Difference between Integral and trapz

Hi, Thank you for viewing the question.

I have tried to do a comparison between Integral and trapz for a simple function of x.^2. with upper and lower limits as 9 and 0 its a simple sanity check I am trying to do

Intergal value for this function is 243.

To check trapz accuracy I ended up giving x and y values taken from the above function, like x = [0:1:9] and y = [0,1,4,9,16,25,36,49,64,81] and trapz is showing a value of 244.50.

May I know why there is such a difference and what other functions are available for me to get the integral of the above discrete data correct.

Answer 1

[Geoff Hayes]

Kushal - from trapz, this function computes the approximate integral of Y via the trapezoidal method with unit spacing. So it is an approximation only. If you increase the number of points, then you will achieve a better result...but it will still be an approximation.

>> x = linspace(0,9,1000);
>> y = x.^2;
>> trapz(x,y)

ans =

  243.0001

[Laura B.]

Hello,

I have the same problem, but the difference between the two computations is higher :

Let W2 the numerical values of an analytic function :

W2 = objfun(x,txp)

where x is a vector of parameters, and txp a vector of time coordinates

Q = trapz(W2) = 2.1616

Q1 = integral(@(t) objfun(x,t),txp(1),txp(end) = 0.0076

Why su difference ?

And what numerical value is correct ?

Thank you in advance for your feedback

Laura Brocco

[Geoff Hayes]

Laura - you probably should include the code for the objfun, the data for x, txp, etc.

[Laura B.]

Hello Geoff,

Thanks for your answer !

The Code :

Apower = 2;
D0 = 1.6;   A0 = 0.3;  phi0 = 0;  alpha_A0 = 1;  alpha_phi0 = 1;
x0 = [D0 t0 A0 phi0 alpha_A0 alpha_phi0];

    objfun = @(x,txp) x(3)./power((x(5)*txp-x(2)),Apower).*...
        sin(2*pi*x(1)*x(1)./power((txp-x(2)),2).*(txp-x(2)) + 2*pi*x(4)*(x(6)*txp-x(2)));

UB = [];
LB = [];
options = optimoptions('lsqcurvefit','MaxFunEvals',1000);
 [x,~,residual,~,~] = lsqcurvefit(objfun,x0,txp,Fxp,UB,LB,options);

W2 = objfun(x,txp);

I joint the data (txp and Fxp) to my answer.

I apologize for the dalay, and I thank you again for your help !

Laura

[Walter Roberson]

t0 is not defined.

[Laura B.]

Sorry,

t0 = -0.1357 s

[Walter Roberson]

load MyData.mat
Apower = 2;
D0 = 1.6;   A0 = 0.3;  phi0 = 0;  alpha_A0 = 1;  alpha_phi0 = 1;
t0 = -0.1357;

x0 = [D0 t0 A0 phi0 alpha_A0 alpha_phi0];


    objfun = @(x,txp) x(3)./power((x(5)*txp-x(2)),Apower).*...
        sin(2*pi*x(1)*x(1)./power((txp-x(2)),2).*(txp-x(2)) + 2*pi*x(4)*(x(6)*txp-x(2)));

UB = [];
LB = [];
options = optimoptions('lsqcurvefit','MaxFunEvals',1000);
 [x,~,residual,~,~] = lsqcurvefit(objfun,x0,txp,Fxp,UB,LB,options);

W2 = objfun(x,txp);

Q = trapz(W2);
Q1 = integral(@(t) objfun(x,t),txp(1),txp(end));

disp(Q - Q1)

TT = linspace(txp(1),txp(end),1000);
YY = objfun(x,TT);
plot(TT,YY,'b--', txp, W2, 'r*', txp, Fxp, 'k+')

Now look at the plot, especially near the beginning. You can see that there are complete rises in between W2 values. Those values appear to be broadly justified according to the original values (+)

The fit is just not that good.

[Walter Roberson]

Are there any constraints on the parameters? I am finding several solutions with fairly different parameter values... as might be expected considering the sin(). There is also a symmetry for negative and positive t0

[Walter Roberson]

For example I have here 46 different solutions all with residue < 1e-35 and negative t0; they are pretty different

-47.95089469474497	-44.16157387672567	-4.371775917311408e-17	-6.079276847115207	4.943729409719964	1.525296255775328
-44.66652754395707	-40.99039637120786	3.219072511060688e-16	-35.07524122407148	-1.485631988687895	-37.3267490861478
-42.49369284069891	-33.36824153438488	-5.876084484599276e-16	-20.7220970568693	-6.631130369829794	-22.5200415144427
-39.60246938430808	-29.00547040120762	-9.008783210618455e-18	-44.37017392200255	37.51773697209195	20.64900789748739
-38.76733218357975	-27.53736287179044	1.95633494448283e-16	-27.04420347526322	38.3427893611626	29.50731063091068
-32.35030236700637	-44.22342446981496	7.696124218853795e-15	22.75562555373097	-28.81354322878946	38.47368113696579
-30.79096259697437	-39.31240264917558	-3.802437871397855e-17	30.09061096642649	33.14764502396599	-6.400996022989665
-28.6835626333566	-37.79637250183026	-7.863324162590477e-17	-8.325892219584183	23.39211157786728	-9.366610674357929
-27.18398939298427	-41.47949785723394	-7.307779354428176e-15	10.92562159387612	49.32891305205408	-34.28578267885976
-26.2630570676281	-41.32702114061822	2.899910469426467e-18	11.97516454846699	-0.1204558920673215	-15.44979170734793
-25.56376551258081	-41.20297330998476	-3.879057950853999e-16	34.31111433937419	-6.948801788373169	13.05008982760977
-21.09104489223759	-45.55670987515589	1.337713117534706e-15	46.07473728167103	30.10215154930555	-27.4941347444438
-20.66314154632227	-40.81783439430432	8.360680221621128e-17	7.115833060872347	-15.72781568903134	-37.48636675718868
-18.88399419417758	-21.01165722499711	7.878012455897127e-17	10.52232330824396	10.11113999687696	41.99995196407257
-18.85146902477141	-43.86053189465629	-3.733897662776094e-16	41.4169990417417	19.20895301694181	10.81056419661596
-17.58704378417622	-44.54512414505932	9.901140003394243e-18	-8.240610079726025	-8.606502456372805	29.94939480735483
-14.62213860603608	-39.07607120988833	6.061475975241885e-19	-19.13992953599639	7.311584594684017	48.32465144217963
-14.116587000359	-31.70843071280881	5.458746378811086e-17	-39.02076642249535	-4.209470305827821	-32.78342364806393
-13.94906263395042	-34.14047786354747	1.814247092464174e-20	-19.13734706566127	-0.6575426312010795	-34.07058063923574
-11.33832518581703	-47.61628978204907	-4.087467080660635e-16	7.377947886194208	-22.80215379729776	-34.17090547585008
-7.141132296424756	-42.59101911843941	-1.974276536125174e-17	-36.92908790787328	19.35219211044549	-33.74913795982584
-6.615278770910967	-11.67117383782504	1.71713275908378e-18	44.86253616579738	36.93601528487618	-42.85221771017089
-4.784741763825965	-31.74533170975311	-2.547280931346719e-19	1.99415587813591	-3.535305579321786	-1.525212988475015
-2.033138146692584	-49.29328043736207	-1.699556749705114e-17	-39.52741899514895	-0.7354480896091764	-41.31830628989966
-1.72572743839612	-49.81630755186946	-7.721634246451934e-15	-6.073369146843959	32.3210046466435	-41.16002879357249
1.088568230546769	-46.16264791890128	-2.731544944591147e-18	-7.144532923379474	4.93555211005385	-21.92310385920268
2.135102195658865	-47.11213549457229	7.798874518070057e-16	-46.6690414048627	-8.833919384687489	36.70807052907853
9.140373076117212	-35.96652732219512	-1.584900825845564e-16	2.586247179487001	-6.091027852139672	23.05532149934864
11.63814502658233	-44.00326128264724	-1.110180417589272e-16	-14.90637324973984	-8.406055711676384	-32.2512228552188
13.27488353827509	-33.70957936198759	3.511506537473809e-17	-28.54061172143811	17.62530237819364	-6.91878920002956
14.06921199707027	-27.2805934249703	2.999351415451131e-18	19.46689055536654	21.32776315293494	12.95742161358538
15.25929389392785	-26.54127555037479	2.889740046490772e-17	-12.85582751642021	-25.55018982923392	7.745501376908562
21.12087578244763	-32.8827316184443	2.245295062740123e-16	-45.80200926204462	-3.629914615216504	-31.76401510335888
21.15477829965959	-40.29292321223875	-3.025337565359654e-17	-2.264513604872063	-12.24001351897262	-7.123354087992601
22.04404343686652	-49.02299250830714	2.056397134640879e-16	-41.16718115133767	13.00497428557341	-7.038357782753925
23.22819701570493	-39.64721006849439	-2.868355240527429e-16	-33.05198553053757	44.2947341073166	-35.78347486499941
24.33482480185546	-47.36089989888285	-1.650901582034252e-16	40.50810939904423	39.15977121162014	-17.6588008327631
27.51208693277401	-35.92979712636892	1.441930129744566e-16	22.56754202729954	30.95710983410491	14.48949043902964
31.65526804343262	-40.48589155541801	-2.167506016113711e-16	13.68188229315157	45.45685585537677	-41.6738308551631
35.16997203781847	-39.14254710437893	1.026874825141096e-15	-3.809446159716262	47.6458077616564	-42.73097582180128
35.44205060306072	-24.8487185871428	-2.513978647861074e-16	-41.50254282593841	-4.831122920315394	32.41990274589189
39.51773491161424	-46.07703456319031	1.096582375254929e-16	-31.80977027629155	4.2894386737609	-18.24318607233912
41.53091425697731	-30.6618715438338	2.761911410748264e-16	-41.95851345397433	15.73575587891305	37.85200983109226
43.28175010459749	-45.80110014206731	-7.97962522378623e-17	41.27390133247491	-9.615874436055448	11.58684299485161
43.6499302816635	-34.19046136891242	-2.003832586568789e-17	31.75055731961091	24.49637898981904	19.95624478802585
44.28547380793928	-26.7597154467868	-2.637435267348304e-18	-9.563851502544852	-5.365026941691676	-28.26884625771725

[Laura B.]

Hello Walter,

First of all, Thank you !

If I understand correctly, the fit is not perfect, then the curves (theoretical and experimental) are different and therefore the area (and the two values of the integral) is different.

Unfortunately, there is no "optimal" predefined theoretical function for these experimental data and what I am looking for is the better solution.

There are no constraints on the parameters, except, maybe for D0 and t0 : the solutions should be as close as possible to the original ones.

Would you have any idea how to improve this fit?