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How to discover the integral of a polynomial expression in Scilab ?
I have that code:
clc;
clear;
x=poly(0,"x");
printf("Digite a expressão para calcular:");
mfun=input("");
printf("MENU DE OPÇÕES\n");
printf("1-DERIVADA\n");
printf("2-INTEGRAL\n");
printf("3-MAX E MIN RELATIVOS\n");
printf("4-EXTREMOS\n");
printf("5-DOMINIO\n");
printf("6-IMAGEM\n");
printf("0-SAIR\n");
opcao=input("Digite um numero:");
if(opcao==0) then
printf("\nVocê optou por sair do programa!");
end
while(opcao<>0 )
select opcao
case 1 then
if(i==0) then
z=derivat(mfun);
i=i+1;
end
if(i<>0)then
z=derivat(z);
end
disp(z);
case 2 then
printf("INTEGRAL");
case 3 then
printf("MAX MIN");
case 4 then
printf("EXTREMOS");
case 5 then
printf("DOMINIO");
case 6 then
printf("A imagem de sua função é: %d",opcao);
end
printf("\n MENU DE OPÇÕES\n");
printf("1-DERIVADA\n");
printf("2-INTEGRAL\n");
printf("3-MAX E MIN RELATIVOS\n");
printf("4-EXTREMOS\n");
printf("5-DOMINIO\n");
printf("6-IMAGEM\n");
printf("0-SAIR\n");
opcao=input("Digite um numero:");
if(opcao==0) then
printf("\nVocê optou por sair do programa!");
end
end
So far I’ve been able to make the polynomial derivative work, I can extract the derivative from the typed expression. I wonder if there is any way to do the same with integral. Someone can help me?
I’ve been through all the documentation and found some integral commands like intg, etc. But these do not make integral with polynomials (you have to put the value of the variable). If anyone has any job suggestions or any commands that might help me, I’d appreciate it!
The expected input:
f(x)=2x²+4x
The expected to leave:
x³ + 2x²
The person who denied me could explain to me what is wrong with my question?
– Maurício Z.B
i do not understand this software and not so much of mathematics to help you, but I found this http://x-engineer.org/graduate-engineering/programming-languages/scilab/define-solve-polynomials-scilab/ and this https://x-engineering.org/graduate-engineering/programming--Languages/Scilab/Solve-Definite-integral-Scilab/,.
– Guilherme Nascimento
Thanks @Guilhermenascimento I’ll try to see!
– Maurício Z.B