Question

Find the formula for the Riemann sum obtained by dividing the interval [a,b] into n equal subintervals and using the right hand endpoint for each ck. then take a limit of these sums as n->approaches infinity to calculate the area under the curve over [a,b].

f(x) = 3x^2+2 over the interval [0,3]

Answer 1

got an hour?

Answer 2

sure

Answer 3

the interval if broken up in to n parts, will each have length \(\frac{3}{n}\) so that is your \(\Delta x\)

Answer 4

can we split the sum up in to two parts? that will make it tons easier

Answer 5

well, I would prefer not. It is up to you.

Answer 6

\[f(x) = 3x^2 +2 \ , \ [0.3] \] for reimann's sums, you have to take the upper and lower limit of the function to get the most approximate area.

Answer 7

lets to \(x^2\) and we can multiply the result by 3 at the end

Answer 8

In general \(\Delta x = (b-a)/n\) and \(x_{iR} =a+\Delta x \times i \)

Answer 9

\[x_0=0,x_1=\frac{3}{n},x_2-\frac{6}{n},x_3=\frac{9}{n}\] and in general
\[x_k=\frac{3k}{n}\]

Answer 10

then
\[f(x_k)=(\frac{3k}{n})^2=\frac{9k^2}{n^2}\]

Answer 11

When using right endpoint, \(x=0\) will not be used. You will have \(3/n \leq x \leq 3\)

Answer 12

makes no difference

Answer 13

\[\ a = 0, \ b = 3\]\[\ \Delta x_i = x_i - x_{i-1} = \frac{b-a}{n}= \frac{3-0}{n} = \frac{2}{n}\]

Answer 14

then
\[\sum f(x_k)\Delta x=\sum (\frac{9k^2}{n^2})\frac{3}{n}=\frac{9}{n^3}\sum k^2\]

Answer 15

ooops typo there
\[\frac{27}{n^3}\sum k^2\]

Answer 16

use the closed form of
\[\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}\]

Answer 17

let me know when I can show you what I got and tell me if it is correct

Answer 18

\[\ x_i = a+i \left( \frac{b-a}{n}\right) =\frac{3i}{n}\]\[\ Area_R=\sum_{i=1}^{n}f(x_i)(\Delta x_i) = \sum_{i=1}^{n}(3x^2+2)\left(\frac{3}{n}\right)=\sum_{i=1}^{n} \left[ 3\left(\frac{3i}{n}\right)^2 +2\right]\left(\frac{3}{n}\right)\]

Answer 19

Sure go ahead.

Answer 20

I'm getting \[\ \sum_{i=1}^{n} \left[ \frac{27i^2}{n^2} +2 \right]\left(\frac{3}{n}\right)\]\[\ \sum_{i=1}^{n} \left[ \frac{81i^2}{n^3}+ \frac{6}{n}\right]\]\[\ \frac{81}{n^3}\sum_{i=1}^{n} i^2 + \frac{6}{n}\sum_{i=1}^{n} 1\]

Answer 21

\[\frac{81}{n^3} \left[ \frac{n(n+1)(2n+1)}{6}\right] + \frac{6}{n}(n)\]

Answer 22

Whatever this reduces down to... one min

Answer 23

\[\ \frac{81(2n^3+3n^2+n)}{6n^3}\] this is what i'm getting.. but this is not fully reduced.

Answer 24

ok tell me if this is right what I have here

Answer 25

Can you tell me how you got \(\ \frac{k}{n}\) as opposed to getting \(\ \frac{3k}{n}\)?

Answer 26

what do you mean? i square k/n and then multiply it by 3

Answer 27

I meant to ask, where did the k/n come from in your equation? Can you tell me how you found it?

Answer 28

@satellite73 found the \(\ x_i\) the same method I had, which is \(\ \large\frac{b-a}{n}\). In your case, \(\ a= 0\) and \(b= 3\)

Answer 29

I got it from the formula in the textbook

Answer 30

\[(\frac{ 27i }{ n })+2*(3/n)\]

Answer 31

Yes, and the 3/n multiplies throughout.

That is how I got the 81i/n^3.

Understand? :o

Answer 32

yes

Answer 33

Awesome \(^_^)/

Answer 34

wait but I am confused about 6/n(n)

Answer 35

\[\ \sum_{i=1}^{n} \left[ \frac{27i^2}{n^2} +2 \right]\left(\frac{3}{n}\right)\]\[\ \sum_{i=1}^{n} \left[ \frac{81i^2}{n^3}+ \frac{6}{n}(1)\right]\]\[\ \frac{81}{n^3}\sum_{i=1}^{n} i^2 + \frac{6}{n}\sum_{i=1}^{n} 1\]\[\frac{81}{n^3} \left[ \frac{n(n+1)(2n+1)}{6}\right] + \frac{6}{n}(n)\]

Answer 36

ok now what happens to the 6/n*n it reduces to 6? then what happens to the 6

Answer 37

Then you get \[\ \ \frac{81(2n^3+3n^2+n)}{6n^3}+6\]

Answer 38

You simplify the fraction then take the limit as n goes to \(\ \infty\)

Answer 39

so when i take the limit don't I just substitute b?

Answer 40

I believe it's just \[\ \lim_{n \rightarrow \infty} \frac{81(2n^3+3n^2+n)}{6n^3}+6\] reduce the fraction, of course.

Answer 41

\[\frac{ 54n^3+81n^2+27n }{ 2n^3 }+6\]

Answer 42

840?

Answer 43

\[\ \lim_{n \rightarrow \infty} \frac{162n^3 + 243n+81n}{6n^3} + 6\]\[\lim_{n \rightarrow \infty} \left(\frac{162}{6} + \frac{243}{6n^2} +\frac{81}{6n^2} +6 \right)\] all the ones that have n's reduce to 0, and you add up all the remaining numbers.

Answer 44

so it is not my simplification above?

Answer 45

your way is right as well, you just simplified \(\ \large \frac{81}{6} = \frac{27}{2}\) which is fine as well

Answer 46

ok so does that simplify to\[30+81n^2+27n\]

Answer 47

So then you'd have \[\ \lim_{n \rightarrow \infty} \left(\frac{ 54n^3+81n^2+27n }{ 2n^3 }+6\right)\]\[\ \lim_{n \rightarrow \infty} \left( 27 + \frac{81}{n^2} + \frac{27}{2n^2} +6\right)\]and taking the limi of this you'd be left with 27+6 \[\ \ \lim_{n \rightarrow \infty} \left( 27 +\cancel{ \frac{81}{n^2}} + \cancel{\frac{27}{2n^2}} +6\right)\]\[\ = 27+6\]

Answer 48

Which is what I got in my previous method.

Answer 49

You need to understand if n is approaching infinity, the fractions with n in their denominator will approach 0.

Answer 50

As the numerator stays constant, the denominator gets exponentially large, therefore the fraction eventually becomes 0.

Answer 51

arent I supposed to sub 3? and my simplification seems different.