Question

need to calculate the upper Reimann sum for g(x) x^4,,,,anyone?? Not sure I am even asking the question correctly....I have 1. Consider g(x) = x^4 on the interval [0,1]...a) calculate the Upper Reimann Sum U=(f,Psub n) where Psub n = {j/n}

Answer 1

What is the interval in question? :-)

Answer 2

...where Psub n = {j/nI j= 0,1,2,33..., n} and we were given the formula for the sum....

Answer 3

....sigma j=1 to n j^4 = n9n+1)(2n+1)(3n^2 + 3n -1) /30

Answer 4

Not sure where I am supposed to start..

Answer 5

bmp....are you there?

Answer 6

I'm having a hard time understanding your question
you want to approximate the integral with of g(x)=x^4 in the interval [0,1] with a reimann sum, correct?

Answer 7

Sorry, got distracted. But @TuringTest can pick up from here :-)

Answer 8

thanks!

Answer 9

how many pieces do you want to break it up into
are you told to use a particular value for n?

Answer 10

yes, think he wants us to do the long way with the Upper Reimann Sum...

Answer 11

ahm...I eidited my question, but don't see it all there.......here goes again...

Answer 12

Psub n ={j/nI j=0, 1, 2, ..., n} does this tell be how many pieces to use, etc?

Answer 13

would this just result in the summation formula he gave ? n(n+1)(2n + 1)(3n^2 + 3n -1) /30?

Answer 14

\[\int_0^1x^4dx\approx\sum_{j=1}^nf(\Delta x^*)\Delta x\]it doewsn't say how many pieces to use, so I suppose they want n to go to infinity, which is the definition of the integral
and yeah I think it will just wind up being that formula, but you need to make sure to keep \(\Delta x\) in there

Answer 15

yes, I see that he says j=1 so height is a function of x^4 and each rect. has width 1/n

Answer 16

ok, I did this same thing then a couple of weeks ago for x^3...but can you continue to help me make sure I'm correct?? :-)

Answer 17

\[\int_0^1x^4dx=\lim_{n\to\infty}\sum_{j=1}^nf(\Delta x_j^*)\Delta x=\lim_{n\to\infty}\sum_{j=1}^n\frac{j^4}{n^4}\cdot\frac1n=\lim_{n\to\infty}\sum_{j=1}^n\frac{j^4}{n^6}\]and now we get to use that hideous formula of yours for the sum of j^4

Answer 18

typo*\[\int_0^1x^4dx=\lim_{n\to\infty}\sum_{j=1}^nf(\Delta x_j^*)\Delta x=\lim_{n\to\infty}\sum_{j=1}^n\frac{j^4}{n^4}\cdot\frac1n=\lim_{n\to\infty}\sum_{j=1}^n\frac{j^4}{n^5}\]

Answer 19

last time I subbed in i/n^3 because ?? my summation ended up like 1/4 summation j=1 to n j^3??

Answer 20

\[\lim_{n\to\infty}\frac1{n^5}\sum_{j=1}^ni^4=\lim_{n\to\infty}\frac1{n^5}{n(n+1)(2n+1)(3n^2 + 3n -1)\over30}\]simplify all that nastiness on the right

Answer 21

where did you get cubed from?
I thought we were doing ^4th power

Answer 22

ok, it's a 1/n^5 outside the summation and NOT just the fraction 1/5? (the cubed is from the problem I did a few weeks ago that muxt have been simial...

Answer 23

yeah we can take the whole n outside the summation sign because it is constant with respect to the sum
it still has to stay inside the limit sign though

Answer 24

in this case the n term is 1/n^5
now we only need to sum j, for which we can use your formula

Answer 25

ok, in the earlier problem I had a 1/4 out front of the summation but started simpl,ifying with a 1/n^4

Answer 26

well a lot of stuff should cancel
now comes the awful algebra part where you need to expand the numerator

Answer 27

ok, I must have just left the n in n^4 out and the prof didn't catch it!

Answer 28

yes, it did in the last formula he gave us for n^3...simpl;ied down with the 1/n^4 that was out front

Answer 29

just loot at the highest power of n in the numerator...\[n\cdot n\cdot 2n\cdot3n^2\]

Answer 30

*look

Answer 31

let me see what the prof gave us again...maybe I mistyped...

Answer 32

n(n+1)(2n+1)(3n^2+3n_1) all over 30

Answer 33

@Zarkon oh, so we don't have to expand after all :)
as usual: Hail Zarkon!

Answer 34

never work harder than you have too ;)

Answer 35

of course.
work smarter, not harder

Answer 36

yes...we can cancel some things.....the prof didn't expand the last portion for the previous prob either, I now remember....

Answer 37

oh, thanks guys!! think I have it...now I just have to evealuate a few values in....THEN use Leibnitz'value twice to integrate HIS way.... lvoely! I may be getting back to you later!!!! :-):-)

Answer 38

welcome, good luck :)