Why I get different values when I integrate a function (by my hand-Displacement Positive & Negative, cumtrapz-Displacement Positive)?

Question

Bob 2016-4-12

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https://ww2.mathworks.cn/matlabcentral/answers/278631-why-i-get-different-values-when-i-integrate-a-function-by-my-hand-displacement-positive-negative

评论： Star Strider 2016-4-13

采纳的回答： Star Strider

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Why I get different values when I integrate a function (by my hand, cumtrapz)?

As you can see below, there is a very big difference when I integrate the function with cumtrapz.

%% Simple

t = 0:1:5; % Time (s)
a = 10*cos(20*t); % Acceleration (m/s^2)
VALUES:
10
4,08082061813392
-6,66938061652262
-9,52412980415156
-1,10387243839048
8,62318872287684
v = sin(20*t)/2; % Velocity (m/s)
VALUES:
0
0,456472625363814
0,372556580239674
-0,152405310551108
-0,496944326961688
-0,253182820554879
x = -cos(20*t)/40; % Displacement (m)
VALUES:
-0,0250000000000000
-0,0102020515453348
0,0166734515413065
0,0238103245103789
0,00275968109597619
-0,0215579718071921

%% Cumtrapz

t = 0:1:5; % Time (s)
a_cumtrapz = 10*cos(20*t); % Acceleration (m/s^2) with Cumtrapz
VALUES:
10
4,08082061813392
-6,66938061652262
-9,52412980415156
-1,10387243839048
8,62318872287684
v_cumtrapz = cumtrapz(t,a_cumtrapz); % Velocity (m/s) with Cumtrapz
VALUES:
0
7,04041030906696
5,74613030987261
-2,35062490046448
-7,66462602173550
-3,90496787949232
x_cumtrapz = cumtrapz(t,v_cumtrapz); % Displacement (m) with Cumtrapz
VALUES:
0
3,52020515453348
9,91347546400327
11,6112281687073
6,60360270760734
0,818805756993430

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Answer 1

Star Strider 2016-4-12

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When you evaluated your hand-calculated analytic integrals, you did not subtract the initial value of the function, as is necessary in evaluating integrals.

If you give cumtrapz a vector of smaller intervels, it performs quite well:

t = linspace(0, 5, 1E+6);
ad = 10*cos(20*t);                              % The ‘cumtrapz’ Evaluations
ade = ad(end)
vd = cumtrapz(t, ad);
vde = vd(end)
xd = cumtrapz(t, vd);
xde = xd(end)
va = sin(20*t)/2;                               % The Analytical Evaluations
vae = va(end)-va(1)
xa = -cos(20*t)/40;
xae = xa(end)-xa(1)

These yield:

vde =
  -253.1828e-003
xde =
     3.4420e-003
vae =
  -253.1828e-003
xae =
     3.4420e-003

Otherwise, it’s not a fair comparison. in your initial post Your analytic solutions assume infinitesimal intervals, and you’re asking cumtrapz to use very large intervals. Give cumtrapz a level playing field, and evaluate your analytic solutions correctly, and the two are comparable.