Mathematica Mathematica - using rules to assign integration limits

[izzy93]

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

 

[SteamKing]

How about 1 <= x <= 4?

 

[Bill Simpson]

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?

 

[izzy93]

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

 

[Bill Simpson]

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/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.

 

[izzy93]

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

 

[Bill Simpson]

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.

 

[izzy93]

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