Integration with arbitrary constant

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YOGESHWARI PATEL on 27 Nov 2021
Commented: DGM on 28 Nov 2021
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 on 27 Nov 2021
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.

DGM on 27 Nov 2021
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.
DGM on 28 Nov 2021
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.