Integral help (substitution and boundaries)

In summary: Or, you can use the substitution ##u = \ln x##, which turns the integral into$$\int_{\ln h}^0 u \, du = -\frac12 \ln^2 h.$$In summary, the conversation discusses the use of substitution in evaluating an improper integral and whether or not the bounds should be changed in the process. It is noted that the image provided may be using sloppy reasoning and that the proper definition of the integral should be taken into account. Ultimately, the conversation suggests using the substitution method and the resulting solution is -1/2 ln^2 h.
  • #1
Kqwert
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3

Homework Statement


Hello,
I´ve added an image where the task is shown. I am wondering about the integral bounds here. When doing the substitution, shouldn't the bounds be inf and 0? When using that u = ln x, i.e. u(0) =inf, u(1) = 0.

Homework Equations

The Attempt at a Solution

 

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  • #2
Kqwert said:

Homework Statement


Hello,
I´ve added an image where the task is shown. I am wondering about the integral bounds here. When doing the substitution, shouldn't the bounds be inf and 0? When using that u = ln x, i.e. u(0) =inf, u(1) = 0.

If they are simply going step-by-step and changing the limits, then you are correct. Otherwise, there may be some intermediary steps to get to that point where they have further manipulated the integral.

Just wondering why use substitution when you can use reverse chain rule? Then you don't have to worry about the limits changing...

[tex] \int_{0}^{\infty} \frac{ln(x)}{x} = \frac{1}{2} ( ln(\infty)^2 - ln(0)^2)[/tex]
 
  • #3
Kqwert said:

Homework Statement


Hello,
I´ve added an image where the task is shown. I am wondering about the integral bounds here. When doing the substitution, shouldn't the bounds be inf and 0? When using that u = ln x, i.e. u(0) =inf, u(1) = 0.

Homework Equations

The Attempt at a Solution


The image is guilty of sloppy reasoning. The actual definition of the improper integral involved is
$$\int_0^1 \frac{\ln x}{x} \, dx \equiv \lim_{h \to 0+} \int_h^1 \frac{\ln x}{x} \, dx.$$
You can evaluate the integral ##\int_h^1 \ln(x)/x \, dx## and then examine the limit.
 

FAQ: Integral help (substitution and boundaries)

What is integral help (substitution and boundaries)?

Integral help (substitution and boundaries) is a method used in calculus to solve integrals. It involves substituting a variable in the integral and applying appropriate boundaries to the integral to make it easier to solve.

How do I know when to use substitution and boundaries in integrals?

Substitution and boundaries are useful techniques when the integrand (the function inside the integral) is complex or involves multiple variables. It can also be used to simplify the integral and make it easier to solve.

What is the process for using substitution and boundaries in integrals?

The first step is to identify a suitable substitution for the variable in the integral. Then, apply the appropriate boundaries to the integral based on the substitution. This will result in a new integral that is easier to solve. Finally, apply the reverse substitution to get the final answer.

Can substitution and boundaries be used in all types of integrals?

Yes, substitution and boundaries can be used in both definite and indefinite integrals. However, the choice of substitution and boundaries may vary depending on the type of integral and the complexity of the integrand.

Are there any common mistakes to avoid when using substitution and boundaries in integrals?

One common mistake is forgetting to apply the appropriate boundaries when using substitution. Another mistake is choosing an incorrect substitution, which can make the integral more difficult to solve. It is important to carefully choose the substitution and apply the boundaries correctly to ensure the correct solution.

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