The slope multiplied by the distance would be, 
using real-valued root (the Wolfram Alpha assumption):

integral_0^(2 π) 1/100 (2 + cos(t)) cos(t)^(1/3) dt = 0.0364298
https://www.wolframalpha.com/input?i=integ_0%5E%282*pi%29+cos%28t%29%5E%281%2F3%29+%282%2Bcos%28t%29%29+%2F100+dt&assumption=%22%5E%22+-%3E+%22Real%22

But I am not sure whether the interval - π/2 to  π/2, 
as you suggest, would give me the slope. The factor
cos(t)^(1/3)  is indeed positive in this  interval, but 

the integrand itself is not symmetric, for example the 
range  π/2 to π is not a  mirror of the range 0 to π/2. Also 
the smaller interval  doesn't improve integrability:

(1) -> integrate(cos(t)^(1/3)*(2+cos(t))/100, t = -%pi/2..%pi/2, "noPole")

   (1)  "failed"

Maybe there is a function in FriCAS for some 
numerical integral algorithm? Have to check the FriCAS
manual, maybe I find something.

Maybe we can validate the Wolfram Alpha result,
show some agreement?

Waldek Hebisch schrieb am Sonntag, 22. Oktober 2023 um 00:07:38 UTC+2:

> On Sat, Oct 21, 2023 at 02:08:32PM -0700, Mild Shock wrote:
> > I think cubic root can be extended to negative real numbers,
>
> Some people define odd roots of real numbers to be negative.
> This is reasonable if you deal with equation solving as
> ambiguity is already there. However, this is problematic
> for symbolic computation. Namely in symbolic computation
> we deal with expressions and normal expressions give you
> _complex_ functions that are analytic except for at most
> countable set of singular points. Computing integrals
> is based on differential fields, that is we want to divide
> by functions. Division is fine for analytic functions
> as set of zeros is at most countable and zeros are isolated.
> But field implies that equation
>
> s^3 = x
>
> has exaxtly 3 solutions. One is usual principal root
> solution, and two other are obtained from principal
> root multiply by third degree roots of 1. To avoid
> misunderstanding: above x means expression defininig
> identity function on complex plane.
>
> Once you introduce your real root of degree 3, we no
> longer have a field, but ring with zero divisors.
> In general this causes unsolvable problems (see
> Richardson), but there is a lot of simple examples
> that can be easily handled by splitting into parts,
> each part analytic (living in its own filed).
>
> In FriCAS currently any splittings needed to handle
> such things must be introduced by user. So instead
> of
>
> integrate(sin(x)*(cos(x))^(1/3), x =0..2*%pi)
>
> you need
>
> integrate(sin(x)*(cos(x))^(1/3), x = -%pi/2..%pi/2, "noPole")
>
> and
>
> integrate(sin(x)*(cos(x))^(1/3), x =%pi/2..(3/2)*%pi, "noPole")
>
> Also, you may need to mangle one or both to get desired
> branch. Note "noPole" is really a statement that no singularities
> prevent correct computation of integral. If you use
> it in cases when there are singulatities you may get wrong
> result.
>
> > Wolfram Alpha asks me whether I want principal root or
> > real-valued root. Maybe FriCAS has also such a feature somewhere?
>
> No.
>
> > Maybe should read the FriCAS manual.
> > 
> > Quick gloss over the FriCAS manual gives me, showing
> > the difference between principal and real-valued root:
> > 
> > (2) -> complexSolve(x^3 = -1/8, 0.0001)
> > 
> > (2)
> > [x = 0.25 - 0.4330127025_0231027603 %i,
> > x = 0.25 + 0.4330127025_0231027603_1494140625 %i, x = - 0.5]
> > 
> > (3) -> solve(x^3 = -1/8, 0.0001)
> > 
> > (3) [x = - 0.5]
> > 
> > But how tell integral what root to use?
>
> In general that is extremaly messy. If roots are independent,
> then in a sense it does not metter, result is valid for
> each choice of roots. But once you get to things like
>
> sqrt(x*y) - sqrt(x)*sqrt(y)
>
> situation gets more tricky. Also "obvious" properies fail:
> "compute integral and then plug in numbers for parameters" is not
> equivalent to "plug in numbers for parameters and then compute
> integral". As indicated, in FriCAS those troubles are left
> to user. User may split integral into parts and resolve
> dependencies between roots.
>
> -- 
> Waldek Hebisch
>

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