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Re: Antiderivatives and Definite Integrals


It's really just an example of getting the wrong general antiderivative 
when x isn't specified, but the right integral when x *is* specified.

f[x_] = Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
g[x_] := Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
Plot[{Sqrt[1 + Cos@t], f@t, g@t}, {t, 0, 2Pi}]

Bobby

On Sat, 1 Mar 2003 02:47:47 -0500 (EST), Garry Helzer <gah at math.umd.edu> 
wrote:

> The antiderivative of Sqrt[1+Cos[x]] discussed here recently (sorry, I 
> lost the thread) provides an amusing illustration of the fact that 
> Mathematica does not always evaluate definite integrals by first finding 
> an antiderivative and then substituting in the upper and lower limits. 
> (See the Mathematica book A.9.5) Make the definitions
>
> f[x_] = Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
> g[x_] := Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
>
> Then f[2Pi] is 0 (wrong) and g[2Pi] if 4Sqrt[2] (correct). Of course 
> f[x]==g[x] returns True.
>
> Garry Helzer
> Department of  Mathematics
> University of Maryland
> 1303 Math Bldg
> College Park, MD 20742-4015
>
>
>



-- 
majort at cox-internet.com
Bobby R. Treat



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