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Mathematica  using rules to assign integration limits 
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#1
Feb1813, 12:18 PM

P: 30

I'm trying to find the integration of my function f[x] from 1 to 4 using the indefinite integral and inserting limits using rules. I am not sure how to insert the limits with rules, have been playing about with the following
Integrate[f[x], x] /. x > {1 >= x <= 4} which ain't working, any help much appreciated 


#2
Feb1813, 01:57 PM

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How about 1 <= x <= 4?



#3
Feb1813, 02:24 PM

P: 1,028

Try using some undefined function q, instead of Integrate, that will do nothing at the moment, just to see what actually happens with your substition.
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
Feb1813, 03:28 PM

P: 30

Mathematica  using rules to assign integration limits
Hello Bill,
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
Feb1813, 07:39 PM

P: 1,028

The reason for your latest error is you left out the "x>" that I had. Compare these:
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/mathema...f/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
Feb1913, 11:44 AM

P: 30

In[32] = q[f[x], x] /. x > {1 <= x <= 4}
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
Feb1913, 02:18 PM

P: 1,028

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. 


#8
Feb1913, 03:45 PM

P: 30

Ok it's odd but I think I see,
Thanks very much for your help 


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