I am struggling with integrating a function.

I have this function that I have to integrate:

And I added inline function to matlab as such:

f=inline('sqrt(1+((66*x)^2/((1.2*x^2)+3))','x')

f =

Inline function:

f(x) = sqrt(1+((66*x)^2/((1.2*x^2)+3))

However, it gives me this error:

> int(f(x),x)

Error using inlineeval

Error in inline expression ==> sqrt(1+((66*x)^2/((1.2*x^2)+3))

Error: This statement is incomplete.

Error in indexing (line 23)

INLINE_OUT_ = inlineeval(INLINE_INPUTS_, INLINE_OBJ_.inputExpr, INLINE_OBJ_.expr); %#ok<DILEVAL>

How can I fix this? I would be so grateful for a help!

[Torsten]

Integration variable and upper limit of the integral both have the same name (x). That doesn't make sense.

[Dyuman Joshi]

As far as I know, that's a common occurence; basically obtaining another function with the upper limit as the independent variable.

syms x z

y(x) = x^2+x+1;
i1 = int(y,x,0,x)

i1 = 

i2 = int(y,x,0,z)

i2 = 

[Torsten]

Maybe common, but often misleading from my experience.

Especially if you want to define a function of the upper limit:

F = @(x)integral(@(x)x.^2,0,x) 

It works, but it's quite confusing in my opinion.

[Dyuman Joshi]

I'd say rather confusing than misleading.

What you have wrote is the numerical version of the symbolic integration; idk what's your point with that but it does work.

[Torsten]

In MATLAB, many things "work", also the unintended ones.

Coming from the FORTRAN generation, I often wonder if this is help or curse for the programmers.

But that's an old question discussed many times in the forum.

Answer 1

[John D'Errico]

Don't use inline functions. They are slow, inefficient things, introduced long ago, in a galaxy far, far away. Instead use function handles, or anonymous functions. (They remain unnamed. Sort of the Voldemort of functions.)

Next, decide if you are going to use a numerical integration utility, thus integral, or a symbolic one, so int. The two are not the same. You used int, but then used it on a function handle.

f = @(x) sqrt(1+(66*x./(1.2*x.^2+3)).^2);

Note my use of the dotted oeprators. They are important to make a numerical code work. Also, see this is a function handle. We can evaluate f at any numerical value for x.

f(2)

ans = 16.9526

Or, we can substitute in a symbolic variable...

syms X
f(X)

ans = 

Since you have not specified limits to the integral, I will presume you want a symbolic result.

int(f(X),[0,X])

ans = 

A problem is that MATLAB did not find an analytical solution for that integral, so it just returned the original expression. That may force you to use numerical integrations, but then you won't get a pretty result.

Had you wanted to perform a numerical integration, you would do it like this:

integral(f,0,3)

ans = 42.1297

There, with limits of [0,3] for the integral, or

vpaintegral(f(X),0,3)

ans = 

42.1297

Finally, you might have decided to try other methods. For example, if we do a Taylor series on f, we get this:

taylor(f(X),'order',20)

ans = 

Now, we might have decided to integrate the Taylor series, we would expect it would not be viable except for small upper limits on X. For example...

vpa(int(taylor(f(X),'order',20),0,.01))

ans = 

0.010080087111363533488105224583383

vpaintegral(f(X),0,0.01)

ans = 

0.0100801

And that did reasonably well. But it will fail for even moderately sized upper limits.

[Betul]

I see thank you so much! It seems like it is not able to further solve the integration, as it does not take the funciton out.

[Walter Roberson]

Now, we might have decided to integrate the Taylor series, we would expect it would not be viable

except for small upper limits on X

format long g
f = @(x) sqrt(1+(66*x./(1.2*x.^2+3)).^2);
syms X UB
T = taylor(f(X),'order',20)

T = 

I = int(T, X, 0, UB)

I = 

ub = solve(I == realmax(), 'real', true)

ub = 

double(ub)

ans = 
      1.32548992166858e+15

Not such a small upper bound.

... Though you lose precision badly at much much smaller values.

Answer 2

[Star Strider]

‘How can I fix this?’

Be certain that tthe expression matches the symbolic expression in the figure. (It currently does not.) Then, be sure that the parentheses enclose the correct sub-expressions and the entire expression since you are taking the square root of all of it.

Also, see the documentation section on Anonymous Functions.

[Walter Roberson]

f=inline('sqrt(1+((66*x)^2/((1.2*x^2)+3))','x')
%       1     2  34    3   45       4  32     1

... should be 0 at the end of the line.