integral

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Nasir Qazi
Nasir Qazi 2012 年 5 月 5 日
can I able to integrate a function with some limits a, b but without define the values to the limit , e.g f(x) = x^3 -3*x , having limits x1, x2 but I don't want to assign the values to it . can I integrate it in Matlab

回答 (2 件)

Walter Roberson
Walter Roberson 2012 年 5 月 5 日
If you have the symbolic toolbox you can use int()
  1 件のコメント
Nasir Qazi
Nasir Qazi 2012 年 5 月 5 日
can you solve the above example just to have an idea , I knw about syms but don't know, we can use it for limit integrals

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bym
bym 2012 年 5 月 5 日
syms x a b
>> f = x^3-3*x;
>> int(f,a,b)
ans =
-((a + b)*(a - b)*(a^2 + b^2 - 6))/4
  5 件のコメント
Nasir Qazi
Nasir Qazi 2012 年 5 月 6 日
% similarly here the answer is wrong
syms H R T a b
f = int((H/R*T^2),T,a,b)
f =
-(H*(a^3 - b^3))/(3*R)
% where 'H' , 'R', are constants and 'a', 'b' are limits and integrate it with respect to 'T' which on hands calculation is different and here it is wrong , why ?
Walter Roberson
Walter Roberson 2012 年 5 月 11 日
That integral for f is certainly correct, as can be seen by removing the constant and realizing one is really only dealing with the integral of T^2, which of course yields 1/3*T^3
((T - T0)*(2*cp0 + T*b - T0*b))/2 is correct for the integral of cp with respect to T over T0 to T, but only provided that b(T-T0) is interpreted as meaning b*(T-T0) rather than b indexed at T-T0.

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