# area using riemann sums

• May 30th 2011, 12:42 PM
area using riemann sums
I am learning about using riemann sums to calculate area. I came across this example and it doesn't make much sense to me.

the example is: find the area of the region bounded by y = 16 - x^2 and the x-axis for
1<x<3

where I am getting confused is at the end.....

(30/n)(n)-[8/n^2][1/2(n)(n+1)]-[8/n^3][1/6(n)(n+1)(2n+1)]

= 70/3 - 8/n -4/3n^2

I don't understand where the 70/3 comes from. and I predict that -8/n and -4/3n^2 just coming from simplifying the corresponding -[8/n^2][1/2(n)(n+1)] and -[8/n^3][1/6(n)(n+1)(2n+1)]. thank you.
• May 30th 2011, 12:56 PM
TheEmptySet
Quote:

I am learning about using riemann sums to calculate area. I came across this example and it doesn't make much sense to me.

the example is: find the area of the region bounded by y = 16 - x^2 and the x-axis for
1<x<3

where I am getting confused is at the end.....

(30/n)(n)-[8/n^2][1/2(n)(n+1)]-[8/n^3][1/6(n)(n+1)(2n+1)]

= 70/3 - 8/n -4/3n^2

I don't understand where the 70/3 comes from. and I predict that -8/n and -4/3n^2 just coming from simplifying the corresponding -[8/n^2][1/2(n)(n+1)] and -[8/n^3][1/6(n)(n+1)(2n+1)]. thank you.

To Be honest you notation is hard to read but it looks like you just need to simplify

If you just extract the constant part from each term you will get

$30-\frac{8}{2}-8\cdot \frac{2}{6}=\frac{70}{3}$

Just carefully simply the whole expression.

http://www.wolframalpha.com/input/?i=%2830%2Fn%29%28n%29-[8%2Fn^2][%281%2F2%29*%28n%29%28n%2B1%29]-[8%2Fn^3][%281%2F6%29*%28n%29%28n%2B1%29%282n%2B1%29]

the full result is halfway down the page
• May 30th 2011, 01:13 PM
Deveno
the trick is to "cancel the n's" so you have constant terms and things like: (something)/(a power of n) <-- all these types of terms go to 0 as n gets very large, so only the constant terms survive in the limit.

so (30/n)n = 30, -(8/n^2)(n(n+1)/2) = -(4n^2 + 4n)/n^2 = -4 - 4/n <--- the 4/n term will disappear in the limit.

and -(8/n^3)(n(n+1)(2n+1)/6) = -(8n^3 + 12n^2 + 4n)/(3n^3) = -8/3 - 4/n - 4/(3n^2) <--- the last two terms will disappear in the limit.

this leaves us with: 30 - 4 - 8/3 = (90 - 12 - 8)/3 = 70/3.

(if you carefully add the above, you get 70/3 - 8/n - 4/(3n^2), as your text indicates).
The text passage contains a mathematical question about calculating the area using Riemann sums, and it also provides an explanation and answer to the question.
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First, describe the content structure you found in 1-2 sentences.

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# area using riemann sums

• May 30th 2011, 12:42 PM
area using riemann sums
I am learning about using riemann sums to calculate area. I came across this example and it doesn't make much sense to me.

the example is: find the area of the region bounded by y = 16 - x^2 and the x-axis for
1<x<3

where I am getting confused is at the end.....

(30/n)(n)-[8/n^2][1/2(n)(n+1)]-[8/n^3][1/6(n)(n+1)(2n+1)]

= 70/3 - 8/n -4/3n^2

I don't understand where the 70/3 comes from. and I predict that -8/n and -4/3n^2 just coming from simplifying the corresponding -[8/n^2][1/2(n)(n+1)] and -[8/n^3][1/6(n)(n+1)(2n+1)]. thank you.
• May 30th 2011, 12:56 PM
TheEmptySet
Quote:

I am learning about using riemann sums to calculate area. I came across this example and it doesn't make much sense to me.

the example is: find the area of the region bounded by y = 16 - x^2 and the x-axis for
1<x<3

where I am getting confused is at the end.....

(30/n)(n)-[8/n^2][1/2(n)(n+1)]-[8/n^3][1/6(n)(n+1)(2n+1)]

= 70/3 - 8/n -4/3n^2

I don't understand where the 70/3 comes from. and I predict that -8/n and -4/3n^2 just coming from simplifying the corresponding -[8/n^2][1/2(n)(n+1)] and -[8/n^3][1/6(n)(n+1)(2n+1)]. thank you.

To Be honest you notation is hard to read but it looks like you just need to simplify

If you just extract the constant part from each term you will get

$30-\frac{8}{2}-8\cdot \frac{2}{6}=\frac{70}{3}$

Just carefully simply the whole expression.

http://www.wolframalpha.com/input/?i=%2830%2Fn%29%28n%29-[8%2Fn^2][%281%2F2%29*%28n%29%28n%2B1%29]-[8%2Fn^3][%281%2F6%29*%28n%29%28n%2B1%29%282n%2B1%29]

the full result is halfway down the page
• May 30th 2011, 01:13 PM
Deveno
the trick is to "cancel the n's" so you have constant terms and things like: (something)/(a power of n) <-- all these types of terms go to 0 as n gets very large, so only the constant terms survive in the limit.

so (30/n)n = 30, -(8/n^2)(n(n+1)/2) = -(4n^2 + 4n)/n^2 = -4 - 4/n <--- the 4/n term will disappear in the limit.

and -(8/n^3)(n(n+1)(2n+1)/6) = -(8n^3 + 12n^2 + 4n)/(3n^3) = -8/3 - 4/n - 4/(3n^2) <--- the last two terms will disappear in the limit.

this leaves us with: 30 - 4 - 8/3 = (90 - 12 - 8)/3 = 70/3.

(if you carefully add the above, you get 70/3 - 8/n - 4/(3n^2), as your text indicates).
user
The text passage contains a mathematical question about calculating the area using Riemann sums, and it also provides an explanation and answer to the question.

TYPE: SINGLE_QUESTION_WITH_ANSWER