Riemann Integral

[Jason76]

Note: I know there is an easier way to do this (the traditional integral way), but CLEP Calculus demands me to know this:

\(\displaystyle \int^{8}_{4} x dx \)

Given lower bound \(\displaystyle a (4)\) and upper bound \(\displaystyle b (8)\)

Formula: \(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{(b - a)}{n} * f(\dfrac{a + i(b - a)}{n})\)

\(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{(8 - 4)}{n} * f([4 + \dfrac{4 + i(8 - 4)}{n})]\)

\(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{4}{n} * f([4 + \dfrac{4 + 4i)}{n}])\) How did this line become the next line?

\(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n} 16 \dfrac{n + i}{n^{2}} \)

 

[JeffM]

Note: I know there is an easier way to do this (the traditional integral way), but CLEP Calculus demands me to know this:

Given lower bound \(\displaystyle a (4)\) and upper bound \(\displaystyle b (8)\)

Formula: \(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{(b - a)}{n} * \dfrac{a + i(b - a)}{n}\)

\(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{(8 - 4)}{n} * [4 + \dfrac{4 + i(8 - 4)}{n}]\)

\(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{4}{n} * [4 + \dfrac{4 + 4i)}{n}]\) How did this line become the next line?

\(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n} 16 \dfrac{n + i}{n^{2}} \)

What is the exact statement of the problem?

Given a = 4 and b = 8 and assuming that the proper starting place is indeed

\(\displaystyle \displaystyle \lim_{n \rightarrow \infty}\left(\sum_{i=1}^n\left\{\dfrac{b - a}{n} * \dfrac{a + i(b - a)}{n}\right\}\right) =\)

\(\displaystyle \displaystyle \lim_{n \rightarrow \infty}\left(\sum_{i=1}^n\left\{\dfrac{8 - 4}{n} * \dfrac{4 + i(8 - 4)}{n}\right\}\right) =\)

\(\displaystyle \displaystyle \lim_{n \rightarrow \infty}\left(\sum_{i=1}^n\left\{\dfrac{4}{n} * \dfrac{4 + 4i}{n}\right\}\right) =\)

\(\displaystyle \displaystyle \lim_{n \rightarrow \infty}\left(\sum_{i=1}^n\left\{4 * \dfrac{1}{n} * 4 * \dfrac{1 + i}{n}\right\}\right) =\)

\(\displaystyle \displaystyle 16 * \lim_{n \rightarrow \infty}\left(\sum_{i=1}^n\dfrac{1 + i}{n^2}\right).\)

 

[Jason76]

What is the exact statement of the problem?

Given a = 4 and b = 8 and assuming that the proper starting place is indeed

\(\displaystyle \displaystyle \lim_{n \rightarrow \infty}\left(\sum_{i=1}^n\left\{\dfrac{b - a}{n} * \dfrac{a + i(b - a)}{n}\right\}\right) =\)

\(\displaystyle \displaystyle \lim_{n \rightarrow \infty}\left(\sum_{i=1}^n\left\{\dfrac{8 - 4}{n} * \dfrac{4 + i(8 - 4)}{n}\right\}\right) =\)

\(\displaystyle \displaystyle \lim_{n \rightarrow \infty}\left(\sum_{i=1}^n\left\{\dfrac{4}{n} * \dfrac{4 + 4i}{n}\right\}\right) =\)

\(\displaystyle \displaystyle \lim_{n \rightarrow \infty}\left(\sum_{i=1}^n\left\{4 * \dfrac{1}{n} * 4 * \dfrac{1 + i}{n}\right\}\right) =\)

\(\displaystyle \displaystyle 16 * \lim_{n \rightarrow \infty}\left(\sum_{i=1}^n\dfrac{1 + i}{n^2}\right).\)

\(\displaystyle \displaystyle \lim_{n \rightarrow \infty}\left(\sum_{i=1}^n\left\{\dfrac{4}{n} * \dfrac{4 + 4i}{n}\right\}\right) =\)

You get \(\displaystyle \dfrac{4(4 + 4i)}{n^{2}} = \dfrac{16 + 16i}{n^{2}} = \dfrac{16(1 + i)}{n^{2}}\) This factored out \(\displaystyle 16\) is put to the left of the limit sign.

Finishing it off

Just for fun, and become internet etiquette demands, let's see where this goes :

\(\displaystyle \dfrac{(1 + i)}{n^{2}} = 1 + \dfrac{1}{2} = \dfrac{3}{2}\)

Finally we multiply our \(\displaystyle 16 * \dfrac{3}{2} = 24\) ANSWER

I suppose integral problems (done this way) would always come out like this (using this method), assuming you got \(\displaystyle \dfrac{(1 + i)}{n^{2}}\) Somewhere toward the end. Is this right?

Check

\(\displaystyle \int^{8}_{4} x dx = \dfrac{x^{2}}{2}\)

Evaluate at \(\displaystyle 8\) and \(\displaystyle 4\)

\(\displaystyle \dfrac{x^{2}}{2} - \dfrac{x^{2}}{2}\)

\(\displaystyle \dfrac{8^{2}}{2} - \dfrac{4^{2}}{2}\)

\(\displaystyle \dfrac{64}{2} - \dfrac{16}{2} = \dfrac{48}{2} = 24 \) ANSWER - same as with other method.

 

[lookagain]

Note: I know there is an easier way to do this (the traditional integral way), but CLEP Calculus demands me to know this:Given lower bound \(\displaystyle a (4)\) and upper bound \(\displaystyle b (8)\) The function is missing. What is f(x)?Formula: \(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{(b - a)}{n} * f\bigg(a + \dfrac{ i(b - a)}{n}\bigg) \ \ \ \)Your lines of work are missing f. \(\displaystyle \ \ \) But I inserted them here. \(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{(8 - 4)}{n} * f\bigg(4 + \dfrac{ i(8 - 4)}{n}\bigg)\)\(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{4}{n} * f\bigg(4 + \dfrac{4i}{n}\bigg)\)\(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{4}{n} * f\bigg(\dfrac{4n}{n} + \dfrac{4i}{n}\bigg)\) \(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{4}{n} * f\bigg(\dfrac{4n + 4i}{n}\bigg)\) > > > Missing later steps would be here. < < < \(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n} 16 \dfrac{n + i}{n^{2}} \)\(\displaystyle \ \ \ \)Source: http://www.qc.edu.hk/math/Resource/AL/Examples of Riemann Integration.pdf

Finishing it off

Just for fun, and become internet etiquette demands, let's see where this goes :

\(\displaystyle \dfrac{(1 + i)}{n^{2}} = 1 + \dfrac{1}{2} \ \ \ \ \) <----- Wrong. \(\displaystyle \ = \dfrac{3}{2}\)

Finally we multiply our \(\displaystyle 16 * \dfrac{3}{2} = 24\)

\(\displaystyle \int^{8}_{4} x dx = \dfrac{x^{2}}{2}\)

Jason, you never stated what f(x) is in the initial post!

 

[Jason76]

Whole Problem

\(\displaystyle \int^{8}_{4} x dx\)

Given lower bound \(\displaystyle a (4)\) and upper bound \(\displaystyle b (8)\)

Formula: \(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{(b - a)}{n} * f(\dfrac{a + i(b - a)}{n})\)

\(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{(8 - 4)}{n} * f([4 + \dfrac{4 + i(8 - 4)}{n})]\)

\(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{4}{n} * f([4 + \dfrac{4 + 4i)}{n}])\)


\(\displaystyle \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{4(4 + 4i)}{n^{2}} = \dfrac{16 + 16i}{n^{2}} = \dfrac{16(1 + i)}{n^{2}}\)


\(\displaystyle 16 \lim_{x\to\infty}\sum_{i=1}^{n}\dfrac{1 + i}{n^{2}} \)

Finishing it off

Just for fun, and become internet etiquette demands, let's see where this goes :grin::

\(\displaystyle \dfrac{(1 + i)}{n^{2}} = 1 + \dfrac{1}{2} = \dfrac{3}{2}\)

Finally we multiply our \(\displaystyle 16 * \dfrac{3}{2} = 24\) ANSWER

I suppose integral problems (done this way) would always come out like this (using this method), assuming you got \(\displaystyle \dfrac{(1 + i)}{n^{2}}\) Somewhere toward the end. Is this right?

Check

\(\displaystyle \int^{8}_{4} x dx = \dfrac{x^{2}}{2}\)

Evaluate at \(\displaystyle 8\) and \(\displaystyle 4\)

\(\displaystyle \dfrac{x^{2}}{2} - \dfrac{x^{2}}{2}\)

\(\displaystyle \dfrac{8^{2}}{2} - \dfrac{4^{2}}{2}\)

\(\displaystyle \dfrac{64}{2} - \dfrac{16}{2} = \dfrac{48}{2} = 24 \) ANSWER - same as with other method.

 

[lookagain]

Whole Problem

\(\displaystyle \int^{8}_{4} x dx\)

Given lower bound \(\displaystyle a = 4\) and upper bound \(\displaystyle b = 8\)

Formula: \(\displaystyle \lim_{n\to\infty}\sum_{i=1}^{n}\dfrac{(b - a)}{n} * f(\dfrac{a + i(b - a)}{n}) \ \ \ \ \) <-------- The variable constant, a, is *not* part of the numerator. For this step to show how the variable a is separate from the fraction and for the next couple of steps, see post # 4.

\(\displaystyle \lim_{n\to\infty}\sum_{i=1}^{n}\dfrac{(8 - 4)}{n} * f([4 + \dfrac{4 + i(8 - 4)}{n})]\)

\(\displaystyle \lim_{n\to\infty}\sum_{i=1}^{n}\dfrac{4}{n} * f([4 + \dfrac{4 + 4i)}{n}])\)


\(\displaystyle \lim_{n\to\infty}\sum_{i=1}^{n}\dfrac{4(4 + 4i)}{n^{2}} = \dfrac{16 + 16i}{n^{2}} = \dfrac{16(1 + i)}{n^{2}} \ \ \)No, you can't put equal signs between these, because you're missing the limit and summation notations in front of the second and third expressions in this line.


\(\displaystyle 16 \lim_{n\to\infty}\sum_{i=1}^{n}\dfrac{1 + i}{n^{2}} \)

Just for fun, and become internet etiquette demands, let's see where this goes :grin::

\(\displaystyle \dfrac{(1 + i)}{n^{2}} = 1 + \dfrac{1}{2} \ \ \ \) <------ Wrong. \(\displaystyle \ = \ \dfrac{3}{2} \ \ \ \ \)It is the limit of the summation that equals \(\displaystyle \ \ 1 + \dfrac{1}{2} \ = \dfrac{3}{2}.\)

Finally we multiply our \(\displaystyle 16 * \dfrac{3}{2} = 24\) ANSWER

I suppose integral problems (done this way) would always come out like this (using this method), assuming you got \(\displaystyle \dfrac{(1 + i)}{n^{2}}\) Somewhere toward the end. Is this right?

Check

\(\displaystyle \int^{8}_{4} x dx = \dfrac{x^{2}}{2}\bigg|_4^8 \ \ \ \ \ \)<-------- The limits of integration can be placed here.

Evaluate at \(\displaystyle 8\) and \(\displaystyle 4\)

\(\displaystyle \dfrac{x^{2}}{2} - \dfrac{x^{2}}{2} \ \ \ \) Wrong.

\(\displaystyle \dfrac{8^{2}}{2} - \dfrac{4^{2}}{2}\)

\(\displaystyle \dfrac{64}{2} - \dfrac{16}{2} = \dfrac{48}{2} = 24 \) ANSWER - same as with other method.

** Also, I (lookagain) missed this: It is supposed to be the limit as n approaches infinity.