# Integration with arbitrary constant

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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.

##### 3 Comments

Image Analyst
on 27 Nov 2021

### Accepted 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

% numerically

a = 0; % have to define a

f = @(x) x-a;

integral(f,0,1) % answer is just a constant

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

##### 6 Comments

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.

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