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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

1 Attachment
trap rule eh ...
Yeah, that one...

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

the area of a trap is average of the height * width f(1)+f(2) ------- * 1 if the width is 1 that is 2
do it for each section
\[\frac{f(0)+f(1)}{2}*\frac{b-a}{4}+\frac{f(1)+f(2)}{2}*\frac{b-a}{4}+\frac{f(2)+f(3)}{4}*\frac{b-a}{4}+\frac{f(3)+f(4)}{4}*\frac{b-a}{4}\] \[\frac{b-a}{2*4}(f(0)+f(1)+f(1)+f(2)+f(2)+f(3)+f(3)+f(4))\] \[\frac{b-a}{2*4}(f(0)+2f(1)+2f(2)+2f(3)+f(4))\]
But how? They don't even label the graph,
0, 6, 9.5 .. just count and estimate the best you can with the intersections
0, 6, 9.5 ,4, 2 0 + 12 + 19 + 8 + 2 41/8 = 5*(b-a)
b-a looks to be 16
41/8 * 16 = 41*2 = 82 or so? does that fit an answer?
Possible choices are: 12 10.5 6.9 8.5 7.4 12.5
Kinda accurate for an estimate
i counted each little blip as 4 so im a little off
each blip might be 1/4
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
Which still isn't close to any of their answers, hmm
whered you get this program at; biglots?
what?
biglots is a store that has a rep of subpar merchandise :)
Yeah, lol I know that, I was wondering what you meant by "program"
:) i assumed you were working this from a program
Ah, no I was trying to do it by good ole' paper :P
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
I found one similar online where I believe they got 11.5, http://www.csuchico.edu/~rford/MATH7BWEB/Math7BQuizzes/Math%207B%20Exam%202%20Review
do we have numbers to go with yours?
nope...
then all we can do is assume a basic interval if 1 per "tic" that i know of
but then one box is over 16
4 per box would be 4 tics wide
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
other than that, its just a guessing game to figure our what it is they are scaling hteir tic marks as
hmm, I do believe that I got points taken off in Physics for not labeling my graphs like that, lol
Hey, I called him first :P

Not the answer you are looking for?

Search for more explanations.

Ask your own question