Can substitution help solve this improper integral?

  • Thread starter regnar
  • Start date
  • Tags
    Integrals
In summary, an improper integral is an integral where either the upper or lower limit of integration is infinite or the function being integrated is not defined at a point within the limits of integration. To determine if an integral is improper, we must evaluate it using limits and see if any of the conditions apply. There are two types of improper integrals and they can be evaluated using techniques such as the comparison test, limit comparison test, and integral test. Improper integrals are important in science because they allow us to find the areas under curves that would otherwise be impossible to evaluate using traditional methods and are essential in solving differential equations and modeling real-world phenomena.
  • #1
regnar
24
0
I've been stuck on this problem for quite some time and I get as far as setting up the problem. They are asking me to evaluate this integral:

[tex]\int_0^1{\frac{(6ln4x)}{\sqrt{x}}dx[/tex] and I get as far as:

[tex]\lim_{b \to 0^+}{6\int_b^1{\frac{(ln4x)}{\sqrt{x}}[/tex]
 
Physics news on Phys.org
  • #2
Substituting x = t^2 seems to work...
 
  • #3
I'm not sure what you mean or where [tex]t^2[/tex] came from.
 
  • #4
It's called substitution.

The substitution is [tex]t=\sqrt{x}[/tex] or [tex]t^2 = x [/tex]
 

1. What is an improper integral?

An improper integral is an integral where either the upper or lower limit of integration is infinite, or the function being integrated is not defined at a point within the limits of integration. This means that the integral cannot be evaluated using traditional methods and may require special techniques to find the value.

2. How do you determine if an integral is improper?

An integral is considered improper if any of the following conditions are met:

  • The upper or lower limit of integration is infinite.
  • The function being integrated is not defined at a point within the limits of integration.
  • The function being integrated becomes unbounded within the limits of integration.
To determine if an integral is improper, we must evaluate the integral using limits and see if any of the above conditions apply.

3. How do you evaluate an improper integral?

To evaluate an improper integral, we must first determine the type of improper integral it is. There are two types: Type 1 and Type 2. For Type 1 improper integrals, we must take the limit as the upper or lower limit of integration approaches infinity. For Type 2 improper integrals, we must split the integral into two or more integrals and take the limit as the points of discontinuity approach the limits of integration. Once the limits have been determined, we can evaluate the integral as we would a regular integral.

4. What are some common techniques for evaluating improper integrals?

Some common techniques for evaluating improper integrals include using the comparison test, the limit comparison test, and the integral test. These techniques involve comparing the given integral to a known function or series and using the properties of convergence and divergence to determine the value of the integral.

5. Why are improper integrals important in science?

Improper integrals are important in science because they allow us to find the areas under curves that would otherwise be impossible to evaluate using traditional methods. This is essential in many scientific fields, such as physics and engineering, where functions may be unbounded or have infinite limits. Improper integrals also allow us to solve differential equations and model real-world phenomena, making them a valuable tool in scientific research and analysis.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
287
  • Calculus and Beyond Homework Help
Replies
7
Views
901
  • Calculus and Beyond Homework Help
Replies
8
Views
720
  • Calculus and Beyond Homework Help
Replies
5
Views
872
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
588
  • Calculus and Beyond Homework Help
Replies
3
Views
538
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top