Integrate a 2D griddedInterpolant wrt one variable

2019 7 17
1 个回答
更新时间：2019 8 2
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Hello,
I would like to know how to integrate a 2D griddedInterpolant with respect to one variable.
I include an example:
[x,y] = ndgrid(-5:0.8:5);
z = x.^2 + y.^2;
F = griddedInterpolant(x,y,z);
integral(@(x) F(x,1),0,1)
The actual interpolant is from data and not availbe in a closed analytical form.
This produces an error. Is there a simple alternative?

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Torsten
Torsten 2019-7-17
编辑：Torsten 2019-7-17
x = -5:0.8:5;
y = x';
z = x.^2 + y.^2;
F = griddedInterpolant({x,y},z);
integral(@(x)F(x,1),0,1,'ArrayValued',true)
Anna Thuenen
Anna Thuenen 2019-7-17
@Torsten
Indeed, this does not return an error message.
However, it returns:
1.6080
but it should return in this setting:
1.3333
Torsten
Torsten 2019-7-17
What happens if you choose a finer discretization in x-direction:
x = -5:0.01:5
e.g.
Anna Thuenen
Anna Thuenen 2019-7-17
As mentioned: This is an example. The actual goal is to integrate an interpolant of data.
Torsten
Torsten 2019-7-17
I know, but the quality of the result depends on the denseness of points (x,y) where you can supply z(x,y).
Andrei Bobrov
Andrei Bobrov 2019-8-2
编辑：Andrei Bobrov 2019-8-2
Hi friends!
>> x = -5:0.8:5;
y = x';
z = x.^2 + y.^2;
F = griddedInterpolant({x,y},z,'spline');
integral(@(x)F(x,1),0,1,'ArrayValued',true)
ans =
1.3333
>>

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回答（1 个）

0 个投票

The integral function in MATLAB does Numerical Integration and is the approximate computation of an integral using numerical techniques. As mentioned by the @Torsten the quality depends on the denseness of points (x,y).

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