How to solve an Improper Integral of Type 2?

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SUMMARY

This discussion focuses on solving improper integrals of Type 2, specifically those defined over a finite interval with a discontinuity. The example provided involves the integral from 0 to 2 of the function 1/(x-1)^(1/3). Participants emphasize the importance of applying the theorem of addition to split the integral into two parts: from 0 to 1 and from 1 to 2. The correct approach includes computing limits as the variable approaches the point of discontinuity, specifically using lim_{a→1}∫_{1}^{a} f(x) dx and lim_{b→1}∫_{b}^{2} f(x) dx.

PREREQUISITES
  • Understanding of improper integrals and their classifications
  • Familiarity with the Fundamental Theorem of Calculus (FTC)
  • Knowledge of limit notation and computation
  • Basic integration techniques for rational functions
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  • Study the application of the theorem of addition in improper integrals
  • Practice computing limits for discontinuous functions in integrals
  • Explore examples of improper integrals of Type 1 and Type 2
  • Learn advanced integration techniques, such as integration by parts and substitution
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Students and educators in calculus, mathematicians focusing on integration techniques, and anyone looking to deepen their understanding of improper integrals and their applications.

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I've posted on Homeworks one of the number I did not understand. However, I would like to know the steps to calculate an improper integral of type 2. The type 2 is the one from constant a to constat b, not the one with inifnite.

Please tell me the steps the accomplish it. :smile:

I know that I'm supposed to set the limits, then put it in the integral form, than FTC it. But, I'm not sure.
 
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Could you give example so I could understand question a little better.
 
limitapproaches0 said:
Could you give example so I could understand question a little better.


Let's say something like [int a=0 b=2] 1/(x-1)^(1/3) dx
 
Use the theorem of addition

\int_{0}^{2} f(x) \ dx=\int_{0}^{1} f(x) \ dx+\int_{1}^{2} f(x) \ dx

You have to integrate & compute 2 limits...

Daniel.
 
dextercioby said:
Use the theorem of addition

\int_{0}^{2} f(x) \ dx=\int_{0}^{1} f(x) \ dx+\int_{1}^{2} f(x) \ dx

You have to integrate & compute 2 limits...

Daniel.

The limits would be 1+ and 1- right ?
 
It should be

\lim_{a\nearrow 1}\int_{1}^{a} f(x) \ dx

\lim_{b\searrow 1}\int_{b}^{2} f(x) \ dx


Daniel.
 

Thread 'Ambiguity of the term "indefinite integral"'

I've encountered a few different definitions of "indefinite integral," denoted ##\int f(x) \, dx##. any particular antiderivative ##F:\mathbb{R} \to \mathbb{R}, F'(x) = f(x)## the set of all antiderivatives ##\{F:\mathbb{R} \to \mathbb{R}, F'(x) = f(x)\}## a "canonical" antiderivative any expression of the form ##\int_a^x f(x) \, dx##, where ##a## is in the domain of ##f## and ##f## is continuous Sometimes, it becomes a little unclear which definition an author really has in mind, though...
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