## Finding the area under a polynomial equation in MATLAB

### mr mo (view profile)

on 30 Apr 2018
Latest activity Commented on by mr mo

on 30 Apr 2018

### John D'Errico (view profile)

Hi. I have two vectors like these.
X=[0 0.0300 0.0333 0.0469 0.0575 0.0649 0.0960 0.1262 0.1610 0.1717 0.2017 0.2411 0.2711 0.3000 0.3035 0.3368 0.3711 0.4068 0.4411 0.4758 0.5253 0.5553 0.5853 0.6000]
Y=[0 126.2766 140.0068 190.5906 211.6557 219.3601 236.8709 253.2271 266.4663 269.3085 273.6913 279.3095 283.5479 286.1191 286.2725 282.7424 277.6845 272.1010 266.4556 260.1458 248.5416 241.1381 233.2857 230.3781]
I use the polyfit command to fit a polynomial curve of degree 6 to my data-set.
B=polyfit(X,Y,6)
Now my question is how can I calculate the area under this polynomial curve?
Thanks a lot.

### John D'Errico (view profile)

on 30 Apr 2018
Edited by John D'Errico

### John D'Errico (view profile)

on 30 Apr 2018

How do you doit? Basic calculus?
X=[0 0.0300 0.0333 0.0469 0.0575 0.0649 0.0960 0.1262 0.1610 0.1717 0.2017 0.2411 0.2711 0.3000 0.3035 0.3368 0.3711 0.4068 0.4411 0.4758 0.5253 0.5553 0.5853 0.6000];
Y=[0 126.2766 140.0068 190.5906 211.6557 219.3601 236.8709 253.2271 266.4663 269.3085 273.6913 279.3095 283.5479 286.1191 286.2725 282.7424 277.6845 272.1010 266.4556 260.1458 248.5416 241.1381 233.2857 230.3781];
xinterp = linspace(min(X),max(X),100);
P = polyfit(X,Y,6);
plot(X,Y,'ro',xinterp,polyval(P,xinterp),'b-')
As models go, I'd call it your basic piece of crapola. And DON'T use a higher order polynomial, thinking it will do better. Just use a spline instead, if you ABSOLUTELY insist on a model. Actually, pchip (shape preserving interpolant) would be my choice here.
If I had to guess, that looks like essentially a piecewise linear curve, with 5 linear segments. But why do you need to integrate a poorly fit polynomial at all?
The area under the curve is...
trapz(X,Y)
ans =
149.16
In fact, this is arguably a better estimator of the area than the integral of that crappy polynomial model.
What do I get for the pchip integral? My SLM toolbox has a nice little utility to integrate a spline. But it is easily enough done without that tool.
pp = pchip(X,Y);
integral(@(x) ppval(pp,x),min(X),max(X))
ans =
149.28
That is not too far off from what trapz gave. Bother are about as good as you can get.
Can we integrate the polynomial model? Again, of course. As I said, basic calc suffices. (Or, you could use polyint.)
P = polyfit(X,Y,6);
Pint = [P./(7:-1:1),0];
diff(polyval(Pint,[min(X),max(X)]))
ans =
149.78
Or if you are feeling lazy,
integral(@(x) polyval(P,x),min(X),max(X))
ans =
149.78
In any event, I'd use what pchip or what trapz gave here as better estimators.

mr mo

### mr mo (view profile)

on 30 Apr 2018
Thanks a lot. This works very well.

### James Tursa (view profile)

on 30 Apr 2018

Just integrate the polynomial and evaluate at the endpoints. E.g.,
IB = polyint(B);
area_under_poly = polyval(IB,X(end)) - polyval(IB,X(1));

mr mo

### mr mo (view profile)

on 30 Apr 2018
Thanks but if I have these vectors, the calculated area must be 102.5 but this code return 112.5
x=[0 10 15]
y=[0 10 11]
B=polyfit(x,y,2)
James Tursa

### James Tursa (view profile)

on 30 Apr 2018
By hand (using the fact that x(1)=0):
>> (1/3)*B(1)*x(end)^3 + (1/2)*B(2)*x(end)^2 + B(3)*x(end)
ans =
112.5000
Why do you think the answer must be 102.5?
mr mo

### mr mo (view profile)

on 30 Apr 2018
Because it is obvious and can be seen by using the plot command
plot(x,y,'-rs')
grid on
if the vectors are like these
x=[0 10 15]
y=[0 10 15]
then the area will be equal to 112.5