- #1

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Integrate[f[x], x] /. x -> {1 >= x <= 4}

which ain't working,

any help much appreciated

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- Thread starter izzy93
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- #1

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Integrate[f[x], x] /. x -> {1 >= x <= 4}

which ain't working,

any help much appreciated

- #2

SteamKing

Staff Emeritus

Science Advisor

Homework Helper

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How about 1 <= x <= 4?

- #3

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In[1]:= q[f[x],x]/.x->{1≥x≤4}

Out[1]= q[f[{1≥x&&x≤4}],{1≥x&&x≤4}]

and that doesn't look anything like

q[f[x],{x,1,4}]

that I suspect you are trying to accomplish.

So does that provide any additional information for your next attempt?

- #4

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when I put that in , comes up with an error

ln[106]= q[f[x], x] /. {1 <= x <= 4}

ReplaceAll::reps: {1<=x<=4} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing. >>

out[107] = q[E^(-x/2) x^2 Sin[5 x], x] /. {1 <= x <= 4}

Yes that's what I'm aiming for but I need to do it using rules. Is there another way of going about this?

thanks

- #5

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In[1]:= q[f[x],x]/.x->{1≥x≤4}

ln[106]= q[f[x],x] /.{1<=x<=4}

Now if you need to use replacement rules and you need to start with {1<=x<=4} then I would look at

FullForm[{1<=x<=4}]

and see if you can think of a way to use rules to turn that into

{x,1,4}

If you can use replacement rules to accomplish that then you will be half way to then using that result as another replacement to turn x into {x,1,4}.

But you have two x that you have to worry about, one inside f[x] and one inside q[ ,x]. You only want to replace the second one, not the first one.

You might look at this

http://reference.wolfram.com/mathematica/ref/Replace.html

and pay particular attention to levelspec. Levels are not exactly obvious to a new user, but with the hint that you want to pay attention to that then you might be able to figure out how to have a replacment only change the particular x that you are wanting it to and leave the other x alone.

So look at that FullForm result and see if you can solve that problem first. Then you can study levels.

- #6

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out[30]=q[{E^(-(1/2) (1 <= x <= 4)) (1 <= x <= 4)^2 Sin[

5 (1 <= x <= 4)]}, {1 <= x <= 4}]

which just substitutes in what I want the limits of x to be

In[1] = FullForm[{1 <= x <= 4}]

Out[2] = List[LessEqual[1,x,4]]

then I tried

In[3]= FullForm[{x >= 1, x <= 4}]

out[4] = List[GreaterEqual[x,1], LessEqual[x,4]]

Afraid I can't see how to turn it into {x,1,4}, is quite beyond me atm!

thanks

- #7

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In[1]:= {1≤x≤4}/.{l_<=v_<=h_}->{v,l,h}

Out[1]= {x,1,4}

So that just used your literal input to make a replacement rule.

Sometimes that won't work and sometimes it isn't obvious why.

In those cases you can sometimes use the FullForm of your

expression to make up your replacement rule.

Out[1]= {x,1,4}

So that just used your literal input to make a replacement rule.

Sometimes that won't work and sometimes it isn't obvious why.

In those cases you can sometimes use the FullForm of your

expression to make up your replacement rule.

Last edited:

- #8

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Ok it's odd but I think I see,

Thanks very much for your help

Thanks very much for your help

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