Expressing limit as definite integral
- Thread starter funnytim
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The integral is "the sum expressed with the sigma symbol as "n" is allowed go to infinity".
The "sigma" is then changed to the "music note".
A "definite" integral is an integral that can be evaluated between two specific values of a variable.
"Definate integral" does not mean writing the summation as an integral as n is allowed go to infinity.
The integral itself is that.
The infinite sum is the integral, giving "exact" area as opposed to approximate.
It is expressed as the limit of the summation as n goes to infinity,
(what it would be if n really could go to infinity!).
Evaluating an integral between "definite limits" uses the term "limit" in a new sense.
These limits are the edges of the bounded area on the variable's axis.
Study how integrals are developed from summations.
Typically, this involves summing "rectangle areas" under a continuous curve.
Then how to evaluate a "definate" area between a starting point and end.
The integral expressed with the music note waits for you to choose the boundaries.
Then you can use these to calculate the area, you are evaluating a definate integral.
{The summation you have quoted is a discrete infinite series,
a different ball game to calculating the area under a curve or line.
If you evaluate this with i=1, 2, 3, 4 up to n,
and then let n approach infinity, you can cancel the n's in the numerator positions,
ending up with n's only in the denominator.
Looks like you'd be adding a string of zeros.} Incorrect..... this is corrected below by galactus
The "sigma" is then changed to the "music note".
A "definite" integral is an integral that can be evaluated between two specific values of a variable.
"Definate integral" does not mean writing the summation as an integral as n is allowed go to infinity.
The integral itself is that.
The infinite sum is the integral, giving "exact" area as opposed to approximate.
It is expressed as the limit of the summation as n goes to infinity,
(what it would be if n really could go to infinity!).
Evaluating an integral between "definite limits" uses the term "limit" in a new sense.
These limits are the edges of the bounded area on the variable's axis.
Study how integrals are developed from summations.
Typically, this involves summing "rectangle areas" under a continuous curve.
Then how to evaluate a "definate" area between a starting point and end.
The integral expressed with the music note waits for you to choose the boundaries.
Then you can use these to calculate the area, you are evaluating a definate integral.
{The summation you have quoted is a discrete infinite series,
a different ball game to calculating the area under a curve or line.
If you evaluate this with i=1, 2, 3, 4 up to n,
and then let n approach infinity, you can cancel the n's in the numerator positions,
ending up with n's only in the denominator.
Looks like you'd be adding a string of zeros.} Incorrect..... this is corrected below by galactus
We apparently have a Riemann sum we can express as an integral. Afterall, that is what an integral is. The sum of an infinite
number of rectangular areas under a curve.
\(\displaystyle {\Delta}x=\frac{1-0}{n}=\frac{1}{n}\)
\(\displaystyle i{\Delta}x=i\frac{1-0}{n}=\frac{i}{n}\)
Sub these in \(\displaystyle f(x_{i})\)
We have \(\displaystyle \lim_{n\to \infty}\sum_{i=1}^{n}\frac{\frac{i}{n}}{1+(\frac{i}{n})^{2}}\) equivalent to the integral \(\displaystyle \int_{0}^{1}\frac{x}{1+x^{2}}dx=\frac{ln(2)}{2}\)
number of rectangular areas under a curve.
\(\displaystyle {\Delta}x=\frac{1-0}{n}=\frac{1}{n}\)
\(\displaystyle i{\Delta}x=i\frac{1-0}{n}=\frac{i}{n}\)
Sub these in \(\displaystyle f(x_{i})\)
We have \(\displaystyle \lim_{n\to \infty}\sum_{i=1}^{n}\frac{\frac{i}{n}}{1+(\frac{i}{n})^{2}}\) equivalent to the integral \(\displaystyle \int_{0}^{1}\frac{x}{1+x^{2}}dx=\frac{ln(2)}{2}\)