Question

See attachment, please help

Answer 1

The Reimann sum on the interval [a,b] is given by\[\sum_{i=0}^nf(a+i\Delta x)\Delta x\]where \(\Delta x=\frac{b-a}n\)

Answer 2

what is \(\Delta x\) in your case?

Answer 3

...n is the number of rectangles btw

Answer 4

I just don't know how to use that f(x)=x^2/10 and the interval...[2,6], i mean i don't understand what's f(x) for

Answer 5

f(x) is the curve you are estimating the area under
answer my questions an I will walk you though the process

Answer 6

6-2/8 which is -1/2 for ∆x right?

Answer 7

why - ?
\(\Delta x=1/2\)

Answer 8

oh yes gotcha

Answer 9

so now after subbing in for a, b, n, and \(\Delta x\) we got\[\sum_{i=0}^8f(2+\frac i2)\frac12\]so far so good?

Answer 10

yes, ok so i see where the 8 and the 2 came from and also the 1/2

Answer 11

we are doing the Ln right because we're using 2...?

Answer 12

we are doing the left because we are starting at i=0, which is like starting at x=2
if we wanted the right endpoints we would use i=1, and we would start at x=2.5
make sense?

Answer 13

O.K, yes :)

Answer 14

ok, so now we are going to sub in for f(x)

Answer 15

since our \(x^*=2+\frac i2\) we now have\[\sum_{i=0}^8\frac{(2+\frac i2)^2}{10}\cdot\frac12\]still with me?

Answer 16

yes, i see how you have these substitutions

Answer 17

by using f(x)=X^2/10 :)

Answer 18

you got it :)
now did you know we can pull out the constants in front of the summation?

Answer 19

Yup

Answer 20

\[\frac1{20}\sum_{i=0}^8(2+\frac i2)^2\]now we just expand and sum it up

Answer 21

it is helpful to know that\[\sum_{i=1}^ni=\frac{n(n+1)}2\]and\[\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6\]have you seen these formulas before?

Answer 22

Yeah, just learned about both 2 days ago

Answer 23

great, then this should be a breeze :)

Answer 24

\[\frac1{20}\sum_{i=0}^8(2+\frac i2)^2\]\[\frac1{20}\sum_{i=0}^84+2i+\frac{i^2}4\]can you sum it from here?

Answer 25

hmm so we left with 1/5 that Reimann sign 4 +2i + i^2 ?

Answer 26

why 1/5 ?

Answer 27

trying to pull a 4 out?
can't do that I'm afraid

Answer 28

oh ok is it just 2i + i^2?

Answer 29

and 1/20 infront

Answer 30

where did the 4 go?
what is\[\sum_{i=0}^84\]?

Answer 31

do we cancelled it with the other 4?

Answer 32

no, we so far have\[\frac1{20}\left(\sum_{i=0}^84+\sum_{i=0}^82i+\sum_{i=0}^8\frac{i^2}4\right)\]\[=\frac1{20}\left(4\sum_{i=0}^81+2\sum_{i=0}^8i+\frac14\sum_{i=0}^8i^2\right)\]now you need to do each sum

Answer 33

so what is\[\sum_{i=0}^81\]?

Answer 34

i completely lost you there, sorry :)

Answer 35

you are summing 1 from index 0 to index 8
0th term:1
1st term:1
2nd term:1
etc.

Answer 36

so that's 1+1+1+1+1+1+1+1+1=9
get it?

Answer 37

oh ok, yes!

Answer 38

cool, onward ho...

Answer 39

\[\frac1{20}\left(4\sum_{i=0}^81+2\sum_{i=0}^8i+\frac14\sum_{i=0}^8i^2\right)\]\[=\frac1{20}\left(4(9)+2\sum_{i=0}^8i+\frac14\sum_{i=0}^8i^2\right)\]now what about\[\sum_{i=0}^8i\]what is that equal to?

Answer 40

hmm in this case i=2 right?

Answer 41

oh crap I screwed up a little...
the sum should go to 7 because the last left endpoint is 5.5 not 6

Answer 42

so\[\sum_{i=1}^71=8\]and all the sums should be from 0 to 7
if it were the right endpoints it would be from 1 to 8

Answer 43

oh ok, no worries hehe

Answer 44

2+(0.5)7=5.5 which is the last endpoint
I hope I didn't confuse you, a minor adjustment...

Answer 45

\[\frac1{20}\left(4\sum_{i=0}^71+2\sum_{i=0}^7i+\frac14\sum_{i=0}^7i^2\right)\]\[=\frac1{20}\left(4(8)+2\sum_{i=0}^7i+\frac14\sum_{i=0}^7i^2\right)\]and i does not equal anything, it is an index that changes with the terms of the summation

Answer 46

\[\sum_{i=0}^7i=0+1+2+3+4+5+6+7\]see, i ticks off the natural numvers

Answer 47

numbers*

Answer 48

also note that\[\sum_{i=0}^7i=0+1+2+3+4+5+6+7=28=\frac{7(7+1)}2\]so this is where our formula you learned 2 days ago comes in

Answer 49

yeah from the formular n(n+1)/2

Answer 50

right, so I hope you understand how \[\sum_ni\]works now

Answer 51

so to continue, we now have...

Answer 52

\[\frac1{20}\left(4\sum_{i=0}^71+2\sum_{i=0}^7i+\frac14\sum_{i=0}^7i^2\right)\]\[=\frac1{20}\left(4(8)+2\frac{7\cdot8}2+\frac14\sum_{i=0}^7i^2\right)\]and the last piece of the puzzle is to find\[\sum_{i=0}^8i^2\]which is what?

Answer 53

sorry that should be sum up to 7, not 8

Answer 54

oh ok..

Answer 55

\[\sum_{i=0}^7i^2=?\]

Answer 56

we use 0^2+1^2+2^2+3^2+4^2+5^2+6^2+7^2 ?

Answer 57

yes, or you could use the formula you learned 2 days ago ;)
(note that starting at zero does not matter because the first term is zero here)\[\sum_{i=0}^ni=\sum_{i=1}^ni\]sam goes for the sum over i^2 or any time the first term is zero

Answer 58

same*

Answer 59

what was your formula for\[\sum_{i=1}^ni^2\]?

Answer 60

n(n+1)(2n+1)6

Answer 61

/6

Answer 62

divide by 6 *

Answer 63

hehe yes

Answer 64

and what is n in your case?

Answer 65

140?

Answer 66

/6

Answer 67

yes that is the sum
I was asking about what n is, which is 7, but you got the right sum anyway
...why did you write /6 again?

Answer 68

oh ok nevermind i was using the second formular which is wrong

Answer 69

so now we're adding the sum together ?

Answer 70

\[\sum_{i=0}^7i^2=\frac{7(8)(15)}6=140\]

Answer 71

so yeah, now we just add it all up

Answer 72

140+28+8?

Answer 73

opps nope

Answer 74

careful\[\frac1{20}\left(4\sum_{i=0}^71+2\sum_{i=0}^7i+\frac14\sum_{i=0}^7i^2\right)\]\[=\frac1{20}\left(4(8)+2(28)+\frac14(140)\right)\]

Answer 75

ok ok i see and then we plug them all back into our break up stuff after we did the sums

Answer 76

yeah, always gotta remember where we are in the formula and what it's supposed to look like

Answer 77

so therefore the answer is 463/5 right ?

Answer 78

?
1/20(32+56+35)=123/20
don't fail on the arithmetic at the end and get the wrong answer, that's just plain frustrating :P

Answer 79

AH!! 6.15 haha

Answer 80

yeah, that's what I get :)

Answer 81

Manh that was one long working out for a single question!!

Answer 82

I really appreciate your time and patient

Answer 83

yeah these things can be a pain, but it will go a lot faster with a little practice and knowing what forms to expect

Answer 84

Any suggestions to better my understanding of how become better like you?

Answer 85

or at least like you, i know you're way way beyond

Answer 86

thanks thanks much! I just violate the open study rule by becoming your fun! :)

Answer 87

I am mostly self-taught, but here is a really good set of notes
http://tutorial.math.lamar.edu/
and also MIT OCW has great, but quite difficult classes

Answer 88

http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/index.htm

Answer 89

and you're very welcome :)
good luck, I have to go hunt down some people causing trouble on this site
later!

Answer 90

thanks much!