

on the interval \([0,1]\) you know \(x^3<x\) so the integral will be
\[\int_0^1(x-x^3)dx \]

Answer 7

this is an easy enough integral to compute without using the midpoint rule with n =4 but if that is what we have to do, we can do it
if you divide the unit interval \([0,1]\) in to four equal parts, they would be at \[0,.25,.5,.75,1\]

Answer 8

you with me so far?

Answer 9

yess sorry i was writing it down so i understand it

Answer 10

ok back, i was stuck for a while with the spinning wheel

Answer 11

now find the midpoint of those intervals

Answer 12

hahah its okk

Answer 13

they are
\[.125,.375,.625,.875\]

Answer 14

hope it is clear how i got that, just divided .25 by 2, got .125, then added .25 repeatedly

Answer 15

yes! i get that

Answer 16

the length of each of those intervals is \(.25\) so your unenviable job it so evaluate \(x-x^3\) at those points, add them up and multiply by \(.25\)

Answer 17

\[(.125-(.125)^3+.375-(.375)^3+.625-(.625)^3+.875-(.875)^3)\times .25\]

Answer 18

i believe you get this
http://www.wolframalpha.com/input/?i=%28.125-%28.125%29^3%2B.375-%28.375%29^3%2B.625-%28.625%29^3%2B.875-%28.875%29^3%29*.25

Answer 19

thank you so much!

Answer 20

yw