Help to Integral Calculus Solutions & Examples

Is it true that $\int <!--\; \int \; -->t\frac{dF}{d\ln <!--\; \; -->t}d\ln <!--\; \; -->t=\int <!--\; \int \; -->\frac{dF}{dt}dt$

Consider the initial value problem
$\{\begin{array}{l}{y}^{\prime }=1-<!--\; -\; -->y^2,t>0\\ y(0)=\frac{e-<!--\; -\; -->1}{e+1}.\end{array}$
Is there a way I find a limit on the mesh-size $h$ such that the explicit Euler method is stable? Do I need to solve the IVP to find it?
Thanks in advance.
Edit: After looking through a number of textbooks and online resources I can see that they apply a method to a linear test problem and then solve $|1+h\mathrm{\lambda <!--\; \lambda \; -->}|⩽<!--\; ⩽\; -->1$. Is there any way I can do something similar to solve for $h$ in this nonlinear case.

Finding the counterclockwise nearest point in $\mathcal{O}(\log <!--\; \; -->(n))$ time
Given a point q, direction w and point set P with convex hull $(v_1,\dots <!--\; \dots \; -->,v_n)$ known, give an algorithm to find the point in P that (q,w) touches first when rotated counterclockwise in $\mathcal{O}(\log <!--\; \; -->(n))$ time.
REMARK: We can assume that $(v_1,\dots <!--\; \dots \; -->,v_n)$ is sorted counterclockwise.
My idea would be the following: Since $(v_1,\dots <!--\; \dots \; -->,v_n)$ is sorted we can modify the approach of Binary Search to get the required $\mathcal{O}(\log <!--\; \; -->(n))$ bound. I suppose we have to test the angles between $(q,w,v_i)$ repeatedly, but I do not see how to formulate this clearly.

Graph of $\frac{e^x-<!--\; -\; -->1}{x}$
i am trying to draw the graph of $f(x)=\frac{e^x-<!--\; -\; -->1}{x}$.
its domain is $\mathbb{R}$-0 but its limit is 1 in the neighbourhood of zero.
Also $li{m}_{x\to <!--\; \to \; -->-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}f(x)=0$ so negative X axis is asymptote.
also $\frac{df(x)}{dx}=\frac{(x-<!--\; -\; -->1)e^x+1}{x^2}$
so f(x) is increasing in $(0\mathrm{\infty <!--\; \infty \; -->})$ and decreasing in $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}0)$
now the ambiguity is $f(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->0$ and $f(0^{-<!--\; -\; -->})\to <!--\; \to \; -->1$ but f(x) is decreasing in negative interval.

When applying Richardson Extrapolation to Euler Method, when isn't the local truncation error of h^2 used to get (4*Euler(h/2) - Euler(h)) / 3 instead of using the global error of h to get 2*Euler(h/2) - Euler(h)?

Calculus application of derivatives
Let $g(x)=2f(x/2)+f(2-<!--\; -\; -->x)$ and $f^{″}(x)<0$ for all $x\in <!--\; \in \; -->(0,2)$. If g(x) increases in (a,b) and decreases in (c,d) find the values of a, b, c and d.
What I thought was a little graphical approach. I figured it out that if f′′(x) is less than zero then f′(x) would be decreasing (not considering the concavity) and the behaviour of g(x) would depend upon f(x). But I couldn't figure out the exact intervals.

What is
$${\int <!--\; \int \; -->}_1^3\frac{e^{\frac{1}{x}}}{x^2}dx?$$
How do you approach solving this?

I have been studying the Backward Euler Method, and I am have having some of problems representing the stability of the method of some particular ODEs.
If I am thinking correctly this method will be stable if the solutions of the ODE are stable, which means that $Re(\lambda )<0$ and $h>0$, (h= step size).One of my problems is that I don't know how to calculate the sign of the $\mathrm{\lambda <!--\; \lambda \; -->}$ of this non-linear equations.
a) $y^{\prime }=1/(t+y),y(0)=1$
b) $y^{\prime }=-<!--\; -\; -->(1+t)exp(y),y(0)=1$
If there is any positive λ, then a step size big enough will be stable even though it might not be a good representation of the solution?
Or will the solution be automatically unstable if the $Re(\lambda )>0$?
Thank you very much, any help is very appreciated.

Question about finding where the function increases and decreases on $f(x)=\frac{1}{x}$.
$f(x)=\frac{1}{x},x\ge <!--\; \ge \; -->1$
I have been staring at this equation for a bit. Things I'm confused on.
the derivative of this is: $f^{\prime }(x)=\frac{-<!--\; -\; -->1}{x^2}$ now, how am I supposed to find where this derivative increases/decreases? Do I find the critical points first? by setting the derivative to 0? or do I solve it like $\frac{-<!--\; -\; -->1}{x^2}>0$ cross multiply to make it: $-<!--\; -\; -->1>x^2$ and if so once I square this does it make the result $x=-<!--\; -\; -->1,x=1$? I'm really lost here and it seems like it should be easier.
Does setting the derivative to > or < or = and solving for the x give a critical point?

The question is to prove that error order of backward euler method is $o(h)$ We know that in backward euler method
$f_i^{\prime }=\frac{f_i-<!--\; -\; -->f_{i-<!--\; -\; -->1}}{h}$
By using taylor seris we can get
$f(x)=f(x_i)+(x-<!--\; -\; -->x_i)f^{\prime }(x_i)+\frac{(x-<!--\; -\; -->x_i{)}^2}{2!}{f}^{″}(x_i)+....$
By putting $x=x_{i-<!--\; -\; -->1}$ i get :
$\frac{f_i-<!--\; -\; -->f_{i-<!--\; -\; -->1}}{h}+f^{\prime }i=\frac{h}{2!}{f}_i^{″}+\frac{h^2}{3!}{f}_i^{‴}+...$
Which is not what i want to get Cause the error is
$\frac{f_i-<!--\; -\; -->f_{i-<!--\; -\; -->1}}{h}-<!--\; -\; -->f_i^{\prime }$
Any help ? I could prove that $o(h)$ is the order of error of forward Euler method by using $x=x_{i+1}$ But for the backward method it seems it doesn’t work

How do I get an integer from a polygon equation, which usually returns fractions?
Known: edge length
Unknown: number of edges
Radius should increase of decrease to an interval to ensure number of edges in Polygon is divisible by 1
E.g. edge length is 100mm, number of edges unknown, radius is 6m. How do I find the closest radius to 6m which gives an integer, divisible by 1, for a realistic number of edges.
Previous Question
If a regular polygon has a fixed edge length, can I know how many edges it has by knowing the length from corner to its center?
and it's answer
the radius of the polygon, and it has the formula
$r=\frac{s}{2\sin <!--\; \; -->\left(\frac{180°}{n}\right)}$
where s is the side length of the polygon and n is the number of sides. So given r and s, you can simply solve the above equation for n.
It's worth pointing out that when you solve for n there's no guarantee that it will turn out to be an integer, and hence correspond to a regular polygon
I have no math background, not even enough to know which tags to attach. How do I solve for these intervals of radii?

Finding the intervals of increase and decrease of $\frac{x^4-<!--\; -\; -->x^3-<!--\; -\; -->8}{x^2-<!--\; -\; -->x-<!--\; -\; -->6}$.
I tried to find the derivative by the quotient rule to obtain the critical points but the formula was getting complicated, I know that $D_f=\mathbb{R}\setminus <!--\; \setminus \; -->\{-<!--\; -\; -->2,3\}$ but then what?

If:
$$\mathrm{\lambda <!--\; \lambda \; -->}={\int <!--\; \int \; -->}_0^1\frac{dx}{1+x^3};$$
Then evaluate:
$$p=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\left(\frac{\underset{r=1}{\overset{n}{\prod <!--\; \prod \; -->}}(n^3+r^3)}{n^{3n}}\right)}^{1/n};$$

Vertical asymptotes if any? How do you find it?
$$f(x)=\frac{4x}{4x^2-<!--\; -\; -->2}$$

Interval of monotonicity of the function $f(x)=\frac{\ln <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}+x)}{\ln <!--\; \; -->(e+x)}$.
Find the exact interval in which the function $f(x)=\frac{\ln <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}+x)}{\ln <!--\; \; -->(e+x)}$ is increasing and decreasing.
My try: The domain of the function is $(-<!--\; -\; -->e,\mathrm{\infty <!--\; \infty \; -->})$
The derivative is: $f^{\prime }(x)=\frac{(e+x)\ln <!--\; \; -->(e+x)-<!--\; -\; -->(\mathrm{\pi <!--\; \pi \; -->}+x)\ln <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}+x)}{(\mathrm{\pi <!--\; \pi \; -->}+x)(e+x){\ln }^2<!--\; \; -->(e+x)}$.
Since due to the domain we have $e+x>0,\mathrm{\pi <!--\; \pi \; -->}+x>0$ and trivially ${\ln }^2<!--\; \; -->(e+x)>0$.
Now it is sufficient to check the nature of numerator above.
Let $p=e+x,q=\mathrm{\pi <!--\; \pi \; -->}+x$.
The numerator is in the form $p\ln <!--\; \; -->(p)-<!--\; -\; -->q\ln <!--\; \; -->(q)$ where $p<q,\mathrm{\forall <!--\; \forall \; -->}x$.
Now we know that the function $h(x)=x\ln <!--\; \; -->x$ is increasing in $[\frac{1}{e},\mathrm{\infty <!--\; \infty \; -->})$.
Case 1. If $\frac{1}{e}\le <!--\; \le \; -->p<q$ we have:
$h(p)<h(q)⟹<!--\; ⟹\; -->p\ln <!--\; \; -->p-<!--\; -\; -->q\ln <!--\; \; -->q<0$
Thus $f^{\prime }(x)<0,\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->[\frac{1}{e}-<!--\; -\; -->e,\mathrm{\infty <!--\; \infty \; -->})=[-<!--\; -\; -->2.35,\mathrm{\infty <!--\; \infty \; -->})$
So f(x) must be decreasing in the above interval. But i compared f(-2) and f(-1) It is the other way round, that is $f(-<!--\; -\; -->2)<f(-<!--\; -\; -->1)$
Now what i did is i used graphing calculator to check the graph. Yes it is decreasing, but there is a vertical asymptote. How to identify the equation of vertical asymptote?
Also the graph is decreasing in $[-<!--\; -\; -->2.541,\mathrm{\infty <!--\; \infty \; -->})$ which is not matching with my answer above.

Solve the integral
$$\int <!--\; \int \; -->\frac{1+x+\sqrt{1+x^2}}{\sqrt{x+1}+\sqrt{x}}dx$$

I am having trouble finding the region of absolute stability for modified Euler method:
$\begin{array}{rl}{w}_{i+1}^{\ast <!--\; \ast \; -->}& =w_i+hf(t_i,w_i)\\ w_{i+1}& =w_i+\frac{h}{2}[f(t_i,w_i)+f(t_{i+1},w_{i+1}^{\ast <!--\; \ast \; -->})].\end{array}$
DEFINITION: We define a region R of absolute stability for a one-step method as the region in the complex plane satisfying:
$R=\{h\kappa \in C:|Q(h\kappa )|<1\}.$
I don't fully understand the above definition of region of absolute stability and how to apply it. Clear and step by step help would be much appreciated. Thank you

First derivative test to find where the function is increasing, and decreasing
I have this function: $f(x)=\frac{x^2}{log(x)-<!--\; -\; -->1}$.
and I want to find increasing intervals, and decreasing intervals. Here's what I did:
- I've found domain in which the function is defined, and this is what I got: $f:]0,+\mathrm{\infty <!--\; \infty \; -->}[(without1)-<!--\; -\; -->>R$.
0 and 1 are critical points in which the function isn't defined (i.e if $x=0$, the log is undefined, and if $x=1$, the fraction is undefined).
- I've computed first derivative, and this is what I got (I've used quotient rule):
$\frac{2x\ast <!--\; \ast \; -->(log(x)-<!--\; -\; -->1)-<!--\; -\; -->x^2\ast <!--\; \ast \; -->1/x}{(log(x)-1{)}^2}$
I've rewritten it, as follows: $\frac{2x\ast <!--\; \ast \; -->(log(x)-<!--\; -\; -->1)-<!--\; -\; -->x}{(log(x)-1{)}^2}$.
the denominator is always positive so the sign depends only on the numerator. (I think the problem is in this step) on a line, I've chosen 4 different random numbers, 1/4, 1/2, 3/2, 2, and I've put each number in the first derivative, and this is what I got:
- in the first interval (from 0, to 1/4), the original function is decreasing (minus sign)
- same for the other intervals
I don't have the full solution of the exercise, therefore I don't knwo if my solution is correct, but I think is strange that in this function there aren't increasing intervals.
- I think the problem is in the first derivative, but I'm sure I've used the quotient rule correctly.

Calculate $\int <!--\; \int \; -->5^{x+1}{e}^{2x-<!--\; -\; -->1}dx$

Is there a way to insert a correction term in order to counterbalance the gain of energy due to the explicit Euler method ?
In particular, let's assume I know the total energy $E_0$ of my system at time $t=0$ (due to the initial conditions). And the energy is defined locally at time t on a grid which is of total size $N^2$ : $e(i,j,t)$. Would it make sense to correct at each iteration (time t): $e^{\prime }(i,j,t)=e(i,j,t)-<!--\; -\; -->\frac{\underset{i}{\sum <!--\; \sum \; -->}e(i,j,t)-<!--\; -\; -->E_0}{N^2}$?