Re: TI-M: Re: Integral of x^x
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Re: TI-M: Re: Integral of x^x
Ok, I see inpart what I did wrong. so the integral of x^x I would think
should then be (x^(x+1))/(x+1), but it isn't.
It is the integral of e^(x ln x) which is the integral of x^x.
I don't know yet.
----- Original Message -----
From: "Paul Schippnick" <peschippnick@earthlink.net>
To: <ti-math@lists.ticalc.org>
Sent: Sunday, May 28, 2000 1:51 PM
Subject: Re: TI-M: Re: Integral of x^x
>
> I don't know. But it is my understanding that integral of x in e^(ln x) is
> .5x^2. But it still seems that the integral of x in e^(x ln x) is still
> e^(x ln x) or integral of x^x
> If there is a error here I don't understand it.
>
>
> ----- Original Message -----
> From: <JasonScho@aol.com>
> To: <ti-math@lists.ticalc.org>
> Sent: Sunday, May 28, 2000 1:16 PM
> Subject: Re: TI-M: Re: Integral of x^x
>
>
> >
> > In a message dated 5/28/00 9:24:03 PM W. Europe Daylight Time,
> > peschippnick@earthlink.net writes:
> >
> > > integral of a power number mostly x^n, would become x^(n+1)/n, Yes?
So
> the
> > > integral of x^x is x^x .
> > > Sinse the first part becomes x^(x+1) and x/x is the same as x^(1-1)
So
> > > x^(x+1-1) becomes x^x. Yes? So I don't really understand the
question?
> It
> > > can also be written e^(x ln x). The integral of e^n is e^n.
> > >
> >
> > integral of x^n = x^(n+1)/n holds only for constant values of n. The
> > integral of e^x is e^x, but the same is not true for e^u.
> >
> > Tell me, what is the integral of e^(ln x), using your logic? What about
> the
> > integral of x? And why aren't they the same???
> >
> > ::terminating condesension mode in 5...4...3...::
> >
> >
>
>
>
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