Plotting NIntegrate

Question

Plot[E^(-0.5 x) NIntegrate[Cos[t] E^(Cos[t] + 0.5 t), {t, 0, x}], {x, 0, 40}]

Mathematica evaluates this integral for each point, which takes a long time. It is possible to iterate each time step without going through the whole function. What is the correct way to do this?

Consider using NDSolve[] instead… – J. M.'s missing motivation May 27 '15 at 10:22 — J. M.'s missing motivation, May 27 '15 at 10:22
That is as answered as this question can get. Thanks. – Gappy Hilmore May 27 '15 at 10:26 — Gappy Hilmore, May 27 '15 at 10:26

score 7 · Accepted Answer · answered May 27 '15 at 10:46

7

To flesh out GuessWho's suggestion:

sol[t_] = NDSolveValue[{Cos[t] E^(Cos[t] + 0.5 t) == f'[t], f[0] == 0}, f[t], {t, 0, 40}]

Show[
  Plot[E^(-0.5 x) NIntegrate[Cos[t] E^(Cos[t] + 0.5 t), {t, 0, x}], {x, 0, 40}]
  Plot[E^(-0.5 t) sol[t], {t, 0, 40}, PlotStyle -> {Red, Dashed}]
]

Mathematica graphics

The second plot is about 300 times faster than the first one.

answered May 27 '15 at 10:46

Sjoerd C. de Vries

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…and thanks for following through! :) – J. M.'s missing motivation May 27 '15 at 11:34
my pleasure, as always – Sjoerd C. de Vries May 27 '15 at 11:40
I was am used to Evaluate[y[x]/.sol] (that's how documentation does it). Are there any difference (cause your way is far better it seems)? – Gappy Hilmore May 27 '15 at 15:54
I believe your question is answered in the Properties & Relations section of the NDSolveValue doc page. This relatively new function precisely aims at shortening this usage pattern. – Sjoerd C. de Vries May 27 '15 at 16:02

Plotting NIntegrate

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