I am trying to integrate the sine-function. My goal is to get not just the value of the area inbetween a certain distance but the specific values of the integrated course.
One way to achieve this is by using cumtrapz. I want to get the same result using integral or quad. So I am wondering if there is something like cumquad?
I tried to write something for myself but it works very slow and seems to be even worse than cumtrapz. Later on I want to integrate measured data. So it won't be as simple as a sine.
Here is my current code:
a = 0; b = 10;
x = a:0.1:b;
y = 2*sin(3*x);
pp = spline(x,y);
y2=zeros(1,length(y));
y3=zeros(1,length(y));
y2(1)=integral(@(x)ppval(pp,x),x(1),x(2));
y3(1)=integral(@(x)ppval(pp,x),x(1),x(2));
for a=2:(length(y)-1)
y2(a) = y2(a-1)+integral(@(x)ppval(pp,x),x(a-1),x(a));
y3(a) = y3(a-1)+quad(@(x)ppval(pp,x),x(a-1),x(a));
end
y4=cumtrapz(x,y);
% y5=cumsum(y);
plot(x,y)
hold on
plot(x,y2,'-ro')
plot(x,y3,'-kx')
plot(x,y4,'g')
syms x % compare with analytical result
ya=2*sin(3*x);
ya5=int(ya)+(2/3);
ezplot(x,ya5)
Using
integral
I don't think there is a way to have MATLAB return the integral along the path, so you are correct in performing the integration one Δx at a time. The slowness comes from the loop and subsequent restart of every
integral
call. You can avoid the loop by posing the integral over every interval as a vector-valued function.The Math
Suppose we divide x into N-1 intervals with N total boundaries and denote an interval boundary as xn where n ∈{1,2,3...,N} such that x1 ≤ x2 ≤ x3 ... ≤ xN. Then any integral over the interval would be
Using the u-substitution:
The integral becomes:
where Δxn = xn - xn-1
The Code
So now, we can pose the interval integration of any function by specifying the lower bound xn-1, specifying the interval width Δx, and integrating from 0 to 1. The best part is that if the lower bound and interval widths are vectors, we can create a vector-valued function in terms of u and have
integral
integrate with the option'ArrayValued' = true
.The
cumsum
accounts for the fact that each integral over a given interval needs to have the value of the previous interval added to it.On my machine, this is at least order of magnitude faster than the loop version and gets better as the interval count increases.
Using
ode45
Use can also use
ode45
to perform the integration. It is not nearly as efficient as theintegral
method, but it may be easier conceptually and look cleaner. In fact,ode45
is about 10 times slower than the integral method above when required to return an absolute error on par with that ofintegral
.