how to get a integral vector result very fast with the upper limits which are supplied in an array?

For exmaple，I have a vector x = [1.2, 3.0, 2.5, 4] which is integral upper limit and a function:

tempFunction = @(tVariable)log(1 + 2 * tVariable .^ 3);

Now I want to calculate the definite integral for the function in the intervals:,[0,1.2], [0,3.0] , [0,2.5] , [0,4]。

If I use loops, I know how to do it：

for index = 1 : 4

    result(1, index) = quad(tempFunction, 0, x(1, index))；

end

Now the problem is x has a dimension of 1 by 14556，using "for" loop is too slow. And I am doing optimization problem. So there is a lot of outer loops in fmincon which called this integral loop. It made my program very very slow.

Is there any method that I can get the integral result fast without using loops? I checked quad and quadv commands, It seems they are all used to handle upper limit is a constant number.

Answer 1

[Andrei Bobrov]

try

result = arrayfun(@(x)quad(tempFunction, 0, x),x);

[X]

this is a faster answer, but I think Teja Muppirala's is much faster.

Answer 2

[Teja Muppirala]

Using ODE45 to do the integration might be faster than using quad, particularly if your x is large:

tempFunction = @(tVariable,y)  (log(1 + 2 * (tVariable*x(:)) .^ 3)).*x(:);
[~,result] = ode45(tempFunction ,[0 0.5 1],[0 0 0 0]');
result = result(3,:);

Also, if you can have the tolerance for integration set to be more rough, then it should finish faster as well.

[Teja Muppirala]

That actually worked better than I expected!

tic;
x = rand(1,14556);
tempFunction = @(tVariable,y)  (log(1 + 2 * (tVariable*x(:)) .^ 3)).*x(:);
[~,result] = ode45(tempFunction ,[0 0.5 1],zeros(size(x)));
result = result(3,:);
toc;

[Andrei Bobrov]

+1

[X]

tempFunction = @(tVariable,y) (log(1 + 2 * (tVariable*x(:)) .^ 3)).*x(:);

[~,result] = ode45(tempFunction ,[0 0.5 1],zeros(size(x)));

I don't understand the function

tempFunction = @(tVariable,y) (log(1 + 2 * (tVariable*x(:)) .^ 3)).*x(:);

why you multiply it with an extra "x(:)" at the end of tempFunction

is it be related with definition of deferential equations?

[X]

your answer is the best one.

[Teja Muppirala]

This is the chain rule. I make a change of coordinates to make the limits of integration 0 to 1, and in doing this, I need to multiply by X.

I = integral of f(t) dt from 0 to X

Let X*u = t, then

I = integral of f(X*u) dt from 0 to X

I = integral of f(X*u)*(dt/du) du from 0 to 1

I = integral of f(X*u)*X du from 0 to 1

[X]

Thanks!