110
Skrytý text:First of all, g has to be zero. Indeed, by substitution (for

, otherwise
\end{align*})
), we get
dt=\frac1x\int_0^xg(s)ds\end{align*})
,
hence
ds=0\end{align*})
and by taking the derivative, we obtain
=0\end{align*})
for all

.
Thus, the exercise can be reformulated as a problem of finding a minimum of
=\int_2^4|x^4-4x-k|dx\end{align*})
.
Note that

is a continuous function that satisfies
=\infty\end{align*})
. Moreover, since
\end{align*})
for all

and all

, where

are finite sets (the roots), we deduce that

is continusously differentiable (the integral does not see the finite sets

) and
=-\int_2^4\text{sgn}(x^4-4x-k)dx\end{align*})
.
This can be zero only if the integrand changes sign in the middle of the interval, that is, if

, i.e.,

. Due to the coercivity and continuous differentiability of

, the point

is the global minimum. Hence the answer is
=-\int_2^3(x^4-4x-69)dx+\int_3^4(x^4-4x-69)=\ldots=110\end{align*})
.
Remark: the only property of the polynomial
=x^4-4x\end{align*})
that is used is
=69\end{align*})