Weibull Integral Explained - No Quotes

  • Thread starter Thread starter longrob
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary
SUMMARY

The discussion focuses on the Weibull integral, specifically the expression for the cumulative distribution function P(v) given by P(v)=\frac{\beta}{\eta}\intop_{0}^{v}\left(\frac{v}{\eta}\right)^{\beta-1}\exp\left(-\left(\frac{v}{\eta}\right)^{\beta}\right)dv. Participants highlight the importance of clear variable notation, suggesting the use of v' for the integration variable to avoid confusion. A change of variable substitution is recommended for clarity, leading to the revised expression P(v)=\frac{\beta}{\eta}\intop_{0}^{v}\left(\frac{V}{\eta}\right)^{\beta-1}\exp\left(-\left(\frac{V}{\eta}\right)^{\beta}\right)dV.

PREREQUISITES
  • Understanding of Weibull distribution and its parameters (β and η).
  • Familiarity with integral calculus and variable substitution techniques.
  • Knowledge of exponential functions and their properties.
  • Experience with mathematical notation and clarity in variable representation.
NEXT STEPS
  • Study the properties of the Weibull distribution in reliability engineering.
  • Learn about variable substitution in integrals for clearer mathematical expressions.
  • Explore the applications of the Weibull integral in statistical modeling.
  • Investigate common pitfalls in mathematical notation and how to avoid them.
USEFUL FOR

Mathematicians, statisticians, engineers, and anyone involved in reliability analysis or statistical modeling who seeks to understand the Weibull integral and improve their mathematical notation skills.

longrob
Messages
51
Reaction score
0
Can someone explain this..
P(v)=\frac{\beta}{\eta}\intop_{0}^{v}\left(\frac{v}{\eta}\right)^{\beta-1}\exp\left(-\left(\frac{v}{\eta}\right)^{\beta}\right)dv=\intop_{0}^{x}e^{-x}dx\hphantom{}\; where\phantom{\:}x=\left(\frac{v}{\eta}\right)^{\beta}

Thanks !
 
Physics news on Phys.org
It just seems like a change of variable, but beware your notation.
In the first integral you use v both as the integration variable and in the upper limit of integration. In the second, x plays both those roles too. This is confusing.
I'd call the integration variable v' in the first one, then just make a change of variable substitution.
 
In line with Galileo's comment, this is how you should write it:
P(v)=\frac{\beta}{\eta}\intop_{0}^{v}\left(\frac{V}{\eta}\right)^{\beta-1}\exp\left(-\left(\frac{V}{\eta}\right)^{\beta}\right)dV=\intop_{0}^{x(v)}e^{-X}dX\hphantom{}\; where\phantom{\:}x(v)=\left(\frac{v}{\eta}\right)^{\beta}
 
Thanks !
 
Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

Similar threads

  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K