## 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.

1. anonymous

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

2. anonymous

iv got tons of time ,

3. anonymous

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

4. anonymous

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

5. anonymous

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

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

then comes the annoying computation part

8. anonymous

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

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

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

11. anonymous

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

12. anonymous

yay, got em' right

13. anonymous

okay so know what ?

14. anonymous

now *

15. anonymous

plug and chug ick

16. anonymous

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

17. anonymous

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

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

19. anonymous

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

okay im getting .2578

21. anonymous

seems right :D , thanks for you help