Solve Error in Integrating z^2lnz from 0 to 2

Let u = lnz and dv = z^2dz, then du = dz/z and v = (z^3)/3. So, the integral becomes:\int^{2}_{0}z^{2}lnzdz = (z^3lnz)/3 - \int^{2}_{0}(z^2/3)dz = (z^3lnz)/3 - (z^3/9)|^{2}_{0} = (8ln2)/3 - 8/9 = (8ln2 - 8)/9In summary, the solution for the given integral is (8ln2 - 8)/9. There may have been an error in substituting and using l'H
  • #1
MathewsMD
433
7

Homework Statement



## \int_0^2 z^2lnzdz ##


2. The attempt at a solution

## u = lnz, dz = zdu, lim_{t→-∞}\int_t^{ln2} ue^udu ##
## lim_{t→-∞}ue^u - e^u l^{ln2}_t ##
##2(ln2 - 1) - lim_{t→-∞}\frac {t-1}{e^{-t}} ## Using l'Hospital's rule:
##2(ln2 - 1)##

This is incorrect though. Any pointers on where I went wrong? I believe the domain tom -∞ to ln2 has no discontinuities for the newly defined function and don't seem to see any blatant errors...

All help is welcomed and greatly appreciated!
 
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  • #2
Looks like an error in substituting. When you solve for dz, when substituting in you should get:

[itex]\int^{2}_{0}z^{3}lnzdu[/itex], then with [itex]z = e^{u}[/itex], it should become:

[itex]\int^{ln2}_{-\infty}ue^{3u}du[/itex]
 
  • #3
I would suggest Integration by Parts rather than a substitution.
 

Related to Solve Error in Integrating z^2lnz from 0 to 2

1. What is the error in integrating z^2lnz from 0 to 2?

The error in integrating z^2lnz from 0 to 2 can depend on various factors such as the method of integration used and the precision of the numerical calculation. In general, there is no exact error value and it may vary from one integration to another.

2. How can I solve the error in integrating z^2lnz from 0 to 2?

To solve the error in integrating z^2lnz from 0 to 2, you can try using different methods of integration such as the trapezoidal rule, Simpson's rule, or Gaussian quadrature. You can also increase the precision of your numerical calculation for a more accurate result.

3. What is the meaning of the integral z^2lnz from 0 to 2?

The integral z^2lnz from 0 to 2 represents the area under the curve of the function z^2lnz between the limits of 0 and 2. In other words, it is the accumulation of all the values of z^2lnz within the given interval.

4. Why do I encounter an error when integrating z^2lnz from 0 to 2?

There are several reasons why you may encounter an error when integrating z^2lnz from 0 to 2. It could be due to a mistake in the integration process, the function being non-integrable, or the numerical calculation not being precise enough.

5. Is there a closed-form solution for the integral z^2lnz from 0 to 2?

No, there is no known closed-form solution for the integral z^2lnz from 0 to 2. This means that it cannot be expressed in terms of elementary functions and requires numerical methods for its evaluation.

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