Solve question

[lrdheat]

I discovered that my TI-PRO 36 allows definite integrals to be used in it's solve app. How would I accomplish the same integral on the Prime?

Integral from 1 to x of (x^2+x+1)/x^(1/2) such that the integral produces a value of 10? (the correct value for x is ~2.987)

[Tim Wessman]

Solve app - enter it with the values from 1 to N or something like that (you can't use X as both the variable of integration, and the limit). Solves it instantly as far as I can tell.

I'll put that on the list of items to investigating as to why it doesn't like the X being used in both places. That may be unintentional and we can improve that.

[Manolo Sobrino]

No Tim, that is (or should be) completely intentional, unless you understand that the X in the integrand is a dummy variable, but this is in fact an abuse of notation as a definite integral in which you integrate over a variable which is also a limit makes no sense(*).

You are dealing with functions defined by integrals. The function depends just on the limit(s) of the integral and the proper way to write that is by using different variables for the integrand.

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(*) BTW Mathematica does this... and it's dumb, because imagine that you define the function:

F[y_]:=Integrate[x, {x, 0, y}]

Now you evaluate, for instance at y=2, and you get:

F[2]=2

Which is fine.

As we are happy with Mathematica understanding:

Integrate[x, {x, 0, x}] = x^2/2

Let's do the same

F[x_]:=Integrate[x, {x, 0, x}]

So far, so good, but let's use this at x=2

F[2]

... and of course we get:

Integrate::ilim: "Invalid integration variable or limit(s) in {2,0,2}."

As you are trying to do this:

Integrate[2, {2, 0, 2}]

Maybe abusing notation is convenient for a while, but in a CAS it will end badly.

[CR Haeger]

Hi. I've been able to do the following without much issue in CAS

solve(int(x^2,x,0,a)=10) yields [30^(1/3)]
fsolve(int(x^2,x,0,a)=10,a,0..10) yields [3.107...]

Also works if you replace a with x everywhere.

Does not seem to work if you use math template for the integral. I used toolbox 2 2.

Edited 30^(1/3).

[lrdheat]

Hi

6^(1/3) is not equal to ~3.18...why the different results?

I was using template...will try alternative

[Tim Wessman]

(05-23-2014 08:04 AM)Manolo Sobrino Wrote:  No Tim, that is (or should be) completely intentional, unless you understand that the X in the integrand is a dummy variable, but this is in fact an abuse of notation as a definite integral in which you integrate over a variable which is also a limit makes no sense(*).

Indeed. I am in 100% agreement there. I was thinking more along the lines of making that some sort of error message to enforce/educate that you shouldn't be doing that rather then making it allowed. It was just very poorly worded being written right before bed.

Right now, the X variable doesn't show up as one you can solve for (correctly) but I can see how this would be confusing since there is no explanation as to why it isn't there. Hence a bit more information/error would be a helpful thing. That is more what I was referring to instead.

[CR Haeger]

(05-23-2014 02:02 PM)lrdheat Wrote:  Hi

6^(1/3) is not equal to ~3.18...why the different results?

I was using template...will try alternative

My keying,thumbing error. Should read 30^(1/3) which I just corrected.