How to calculate the limit of an integral?

조회 수: 2 (최근 30일)
Jonathan Mayers
Jonathan Mayers 2016년 6월 23일
편집: madhan ravi 2019년 2월 22일
Is it possible to calculate the value d0 of the below equation? If so, how may I go about doing so?

채택된 답변

Roger Stafford
Roger Stafford 2016년 6월 23일
Assuming you already know the numerical values of rho, P(K-1), alpha, r0, r, and the function f(x,r), you should be able to solve for d0 with the use of matlab’s ‘fzero’. For that purpose you need to create a function for use in ‘fzero’ which computes the difference in both sides of your equation for input values of d0. Using this you can make a plot of this function’s values as d0 varies and note where it crosses zero. Then you can pick two points on either side of the crossing point and set ‘fzero’ to work with those two points as your initial two-point estimate and locate the crossing point with accuracy. Of course the above function will have to call on matlab’s ‘integral’ function three times for each evaluation.

추가 답변 (1개)

Juan Jose Ortiz Torres
Juan Jose Ortiz Torres 2019년 2월 22일
In my case, I need to get the final limit of an line lenght integral. so I put this on matlab:
syms x a1
solve(int(x^2 + 2*x + 3,x,0,a1)==4,a1)
And I got:
ans =
root(z^3 + 3*z^2 + 9*z - 12, z, 1)
root(z^3 + 3*z^2 + 9*z - 12, z, 2)
root(z^3 + 3*z^2 + 9*z - 12, z, 3)
So I use the fzero function:
fzero(solve(int(x^2 + 2*x + 3,x,0,a1)==4,a1))
But it doesn't work
  댓글 수: 1
madhan ravi
madhan ravi 2019년 2월 22일
편집: madhan ravi 2019년 2월 22일
Remove fzero and add 'maxDegree’,3) in solve or just simply use vpasolve() instead of solve()

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