Integration of a Function of 2 Variables with Respect to One Variable

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SUMMARY

The discussion centers on the numerical integration of a function of two variables, specifically integrating the expression \(\int \frac{f(A)}{f(A,B)} dA\) to yield a function of one variable, \(f(B)\). The user seeks clarity on whether to substitute a value for \(B\) before integration or to leave it as a variable. The integration will be performed using MATLAB, emphasizing the need for practical implementation rather than theoretical solutions. The consensus indicates that both approaches can lead to the same outcome.

PREREQUISITES
  • Understanding of numerical integration techniques in MATLAB
  • Familiarity with functions of multiple variables
  • Basic knowledge of calculus, specifically integration
  • Experience with MATLAB syntax and plotting functions
NEXT STEPS
  • Explore MATLAB's 'integral' function for numerical integration
  • Learn about handling symbolic variables in MATLAB using the Symbolic Math Toolbox
  • Research methods for plotting functions in MATLAB
  • Investigate the implications of variable substitution in multivariable calculus
USEFUL FOR

Mathematicians, engineers, and data scientists who are implementing numerical integration in MATLAB, particularly those working with functions of multiple variables and seeking to visualize results effectively.

mea03wjb
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Hi,
I don't know how to describe this properly but I want to integrate a function of 2 variables to end up with a function of one variable.

\int \frac{f(A)}{f(A,B)} dA

I will integrate this function and end up with a function of B, i.e. f(B). Then I can substitue in different values of B and plot a graph.

Is this possible? Or would I have to substitute in a value of B first then integrate it?

Any help would be greatly appreciated.

Thanks,

Note: I will be doing this via numerical integration using MATLAB, not using pen and paper - this is not an algebra problem. I need to know how to implement the method not know the solution.
 
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Either way works; leaving the non-integrated term still in the equation or plugging in a value of B results in the same solution.
 

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