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.
