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

Question

anonymous · Answer

this is going to take like forever, but it is doable
how much time you got?

anonymous · Answer

iv got tons of time ,

anonymous · Answer

ok 
it is not that it is that hard to understand, it is just that it is a bunch of computation

anonymous · Answer

okay i know how to do this when their is only one boundary line but not when there is two

anonymous · Answer

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 · Answer

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 · Answer

then comes the annoying computation part

anonymous · Answer

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 · Answer

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

anonymous · Answer

.125 , .375 , .625 and .875 ?

anonymous · Answer

midpoints are $$.125,.375,.625,.875$$

anonymous · Answer

yay, got em' right

anonymous · Answer

okay so know what ?

anonymous · Answer

now *

anonymous · Answer

plug and chug ick

anonymous · Answer

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

anonymous · Answer

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 · Answer

actually since i wrote it all down, we can use wolfram to get it
wolfram reads latex

anonymous · Answer

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 imes+.25

anonymous · Answer

okay im getting .2578

anonymous · Answer

seems right :D , thanks for you help