Reimann Sum - Another Problem

[Jason76]

\(\displaystyle \left(\sum_{i=1}^{50}(\dfrac{i}{50})^{2} \dfrac{1}{50}\right).\) on the closed interval \(\displaystyle [0, 1]\) is an approximation for what definite integral? Answer: \(\displaystyle \int^{1}_{0} x^{2} dx\)

No idea on this one. Any starting hints? I can see that the interval of \(\displaystyle [0, 1]\) corresponds to the upper and and lower bound of the definite integral, \(\displaystyle [a, b]\) and \(\displaystyle \int^{b}_{a} x^{2} dx\) that's all.

 

[JeffM]

What is the function to be integrated?

 

[lookagain]

\(\displaystyle \left(\sum_{i=1}^{50}(\dfrac{i}{50})^{2} \dfrac{1}{50}\right).\) on the closed interval \(\displaystyle [0, 1]\) is an approximation for what definite integral? Answer: \(\displaystyle \int^{1}_{0} x^{2} dx\)

No idea on this one. Any starting hints? I can see that the interval of \(\displaystyle [0, 1]\) corresponds to the upper and and lower bound of the definite integral, \(\displaystyle [a, b]\) and \(\displaystyle \int^{b}_{a} x^{2} dx\) that's all.

From another thread, I corrected that a formula for the exact area:

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



For this case with a finite n, n = 50, a = 0, and b = 1.

You can reorder it as:



\(\displaystyle \ \ \ \displaystyle\sum_{i = 1}^{50} \ \bigg(\dfrac{1}{50}\bigg)\bigg(\dfrac{i}{50}\bigg)^2 \ = \)


\(\displaystyle \displaystyle\sum_{i = 1}^{50} \ \bigg( \dfrac{1 - 0}{50}\bigg)\bigg(0 \ + \ \dfrac{i(1 - 0)}{50}\bigg)^2\)


The expression in the second pair of parentheses is squared, so that suggests that the function is \(\displaystyle \ f(x) \ = \ x^2.\)

 

[Jason76]

From another thread, I corrected that a formula for the exact area:

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



For this case with a finite n, n = 50, a = 0, and b = 1.

You can reorder it as:



\(\displaystyle \ \ \ \displaystyle\sum_{i = 1}^{50} \ \bigg(\dfrac{1}{50}\bigg)\bigg(\dfrac{i}{50}\bigg)^2 \ = \)


\(\displaystyle \displaystyle\sum_{i = 1}^{50} \ \bigg( \dfrac{1 - 0}{50}\bigg)\bigg(0 \ + \ \dfrac{i(1 - 0)}{50}\bigg)^2\)


The expression in the second pair of parentheses is squared, so that suggests that the function is \(\displaystyle \ f(x) \ = \ x^2.\)

Oh I see now. Awesome :cool:

What is the function to be integrated?

They wanted you to identify a certain "Reimann sum solving stage". After that you pick the definite integral which is the equivalent of the Reimann problem (based on given info).