On Thu, 2010-11-11 at 09:33 -0600, Barton Willis wrote:
> How many users will be satisfied with:
>
> (%i1) integrate(sqrt((x^2-2*x+1)/x),x,0,1);
> (%o1) -4/3
>
> (%i2) integrate(sqrt(x-2+1/x),x,0,1);
> (%o2) 4/3
Notice also that:
(%i3) plot2d(sqrt((x^2-2*x+1)/x),[x,0,1]);
(%i4) plot2d(sqrt(x-2+1/x),[x,0,1]);
Both show the same plot, which in both cases is above the x-axis.
Whatever meaning it is decide to give to sqrt, plot2d(f,[x,0,1]) and
integrate(f,x,0,1) should give consistent results.
Regards,
Jaime