Given $$f(x)=\begin{cases}x^2-1,&x<1,\\\ln(x),&x\geq1,\end{cases}$$ find the area between $f(x)$ and $x$-axis to the right of the absolute extremum.
The $x$-axis is the equation $y=0$. We have the formula of $f$. Also, I have asked about the absolute extremum of $f$ and I have received a proof that $x=0$ is the (unique) absolute minimum of $f$.
With all this, I am not able to understand what "$x$-axis to the right of the absolute extrema" means. Does it mean "The range $x\geq0$"?
Perhaps we can graph all the involved parts:
and see that we need to find $$\int_0^\infty f(x)\,\mathrm{d}x.$$ For $x<1$ we have $f(x)\leq0$, and for $x\geq1$ we have $f(x)\geq0$. Hence, the area is $$A=-\int_0^1(x^2-1)\,\mathrm dx+\int_1^\infty\ln(x)\,\mathrm dx.$$ But $\int_1^\infty\ln(x)\,\mathrm dx$ diverges, so $A=\infty$, but it does not make sense at all.
Where is my mistake or what do you interpret by "$x$-axis to the right of the absolute extrema"?
Thanks!!