Question

Find formula for upper sum, then take limit of sums to calculate area: f(x)=3x^2 +x over the interval [0,1] sum_{i=1}^{n} i= n(n+1)/2, sum_{i=1}^{n}= n(n+1)(2n+1)/6

Answer 1

notice that on [0,1] f(x) is increasing so the max on any sub interval is the right hand endpoint

Answer 2

I'll try my best...thanks for the reminder :)

Answer 3

lol

Answer 4

ok, so i know the correct format would be:\[\sum_{i=1}^{n} i= n(n+1)/2, \sum_{i=1}^{n}= n(n+1)(2n+1)/6\sum\]

Answer 5

uh oh, ignore that last sigma

Answer 6

you forgot your \(i^2\) again

Answer 7

oh no, ok i'll try again :(

Answer 8

\[\sum_{i=1}^{n} i= n(n+1)/2, \sum_{i=1}^{n}i^2= n(n+1)(2n+1)/6\]

Answer 9

better :)

Answer 10

so what steps would I have to actually do to solve the rest of the problem?

Answer 11

you need to calculate

\[\sum_{i=1}^{n}f(x_i^*)\Delta x\]

Answer 12

in this case you can use

\(x_i^*=\frac{i}{n}\) and \(\Delta x=\frac{1}{n}\)

Answer 13

would it be too much to ask if you showed me how to do this problem step by step? I learn best by example :)

Answer 14

do you understand what I have done so far?

Answer 15

i think so

Answer 16

we then need to compute

\[\sum_{i=1}^{n}f(x_i^*)\Delta x=\sum_{i=1}^{n}f\left(i\frac{1}{n}\right)\frac{1}{n}\]

Answer 17

\[=\sum_{i=1}^{n}\left(3\left(i\frac{1}{n}\right)^2+i\frac{1}{n}\right)\frac{1}{n}\]

Answer 18

ok?

Answer 19

yep, makes sense so far

Answer 20

\[=\sum_{i=1}^{n}\left(3\left(i^2\frac{1}{n^2}\right)+i\frac{1}{n}\right)\frac{1}{n}\]
\[=\sum_{i=1}^{n}\left(3i^2\frac{1}{n^3}+i\frac{1}{n^2}\right)\]

Answer 21

\[=3\frac{1}{n^3}\sum_{i=1}^{n}i^2+\frac{1}{n^2}\sum_{i=1}^{n}i\]

Answer 22

\[=3\frac{1}{n^3}\frac{n(n+1)(2n+1)}{6}+\frac{1}{n^2}\frac{n(n+1)}{2}\]

Answer 23

oh, is that the final answer?

Answer 24

no

Answer 25

its just the formula to solve for the area right?

Answer 26

\[\lim_{n\to\infty}\left[3\frac{1}{n^3}\frac{n(n+1)(2n+1)}{6}+\frac{1}{n^2}\frac{n(n+1)}{2}\right]=1+\frac{1}{2}=\frac{3}{2}\]

Answer 27

so 3/2 would be the answer?

Answer 28

yes

Answer 29

thank you. would you mind helping me with an integral problem?

Answer 30

I have to go...Will probably be back on later. Post your problem in a new thread and I'm sure someone will help

Answer 31

ok, well thanks anyways :)