how can i integrate this function ?

Hello everybody , i have the bellow function , i want to integrate it from a to b
but the result is not complete , there is no result for a , the code for this
function is :
syms y
syms bx
syms v
syms a
syms b
% c is constant
f= y / ((bx - y)^2/v + 1)^(v/2)
z = int ( f , a,b , y )
***
the result after i run it in matlabe is :
z =
-(y*(b - y))/((bx - y)^2/v + 1)^(v/2)

Respuestas (1)

Steven Lord
Steven Lord el 10 de Mzo. de 2018

0 votos

You're close. Look at the documentation for the int function. Pay close attention to the order of the input arguments and what each input argument represents.

6 comentarios

elham kreem
elham kreem el 10 de Mzo. de 2018
i want to thanks for you
i applied this code , but no result
z = int ( f , y , [a,b] )
the result was :
z =
int(y/((bx - y)^2/v + 1)^(v/2), y, a, b)
Torsten
Torsten el 12 de Mzo. de 2018
Where did you find this syntax with "[a,b]" for "int" ? It's wrong.
elham kreem
elham kreem el 12 de Mzo. de 2018
from help of matlab , as :
int(expr,var,a,b) computes the definite integral of expr with respect to var from a to b. If you do not specify it, int uses the default variable determined by symvar. If expr is a constant, then the default variable is x.
int(expr,var,[a,b]), int(expr,var,[a b]), and int(expr,var,[a;b]) are equivalent to int(expr,var,a,b).
** If I'm imposed values for a and b , also , the result as the same
elham kreem
elham kreem el 12 de Mzo. de 2018
and if i omit a ,b , the result is :
int(y/((bx - y)^2/v + 1)^(v/2), y)
Torsten
Torsten el 12 de Mzo. de 2018
This indicates that MATLAB was not able to find an analytical expression for the integral. You will have to insert numerical values for b and v, numerical values for the limits a and b and use "integral" for a numerical approximation of the integral.
elham kreem
elham kreem el 14 de Mzo. de 2018
thank you

Iniciar sesión para comentar.

Etiquetas

Preguntada:

el 10 de Mzo. de 2018

Comentada:

el 14 de Mzo. de 2018

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by