Solving an Improper Integral Problem: 1/(4x-2)^5 dx from 2 to infinity

In summary, the conversation is about solving an integral problem that involves finding the value of 1/(4x-2)^5 from 2 to infinity. The initial attempt is to integrate and the answer is given as -(1/(16(2-4x)^4)), top - bottom which comes out to 0-(-.0004822). However, the answer is rejected as incorrect and it is suggested that the online platform may be looking for the exact answer as a fraction instead of a decimal approximation. The correct answer is determined to be 1/20736 and there is frustration expressed about the challenges of online answer checking.
  • #1
loganblacke
48
0

Homework Statement


Find the value of Integral from 2 to infinity of 1/(4x-2)^5 dx.

Homework Equations





The Attempt at a Solution


When I integrated I came up with -(1/(16(2-4x)^4)), top - bottom which ends up being 0-(-.0004822) but it says that is wrong. Help please!
 
Physics news on Phys.org
  • #2
That looks fine to me. Maybe they want the exact answer as a fraction instead of the decimal approximation?
 
  • #3
Dick said:
That looks fine to me. Maybe they want the exact answer as a fraction instead of the decimal approximation?

Good call, that is exactly what they were looking for, 1/20736. Whoever decided calculus homework should be done online should be tortured.
 
  • #4
loganblacke said:
Good call, that is exactly what they were looking for, 1/20736. Whoever decided calculus homework should be done online should be tortured.

Online answer checking in general is a pain in the neck. It has no idea whether you are right or wrong, just whether it fits whatever form it's set to recognize. I sympathize.
 
  • #5
loganblacke said:
When I integrated I came up with -(1/(16(2-4x)^4)), top - bottom which ends up being 0-(-.0004822) but it says that is wrong.

There should be four 0's after the decimal point: 0.000048225308641975306 ≈ 1/20736
So make sure your enter your number correctly when submitting it online. :wink:
 

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 integrand has a vertical asymptote within the bounds of integration. It cannot be evaluated using the standard methods of calculus and requires special techniques to find its value.

2. How do you determine if an improper integral converges or diverges?

To determine convergence or divergence of an improper integral, you must evaluate the limit as one or both of the bounds of integration approach infinity. If the limit is a finite number, then the integral converges. If the limit is infinite or does not exist, then the integral diverges.

3. What is the difference between a type 1 and type 2 improper integral?

A type 1 improper integral has an infinite limit of integration, while a type 2 improper integral has an integrand that approaches infinity within the bounds of integration. Type 1 improper integrals can be evaluated using limits, while type 2 improper integrals require integration by parts or other techniques to be evaluated.

4. Can you use substitution to evaluate improper integrals?

Yes, substitution can be used to evaluate improper integrals, but it may not always be the most efficient method. In some cases, integration by parts or other techniques may be more useful for evaluating an improper integral.

5. Are there any real-world applications of improper integrals?

Improper integrals have many real-world applications in physics, engineering, and economics. For example, they can be used to calculate the area under a curve that has a vertical asymptote, such as the demand curve in economics. They are also used in calculating the total energy of a system in physics, where the bounds of integration may be infinite.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
436
  • Calculus and Beyond Homework Help
Replies
7
Views
704
  • Calculus and Beyond Homework Help
2
Replies
54
Views
8K
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
807
  • Calculus and Beyond Homework Help
2
Replies
44
Views
4K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top