Numerical Integration: Simpson's Rule for 1/x

In summary, the speaker is trying to find the area under the curve of 1/x using Simpson's rule of n = 6, but is unsure how to handle the undefined value at 0. Another speaker suggests using a limit approach to solve the problem, but the final result is that the integral diverges.
  • #1
irony of truth
90
0
I am performing the numerical integration of finding the area of 1/x dx from 0 to 2... using Simpson's rule of n = 6. What will I do in this problem like this since 0 to be evaluated in the f(x) = 1/x is undefined?
 
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  • #2
I'd check the question since the integral doesn't exist.
 
  • #3
I'm sure it does.

The integral(anti-derivative) is ln x.

I'm pretty darn confident that's what it is. I'd prove it, but I'm not in the mood to spend time on here, and last time I tried Latex it didn't work.

y = ln x

dy/dx = 1/x
 
  • #4
if you are so concerned about your time why are you wasting ours?
 
  • #5
JasonRox said:
I'm sure it does.

The integral(anti-derivative) is ln x.

I'm pretty darn confident that's what it is. I'd prove it, but I'm not in the mood to spend time on here, and last time I tried Latex it didn't work.

y = ln x

dy/dx = 1/x

Yes, but what is the value of ln(x) when x = 0? Even if you try to avoid that limit of integration your answer will diverge as you move your endpoint closer to x = 0.
 
  • #6
the integral, as an improper riemann integral (limit h to zero of int from h to 2) does not exist, Jason. The function has an antiderivative, and that can be used to prove this fact.
 
  • #7
mathwonk said:
if you are so concerned about your time why are you wasting ours?

Relative to me, your time is going slow, so you might as well take advantage of it. :biggrin:

Kidding.

Sorry, I was getting ready to go to bed, and Latex in fact didn't work last time.

Should of checked my answer, and yes it is wrong.
 
  • #8
irony of truth said:
I am performing the numerical integration of finding the area of 1/x dx from 0 to 2... using Simpson's rule of n = 6. What will I do in this problem like this since 0 to be evaluated in the f(x) = 1/x is undefined?

You can still use Simpson method,though the integral is improper,by taking the limits of the integral as being from 'a' to 2 (where 'a' tends to 0).Finally take the limit for a->0 from the expression in 'a' obtained after applying Simpson's method.The results is ∞,the integral diverges.
 

FAQ: Numerical Integration: Simpson's Rule for 1/x

1. What is Simpson's Rule for 1/x?

Simpson's Rule for 1/x is a numerical integration method used to approximate the area under the curve of a function that involves the reciprocal of x. It is a more accurate method than the commonly used rectangular or trapezoidal rule, especially when dealing with rapidly changing functions.

2. How does Simpson's Rule for 1/x work?

Simpson's Rule for 1/x works by dividing the area under the curve into multiple smaller sections, approximating each section using a quadratic function, and then summing up the areas of all the sections to get an overall approximation of the total area.

3. When should Simpson's Rule for 1/x be used?

Simpson's Rule for 1/x should be used when dealing with functions that involve the reciprocal of x, such as 1/x, x^-1, or any other function that can be written in the form of 1/x. It is also useful for rapidly changing functions, as it provides a more accurate approximation compared to other methods.

4. What are the advantages of using Simpson's Rule for 1/x?

One advantage of using Simpson's Rule for 1/x is its accuracy. It provides a more precise approximation of the area under the curve compared to other numerical integration methods. Additionally, it can handle functions that involve the reciprocal of x, which cannot be easily integrated analytically.

5. Are there any limitations to using Simpson's Rule for 1/x?

One limitation of Simpson's Rule for 1/x is that it can only be used for functions that involve the reciprocal of x. It is not suitable for other types of functions. Additionally, it may not give accurate results for functions with sharp corners or discontinuities.

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