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pallavi pradeep bhardwaj Grade: 12
        how could we put the limits in definite integration

7 years ago

Answers : (1)

Askiitian.Expert Rajat
24 Points
										

Hi,


Let me explain the concept of limits in Definite Integrals by taking a general example :


 


The following problems involve the limit definition of the definite integral of a continuous function of one variable on a closed, bounded interval. Begin with a continuous function $ y=f(x) $ on the interval $ [a, b] $. Let


$ a=x_{0}, x_{1}, x_{2}, x_{3}, $ ... $ , x_{n-2}, x_{n-1}, x_{n}=b $


be an arbitrary (randomly selected) partition of the interval $ [a, b] $ , which divides the interval into $ n $ subintervals (subdivisions). Let


$ c_{1}, c_{2}, c_{3}, $ ... $ , c_{n-2}, c_{n-1}, c_{n} $


be the sampling numbers (or sampling points) selected from the subintervals. That is,


$ c_{1} $ is in $ [x_{0}, x_{1}] $,


$ c_{2} $ is in $ [x_{1}, x_{2}] $,


$ c_{3} $ is in $ [x_{2}, x_{3}] $, ... ,


$ c_{n-2} $ is in $ [x_{n-3}, x_{n-2}] $,


$ c_{n-1} $ is in $ [x_{n-2}, x_{n-1}] $,


and


$ c_{n} $ is in $ [x_{n-1}, x_{n}] $ .


Define the mesh of the partition to be the length of the largest subinterval. That is, let


$ \Delta x_{i} = x_{i} - x_{i-1} \ \ $


for $ i = 1, 2, 3, ..., n $ and define


$ mesh = \displaystyle{ \max_{1 \le i \le n} \{ x_{i} - x_{i-1} \}} $ .


The definite integral of $ f $ on the interval $ [a, b] $ is most generally defined to be


$ \displaystyle{ \int^{b}_{a} f(x) \, dx}
<br/>= \displaystyle{ \lim_{mesh \to 0} \sum_{i=1}^{n} f(c_{i}) \Delta x_{i} } $ .


For convenience of computation, a special case of the above definition uses $ n $ subintervals of equal length and sampling points chosen to be the right-hand endpoints of the subintervals. Thus, each subinterval has length


equation (*) $ \ \ \ \ \ \ \ \ \Delta x_{i} = \displaystyle{ b-a \over n } $


for $ i = 1, 2, 3, ..., n $ and the right-hand endpoint formula is


equation (**) $ \ \ \ \ \ \ \ \ c_{i} = \displaystyle{ a + \Big( { b-a \over n } \Big) i } $


for $ i = 1, 2, 3, ..., n $ . The definite integral of $ f $ on the interval $ [a, b] $ can now be alternatively defined by


$ \displaystyle{ \int^{b}_{a} f(x) \, dx}
<br/>= \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} f(c_{i}) \Delta x_{i} } $ .


We will need the following well-known summation rules in case od different types of functions :



  1. $ \displaystyle{ \sum_{i=1}^{n} c = c + c + c + \cdots + c } $ (n times) $ = nc $ , where $ c $ is a constant

  2. $ \displaystyle{ \sum_{i=1}^{n} i = 1 + 2 + 3 + \cdots + n
<br/>    = { n(n+1) \over 2 } } $

  3. $ \displaystyle{ \sum_{i=1}^{n} i^2 = 1^2 + 2^2 + 3^2 + \cdots + n^2
<br/>    = { n(n+1)(2n+1) \over 6 } } $

  4. $ \displaystyle{ \sum_{i=1}^{n} i^3 = 1^3 + 2^3 + 3^3 + \cdots + n^3
<br/>    = { n^2(n+1)^2 \over 4 } } $

  5. $ \displaystyle{ \sum_{i=1}^{n} k f(i) }
<br/>    = \displaystyle{ k \sum_{i=1}^{n} f(i) } $ , where $ k $ is a constant

  6. $ \displaystyle{ \sum_{i=1}^{n} (f(i) \pm g(i)) }
<br/>    = \displaystyle{ \sum_{i=1}^{n} f(i) \pm \sum_{i=1}^{n} g(i) } $


Be sure to ask if anything's not clear.


Regards and Best of Luck,


Rajat


Askiitian Expert

7 years ago
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