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how could we put the limits in definite integration

how could we put the limits in definite integration

Grade:12

1 Answers

Askiitian.Expert Rajat
24 Points
15 years ago

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} = \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} = \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 = { n(n+1) \over 2 } } $
  3. $ \displaystyle{ \sum_{i=1}^{n} i^2 = 1^2 + 2^2 + 3^2 + \cdots + n^2 = { n(n+1)(2n+1) \over 6 } } $
  4. $ \displaystyle{ \sum_{i=1}^{n} i^3 = 1^3 + 2^3 + 3^3 + \cdots + n^3 = { n^2(n+1)^2 \over 4 } } $
  5. $ \displaystyle{ \sum_{i=1}^{n} k f(i) } = \displaystyle{ k \sum_{i=1}^{n} f(i) } $ , where $ k $ is a constant
  6. $ \displaystyle{ \sum_{i=1}^{n} (f(i) \pm g(i)) } = \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

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