CODEWITHSUNDEEP
matlabsymbolic-mathnumerical-integrationsymbolic-integration
Nella
Nella

Reputation: 13

integrate function of a function

i am trying to integrate a power sum raised to power of 1.2.

The question is this integrate (((t^1)+(t^2)+(t^3))^(1.2)) from 0 to 1, with respect to t.

x=1:3
syms t
y=sum(t.^x)
fun=@(y) y^(1.2)
integral(fun,0,1)

Output as: Error using ^ Inputs must be a scalar and a square matrix. To compute elementwise POWER, use POWER (.^) instead.

But I'm not trying to compute elementwise.

Any comment/insight would be helpful. Thank you.

Upvotes: 1

Views: 74

Answers (2)

Daniel
Daniel

Reputation: 36710

You are mixing symbolic math (syms and sum) with a numeric function to evaluate the integral (integarl). While it is possible, it is typically not a good idea because you end up with the precision problems of a numeric solution and the bad performance of the symbolic math toolbox. If you want a numeric solution, do not use any function from the symbolic toolbox. If you want to solve it with the symbolic math toolbox, maybe getting a analytic result, use int from the symbolic toolbox.

To explain what happened in your case. integral evaluates your function for multiple y-values to calculate the integral, something like fun([0,.5,1]). Your function calculates y^1.2 which is not possible, you want the element-wise operation in this case.

A further problem is, that the first y you assign is unused. The y in the next line where you define fun is a new variable.

This answer does not contain a solution because I do not know if a symbolic or numeric solution is intended.

Upvotes: 1

brainkz
brainkz

Reputation: 1345

I think, in your last line you refer to the generic y that has nothing to do with y you have specified before. so, instead of fun you need fun(y). Then, since the output of your fun is symbolic expression, then you need to convert this expression to the function handle using matlabFunction. So, the final code would look like:

x=1:3
syms t
y=sum(t.^x)
fun=@(y) y^(1.2)
integral(matlabFunction(fun(y)),0,1)

Output:

1.1857

Hope that helps, good luck!

Upvotes: 3

Related Questions