Getting error message when integrating a simple piecewise function

Question

I reviewed the following questions: question 1 and question 2, but I need to the following integral of a piecewise function:

A[1][t_] = Piecewise[{{1 - 2*t, 0 <= t <= 1/2}}, 0]
Integrate[A[1][x], {x, 0, t}]

I want the following output

Piecewise[{{t - t^2, 0 <= t <= 1/2}}, 0]

but I get the error

Integrate::pwrl: Unable to prove that integration limits {0,t} are real. Adding assumptions may help.

Also, for

A2[t_] = Piecewise[{{2*t, 0 <= t <= 1/2}, {2 - 2*t, 1/2 <= t <= 1}}, 0]

I need the following output:

Any suggestions?

Your expected output for the second case seems to be wrong. You should get 1/2 for t=1 — george2079, Commented Dec 23, 2016 at 20:58
(1) You shouldn't expand questions after people have answered, unless you weren't clear enough to begin with. (2) Your second question doesn't make sense. Guessing you might mean you want the integral of A2, then the needed output is wrong. — Michael E2, Commented Dec 25, 2016 at 14:57
@Michael E2 I need to Integral of all functions in pieceswise, separately, without consider distance. — user45459, Commented Dec 25, 2016 at 16:03
But the way the integral is defined in calculus, and in Mathematica, the result should be continuous, which your desired output is not. This will output what you want: MapAt[Integrate[#, t] &, A2[t], {1, All, 1}]. — Michael E2, Commented Dec 25, 2016 at 16:12
@Michael E2 Many many thanks. Your comment is my answer. If possible for you change comment to answer, and reply about : {1, All, 1}. — user45459, Commented Dec 25, 2016 at 16:17

Nasser · Accepted Answer · 2016-12-23 20:08:45Z

4

This works for me:

ClearAll[x, t]
A0[t_] := Piecewise[{ {1 - 2*t, 0 <= t <= 1/2}}];
Assuming[t > 0, Integrate[A0[x], {x, 0, t}]]

Mathematica graphics

answered Dec 23, 2016 at 20:08

Nasser

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$\begingroup$ I edited my question. Please review if possible. $\endgroup$
– user45459
Commented Dec 23, 2016 at 20:51
2

$\begingroup$ @user45459 I would add that Nasser is only following the advice in the error message. (Literally, it was suggesting Integrate[A[1][x], {x, 0, t}, Assumptions -> t \[Element] Reals], or rather Integrate[A[1][x], {x, 0, t}, Assumptions -> {0, t} \[Element] Reals].) $\endgroup$
– Michael E2
Commented Dec 25, 2016 at 15:00

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cyrille.piatecki · Accepted Answer · 2016-12-23 20:08:28Z

4

Ok it may be strange to put an assumption on t but this

A[1][t_] = Piecewise[{{1 - 2*t, 0 <= t <= 1/2}}, 0]
Assuming[0 <= t <= 1, Integrate[A[1][x], {x, 0, t}]]

works

answered Dec 23, 2016 at 20:08

cyrille.piatecki

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$\begingroup$ I edited my question. Please review if possible. $\endgroup$
– user45459
Commented Dec 23, 2016 at 20:51
1

$\begingroup$ even Assuming[Element[t, Reals], ..] works (including the 0 for t<0 part) $\endgroup$
– george2079
Commented Dec 23, 2016 at 20:53

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corey979 · Accepted Answer · 2016-12-24 01:04:44Z

2

There's also

A[1][t_] = Piecewise[{{1 - 2*t, 0 <= t <= 1/2}}, 0];
FullSimplify @ Normal @ Integrate[Simplify`PWToUnitStep @ A[1][x], {x, 0, t}]

Advantage: no Assumptions/Assuming.

answered Dec 24, 2016 at 1:04

corey979

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3 Answers 3