Integration with arbitrary constant

Question 1 : I want to intgerate f(x)=x-a where a is some artitrary constant

I use the MATLAB code

fun=(2) x-a
ans=intgeral(fun,0,1)

But this code is showing the error due to arbitary constant .I also tried by mentioning syms a .But still error exist.

Question 2: Usng the following code i generated the series solution for k=1:11

series1(x,t)=simplify(series1(x,t)+U(k)*(power(t,k-1)));

series2(x,t)=simplify(series2(x,t)+V(k)*(power(t,k-1)));

end

series1

series2

C1=zeros(1);

C2=zeros(1);

for x=1:11:101

e=(x-1);

for t=1:5:31

f=(t-1)/10;

C1(x,t)=series1(e,f);

end

end

vpa(C1,15)

The answer are displayed, but it is displayed in the different form you can check the attachment. I want continous values.

[Image Analyst]

What we get is

Invalid expression. Check for missing multiplication operator, missing or unbalanced delimiters, or other syntax error. To construct matrices,

use brackets instead of parentheses.

Your fun line is not correct syntax.

Also, the code you gave never defines a value for "a". What did you assign to a, and where did you do that assignment?

Please give the actual code you used.

[DGM]

It depends if you want to do this numerically or symbolically

% symbolically
syms x a
f = x-a;
int(f,0,1) % answer is a function of a

ans = 

% numerically
a = 0; % have to define a
f = @(x) x-a;
integral(f,0,1) % answer is just a constant

ans = 0.5000

The second part with the series stuff, I don't know, since half the code is missing.

[Image Analyst]

Probably right @DGM, so you should put it in the Answers section below so he can accept it to award you reputation points for it. @YOGESHWARI PATEL, click "Show older comments" to see his answer.

Answer 1

[DGM]

It depends if you want to do this numerically or symbolically

% symbolically
syms x a
f = x-a;
int(f,0,1) % answer is a function of a

ans = 

% numerically
a = 0; % have to define a
f = @(x) x-a;
integral(f,0,1) % answer is just a constant

ans = 0.5000

The second part with the series stuff, I don't know, since half the code is missing.

[YOGESHWARI PATEL]

Thank you for your reply.

the second code

syms x t

for k=1:10

U(k)=(-x)^k-1/(factorial(k))

end

for k=1:9

series1(x,t)=simplify(series1(x,t)+U(k)*(power(t,k-1)));

end

series1

C1=zeros(1);

for x=1:11:101

e=(x-1);

for t=1:5:31

f=(t-1)/10;

C1(x,t)=series1(e,f);

end

end

vpa(C1,15)

[DGM]

What is series1?

[YOGESHWARI PATEL]

I put a new code in which i mention the the value of coefficient of series

[DGM]

Where is it defined?

[YOGESHWARI PATEL]

comment section

[DGM]

The comment above and the original post are the only places where the variable 'series1' is mentioned.

syms x t
for k=1:10
    U(k)=(-x)^k-1/(factorial(k))
end 
for k=1:9 
    series1(x,t)=simplify(series1(x,t)+U(k)*(power(t,k-1)));
end
% ...

Here, series1 is used before it's defined. If you're trying to do something with symbolic functions, you might by over my head, but all I know is that this throws an error.