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A:

See discussion here:

There is not a way to upgrade. You need to either purchase another license or just roll back to your previous copy of AutoCAD until a final release of Autodesk’s new release happens.

Q:

Let $f\in L^{\infty}(\mathbb R)$ and define $g(x)=\int_0^xf(t)dt.$ Then $g\in L^{\infty}(\mathbb R).$

Let $f\in L^{\infty}(\mathbb R)$ and define $$g(x)=\int_0^x{f(t)dt}=\int_0^x{(t\,\,f(t))dt}.$$ Then
(i) ${g}^2\in L^{\infty}(\mathbb R)$;
(ii) ${g}^3\in L^{\infty}(\mathbb R)$.

The first part is easy.
We can prove the second part by writing $${g}^3(x)=\int_0^x\int_0^{t_1}f(t_2)dt_2dt_1=\int_0^x\int_0^tf(t_2)dt_2dt_1.$$ Now, integrating by parts, $$\int_0^tf(t_2)dt_2dt_1=\left[t_1f(t_2)-\frac{1}{2}f

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A:

I did some research and it turns out it’s the default behaviour of Windows.
In Ubuntu, you have to edit the /etc/default/grub file and change the default setting to GRUB_CMDLINE_LINUX=”” as below:
GRUB_CMDLINE_LINUX=””
GRUB_DEFAULT=0
GRUB_HIDDEN_TIMEOUT=0
GRUB_HIDDEN_TIMEOUT_QUIET=false
GRUB_TIMEOUT=10
GRUB_DISTRIBUTOR=”`lsb_release -i -s 2> /dev/null || echo Debian`”
GRUB_CMDLINE_LINUX_DEFAULT=”quiet splash”
GRUB_CMDLINE_LINUX=””
GRUB_DISABLE_OS_PROBER=”true”
GRUB_RELOCATE_HOOK=”true”
GRUB_CMDLINE_LINUX_DEFAULT=”quiet splash”

After that, you need to run sudo update-grub and then reboot. This will enable you to have the problem fixed.
Hope this helps 🙂

Q:

How to find partitions of a point?

I have this partition problem but am having trouble in actually doing it.
Here it goes
Given a point $p$ in $\mathbb{R}^2$ find the four parts of $\mathbb{R}^2$ that are partitioned by the line $L$ which passes through $p$. How?
I understand that the line will pass through either the $x$-axis or the $y$-axis but i can’t find a formula (if there is one) to find the two points that divide the partitioned sections. I have done this problem before but it was for a different point and then part of another problem and now i’m thinking of doing it over again for a point.

A:

If $p$ is on the line $x=c$ for some $c\in\mathbb R$ then we have the line segment $[-c,c]$ and its symmetric, $[c,c]$.
Otherwise, if $p$ is on the line $y=c$ for some $c\in\mathbb R$, then we have the line segment $[c,c]$ and its

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