## anonymous 4 years ago I'm lost, Estimate the area of the shaded region in the graph by using the Trapezoidal Rule with n=4.

1. anonymous

2. amistre64

trap rule eh ...

3. anonymous

Yeah, that one...

4. amistre64

the area of a trap is average of the height * width f(1)+f(2) ------- * 1 if the width is 1 that is 2

5. amistre64

do it for each section

6. amistre64

$\frac{f(0)+f(1)}{2}*\frac{b-a}{4}+\frac{f(1)+f(2)}{2}*\frac{b-a}{4}+\frac{f(2)+f(3)}{4}*\frac{b-a}{4}+\frac{f(3)+f(4)}{4}*\frac{b-a}{4}$ $\frac{b-a}{2*4}(f(0)+f(1)+f(1)+f(2)+f(2)+f(3)+f(3)+f(4))$ $\frac{b-a}{2*4}(f(0)+2f(1)+2f(2)+2f(3)+f(4))$

7. anonymous

But how? They don't even label the graph,

8. amistre64

0, 6, 9.5 .. just count and estimate the best you can with the intersections

9. amistre64

0, 6, 9.5 ,4, 2 0 + 12 + 19 + 8 + 2 41/8 = 5*(b-a)

10. amistre64

b-a looks to be 16

11. amistre64

41/8 * 16 = 41*2 = 82 or so? does that fit an answer?

12. anonymous

Possible choices are: 12 10.5 6.9 8.5 7.4 12.5

13. anonymous

Kinda accurate for an estimate

14. amistre64

i counted each little blip as 4 so im a little off

15. amistre64

each blip might be 1/4

16. amistre64

0+2*1.5+2*2.5+2*1+.5 3+5+2+.5 = 10.5 4 ----- 42.0/(2*4) 42/8 = 5.somehting as far as I can tell

17. anonymous

Which still isn't close to any of their answers, hmm

18. amistre64

whered you get this program at; biglots?

19. anonymous

what?

20. amistre64

biglots is a store that has a rep of subpar merchandise :)

21. anonymous

Yeah, lol I know that, I was wondering what you meant by "program"

22. amistre64

:) i assumed you were working this from a program

23. anonymous

Ah, no I was trying to do it by good ole' paper :P

24. amistre64

well, the concept I believe is sound: b-a/n = width and the average of yor heights is the height b*h = area for each section

25. anonymous

I found one similar online where I believe they got 11.5, http://www.csuchico.edu/~rford/MATH7BWEB/Math7BQuizzes/Math%207B%20Exam%202%20Review

26. amistre64

do we have numbers to go with yours?

27. anonymous

nope...

28. amistre64

then all we can do is assume a basic interval if 1 per "tic" that i know of

29. anonymous

but then one box is over 16

30. amistre64

4 per box would be 4 tics wide

31. amistre64

0 to 6 has an avg of 3 6 to 10 has an avg of 8 10 to 4 has an avg of 7 4 to 2 has an avg of 3 4(3+3+8+7) 4(21) = 84 total

32. amistre64

other than that, its just a guessing game to figure our what it is they are scaling hteir tic marks as

33. anonymous

hmm, I do believe that I got points taken off in Physics for not labeling my graphs like that, lol

34. anonymous