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

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

##### 1 Attachment

- amistre64

trap rule eh ...

- anonymous

Yeah, that one...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

- amistre64

do it for each section

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- amistre64

b-a looks to be 16

- amistre64

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

- anonymous

Possible choices are:
12
10.5
6.9
8.5
7.4
12.5

- anonymous

Kinda accurate for an estimate

- amistre64

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

- amistre64

each blip might be 1/4

- 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

- anonymous

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

- amistre64

whered you get this program at; biglots?

- anonymous

what?

- amistre64

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

- anonymous

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

- amistre64

:) i assumed you were working this from a program

- anonymous

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

- 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

- 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

- amistre64

do we have numbers to go with yours?

- anonymous

nope...

- amistre64

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

- anonymous

but then one box is over 16

- amistre64

4 per box would be 4 tics wide

- 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

- amistre64

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

- anonymous

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

- anonymous

i need help on http://openstudy.com/study#/updates/4f3084efe4b0fc09381eed4a asap pretty please!!!!

- anonymous

Hey, I called him first :P

Looking for something else?

Not the answer you are looking for? Search for more explanations.