RIEMANN SUM

[Ryan Rigdon]

i forgot i had a scanner. so i scanned my problem that i am having trouble and i am attaching it. will be doing this more often now. its easier here is my problem.

 

[galactus]

I am not fully understanding that inequality, but I would assume you are finding the area using a Riemann sum?

Integrate -x^2+12x from 1 to 13.

\(\displaystyle {\Delta}x=dx=\frac{b-a}{n}=\frac{13-1}{n}=\frac{12}{n}\)

Use the right endpoint method:

\(\displaystyle x_{k}=a+k{\Delta}x=1+\frac{12k}{n}\)

\(\displaystyle f(x_{k}){\Delta}x=f(x)dx=\left[-(1+\frac{12k}{n})^{2}+12(1+\frac{12k}{n})\right]\frac{12}{n}\)

Perform the hideous algebra and take the sums:

\(\displaystyle =\frac{-1728}{n^{3}}\sum_{k=1}^{\infty}k^{2}+\frac{1440}{n^{2}}\sum_{k=1}^{\infty}k+\frac{132}{n}\sum_{k=1}^{\infty}1\)

Now, there are some identities to know. They are probably in your book. The sum of \(\displaystyle \sum_{k=1}^{\infty}k^{2}=\frac{n(n+1)(2n+1)}{6}\) and \(\displaystyle \sum_{k=1}^{\infty}k=\frac{n(n+1)}{2}\)

Replace the k's in the above summations with those and it is entirely in terms of n.

Doing so, leads to some more messy algebra stuff. Take the limit and we have the area under the curve:

\(\displaystyle \lim_{n\to {\infty}}\left[\frac{-144}{n}-\frac{288}{n^{2}}+276\right]\)

It is easy to see what that limit is because n ---> infinity.

It gives the same answer as \(\displaystyle \int_{1}^{13}(-x^{2}+12x)dx\), only we done it the long way.

See, what this does is illustrate how an integral works by adding up the area of the infinite number of rectangles under the curve. As the number of rectangles becomes unbounded, we get the area. f(x) is the height of the nth rectangle, dx is the width. We add an infinite number of them up. Hence the sums and limit.

 

[Ryan Rigdon]

the final answer was 198.625.

i tried to figure out how they got their answer by

taking the area from 1 to 7.5 and then adding the area from 7.5 to 18 i came up with 276. Hmmm!!

i wonder how they got their answer of 198.625.

any ideas?

 

[galactus]

It's an approximation method. As I said, I am unsure of what they mean with 1<6.5<7.5<13.

Now you mention 18. I used 13.

I showed the Riemann sum method with n rectangles. Thus, as \(\displaystyle n\to \infty\), we get the area under the curve.

I am not sure what they are getting at. It would appear they are just using a couple of rectangles for a rough approximation.

Add up the area of the rectangles they are using.

 

[Ryan Rigdon]

that was a typo on 18 sorry i figured out how they got their answer. doing the quiz again. once i answer this question again i will post my findings. cant believe it was an easy as pie just didnt see it. thanx for the help thus far.

 

[Ryan Rigdon]

here is another problem just like it. and i got it right just thought i would share my findings with you.

 

[galactus]

I see what they are wanting you to do. Here is a graph of varying partitions. This way you can see as the more rectangles we add, the closer to the actual area under the curve.

 

[BigGlenntheHeavy]

\(\displaystyle Simpson \ Anyone?\)

\(\displaystyle An \ (if \ he \ was \ alive) \ would \ be \ the \ first \ to \ say \ "don't \ ask \ me, \ I \ found \ it \ in \ a \ book."\)