Question

Definitely could use somebody to help check my work, see why these answers are not being taken (homework, only a few points but I'm trying to understand the concepts). This is a continuation of this topic: http://openstudy.com/study#/updates/4ff2de95e4b03c0c488b3914

So for this problem I'm asked to find the right, left, midpoint, and trapezoid approximations of a function which is show on a graph, but isn't a function we are given. It's the graph and only the graph. I will post the image and some formulas below in a moment.

Ty in advance to any whom are brave & wise enough to attempt

Answer 1

So here's the graph, which @experimentX has already seen

Answer 2

The Trapezoid rule is (feel free to correct me if I'm incorrect):
\[T_n= \frac{\triangle x}{2} (f(x_1)+2f(x_2)+...2f(x_{n-1})+f(x_n)\]

Answer 3

I used 0.5 as my sample width before, \(\triangle x\)

Answer 4

well ... this is right. calculate all those f(x1)'s

Answer 5

The 0.5 is from using 8 samples, each is a half.
\[\triangle x = \frac{b-a}{n} = \frac{4-0}{8}=\frac{1}{2}\]

Answer 6

2.5 + 2(5.3 + 7.8 + 10.3 + 12.5 + 14 + 16.5) + 17.5 * times * 1/4

Answer 7

roughly ... since we do not know the exact scale ... it should be around that value.

Answer 8

Left endpoint?
f(0)+f(0.5)+f(1)+f(1.5)+f(2)+f(2.5)+f(3)+f(3.5)+f(4)+f(4.5)

Answer 9

Right endpoint?
f(0.5)+f(1)+f(1.5)+f(2)+f(2.5)+f(3)+f(3.5)+f(4)+f(4.5)+f(5)

Answer 10

that's upper sum ...
2.5 + 2(5.3 + 7.8 + 10.3 + 12.5 + 14 + 16.5) + 17.5 * times * 1/4 = 45.7 (from trapezoidal rule)

Answer 11

do you have a scale??

Answer 12

Midpoints?
f(0.25)+f(0.75)+f(1.25)+f(1.75)+f(2.25)+f(2.75)+f(3.25)+f(3.75)+f(4.25)+f(4.75)

Answer 13

Erm, no scale, just the image shown with the x & y axes as labled. Unless I'm misunderstanding your scale.

Answer 14

yep that's right ... but this is quite difficult to do. we usually approximate measurements since function is not given.

isn't there some kind of scaling app??

Answer 15

Umm... Not given, it's literally just an image embedded on the form field

Answer 16

Yeah this one is kind of a pain lol, believe me

Answer 17

well ... best of luck with that. i did for trapezoidal rule. you can do similar for Left and right sums.

Answer 18

for midpoint it is going you have to take complete different measurements.

Answer 19

man this is a pain!

Answer 20

At least the part of which is largest to smaller I have good:
Left < Trapezoid < actual integral < Midpoint < Right

So I get the overall concept here, maybe just have to accept the losses.

Answer 21

And the numbers I have posted (which are counted as wrong) seem to fit that. Stupid computers and their 100% accuracy :P

Answer 22

updated
2.5 + 2 (5.3 + 7.8 + 10.3 + 12.5 + 14 + 16.5 + 17.1) + 17.5 * 1/4 for trapezoidal rule

Answer 23

((5.3 + 7.8 + 10.3 + 12.5 + 14 + 16.5 + 17.1) + 17.5 )* 1/2 for right sums

Answer 24

2.5 + 2 (5.3 + 7.8 + 10.3 + 12.5 + 14 + 16.5 + 17.1) * 1/2 for left sums.

Answer 25

Am I correct in the above for the function input values?

Answer 26

yup ... since do you not know functional values ... that's pretty useless.

Answer 27

Alright, well good I got the concept down at least then :-)

Answer 28

Your left sum is 86. Your right is 50.5 Left endpoint should be the smallest of all

Answer 29

the value i got seems pretty close
41.95 for left
50.5 for right
46.75 for trapezoidal ...
seems i took pretty high value.
this is the most boring part of calculus.
best of luck with mid point work

Answer 30

Hmm ... i guess you forgot to divide it by 2 ... check that expression ... that is not quite correct. i just gave you the data.

Answer 31

Hey good news it took your version of T\(_n\)! :-D The rest of it it was like meh, nope. Even with the corrections to make sure the 5-part inequality still held true

Answer 32

For garnering one more point of out of this mess, a well deserved medal for you ;D

Answer 33

well i hope you will do for midpoints.

Answer 34

@experimentX, finally managed to get the right and left endpoints, just so you know :D

Answer 35

well ... sure it took quite a time.

Answer 36

I went and did the intro to surface integrals first ^_^ and ate some lunch

Answer 37

OH .. well congrats man!! i gotta go to study too.