Help to Integral Calculus Solutions & Examples

Why is 2 double root of the derivative?
A polynomial function P(x) with degree 5 increases in the interval $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1)$ and $(3,\mathrm{\infty <!--\; \infty \; -->})$ and decreases in the interval (1,3). Given that $P^{\prime }(2)=0$ and $P(0)=4$, find P'(6).
In this problem, I have recognised that 2 is an inflection point and the derivative will be of the form: $P^{\prime }(x)=5(x-<!--\; -\; -->1)(x-<!--\; -\; -->3)(x-<!--\; -\; -->2)(x-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$.
But I am unable to understand why $x=2$ is double root of the derivative (i.e. why is $\mathrm{\alpha <!--\; \alpha \; -->}=2$?). It's not making sense to me. I need help with that part.

Which of the following is not a rational number?
a) $\sqrt{25}$
b) $\sqrt{45}$
c) $\sqrt{\frac{256}{225}}$
d) $\frac{3}{4}$

Convert the following numbers in the given base to decimal equivalents:
1) ${174053}_8$ into decimal
2) ${10110101}_2$ into decimal
3) ${34}_8$ into decimal
4) ${1011}_2$ into decimal
5) $FB3D_{16}$ into decimal
6) $3E_{16}$ into decimal

1) For any natural $i,f(i)\le <!--\; \le \; -->(i-<!--\; -\; -->1)$ .
2) If i divides a number of the form $b^{n-<!--\; -\; -->1}+1$ or $b^n-<!--\; -\; -->1$ (for applicable n) then $f(i)\le <!--\; \le \; -->n$ (I believe this to be an equality if n is minimal).
3) If $i=j_1\times <!--\; \times \; -->j_2$ with $j_1$ and $j_2$ coprime, then $f(i)\le <!--\; \le \; -->\text{lcm}(f(j_1),f(j_2))$ .

When multiplying two decimal numbers, you first ignore the decimals, find the product, then count the number of decimal places that need to be in the answer by taking the sum of the original decimal places. Why exactly does this work?

What two kind of decimals you times together to make 1?

How many decimal places should the answer for x show?
$x=\frac{16.5}{253.25}$
a) one decimal place, two decimals places, three decimal places, four decimal places, or five decimal places.

I know the difference between the forward (explicit) Euler method and backward (implicit) Euler method in the context of approaching solutions to ODEs.
However, I have a doubt regarding the following method:
$y_{n+1}=y_n+\frac{h}{2}(f(t_n,y_n)+f(t_{n+1},y_{n+1}))$
I am not sure whether it is an implicit or an explicit method.
I tend to say it is implicit one step methode because we have to solve an equation to calculate $y_{n+1}.$
Thank you for your help.

Asymptotes of a rational function
We have a function
$f(x)={\displaystyle \frac{2x^4+4x^3+3x^2+4x-<!--\; -\; -->4}{x^3-<!--\; -\; -->x^2-<!--\; -\; -->6x}}$
How would I systematically go about finding the asymptotes of this function? I know how to find the asymptotes of for example log functions or functions with a square root in it, but I don't really know how to find them for this function.

Define
$$I(x)={\int <!--\; \int \; -->}_x^{\mathrm{\infty <!--\; \infty \; -->}}{e}^{-<!--\; -\; -->y^2}dy$$
Evaluate
$${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}{e}^{x^2}[I(x){]}^{2}dx$$

Consider the following problem
$y^{\prime }=\ln <!--\; \; -->\ln <!--\; \; -->(4+y^2),x\in <!--\; \in \; -->[0,1],y(0)=1$
We can formulate the problem the approximate the solution
$y_{n+1}=y_n+h\ln <!--\; \; -->\ln <!--\; \; -->(4+y_n^2),n=0,1,...,N-<!--\; -\; -->1,y_0=0$
with mesh points $x_n=nh$. I am tasked with finding the truncation error. Now considering a Taylor series we find that
$|T_n|\le <!--\; \le \; -->\frac{h}{2!}|(\ln <!--\; \; -->\ln <!--\; \; -->(4+{\mathrm{\xi <!--\; \xi \; -->}}^2){)}^{\prime }|\mathrm{\xi <!--\; \xi \; -->}\in <!--\; \in \; -->(x_n,x_{n+1})$
Now
$\ln <!--\; \; -->\ln <!--\; \; -->(4+y^2{)}^{\prime }=\frac{d}{dy}\ln <!--\; \; -->\ln <!--\; \; -->(4+y^2)=\frac{2y\ln <!--\; \; -->\ln <!--\; \; -->(4+y^2)}{\ln <!--\; \; -->(4+y^2)(4+y^2)}\le <!--\; \le \; -->\frac{2y}{4+y^2}\le <!--\; \le \; -->1/2$
$\therefore <!--\; \therefore \; -->|T_n|\le <!--\; \le \; -->\frac{h}{4}$
Is this the correct way to go about obtaining this?

I have a differntial equation $du/dx=f(x,u)$, where
$f(x,u)=\frac{d\mathrm{\psi <!--\; \psi \; -->}}{dx}+p(u-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->}(x){)}^4,$
where $\mathrm{\psi <!--\; \psi \; -->}$ is a given function of $x$ independent of $u$.
The backward Euler Method written for step $i$ is
$u_{i+1}=u_i+hf(x_{i+1},u_{i+1}).$
So, at every step we have a 4th degree polynomial to solve for $u_{i+1}$, which can have more than one root. I am told to employ a root-finding algorithm. But I don't understand how to find the one correct root and reject all others. Bisection, Newton's and other methods seem to only work when there is one root in an interval. Would appreciate any help!

Definite integration evaluation of ${\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}\frac{x}{(a^2{\cos }^2<!--\; \; -->x+b^2{\sin }^2<!--\; \; -->x{)}^2}dx$

How can we evaluate
$${\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}\frac{1}{1+(\tan <!--\; \; -->x{)}^{\sqrt{2}}}dx$$

Let $x>0$ be real. Let $n_0$ be the largest integer such that $n_0⩽<!--\; ⩽\; -->x$ (Note that the existance of $n_0$ depends on the archimedian property of R).
1) Why he write about natural number but he did not defines them?
2) How he uses archimedian property in existance of $n_0$ ?

What would the critical points be of this equation?
${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{-<!--\; -\; -->4x}{(1+x^2{)}^2}}$
I just dont understand how find them with a reciprocal function. If someone could explain how to find it in this this situation that would be fantasic!

Solve. $\int <!--\; \int \; -->\frac{6x}{x-<!--\; -\; -->4}dx$

Definite Integral of $\cos <!--\; \; -->(x)/\sqrt{\sin <!--\; \; -->(x)}$ from $\mathrm{\pi <!--\; \pi \; -->}/6$ to $\mathrm{\pi <!--\; \pi \; -->}/2$

I have the following system:
$y^{\prime }=y+e^x$
$y(0)=0$
The problem asks for applying Euler's method and then finding an expression for the global error. Finally, supposing that
$\underset{h-<!--\; -\; -->>0}{lim}\frac{1-<!--\; -\; -->(\frac{1+h}{e^h}{)}^{\frac{x}{h}}}{\frac{x}{h}(e^h-<!--\; -\; -->(1+h))}=1$
Check that the global error tends to 0 as $h$ tends to 0 as well.
Let's first apply Euler's method.
${\mathrm{\eta <!--\; \eta \; -->}}_0=y_0=0$
$x_i=x_0+ih=ih$
${\mathrm{\eta <!--\; \eta \; -->}}_{i+1}={\mathrm{\eta <!--\; \eta \; -->}}_i+hf(x_i,{\mathrm{\eta <!--\; \eta \; -->}}_i)$
Since $x_i=ih$, we have ${\mathrm{\eta <!--\; \eta \; -->}}_{i+1}={\mathrm{\eta <!--\; \eta \; -->}}_i+h({\mathrm{\eta <!--\; \eta \; -->}}_i+e^{ih}$. Factoring by ${\mathrm{\eta <!--\; \eta \; -->}}_i$:
${\mathrm{\eta <!--\; \eta \; -->}}_{i+1}={\mathrm{\eta <!--\; \eta \; -->}}_i(1+h)+h{e}^{ih}$
And now I don't how should I proceed. Should I solve the iteration? I've done problems where I can express ${\mathrm{\eta <!--\; \eta \; -->}}_{i+1}$ in terms of ${\mathrm{\eta <!--\; \eta \; -->}}_0$ by iterations, but here I can't do so because the $h{e}^{ih}$ prevents that (or at least I haven't managed to do it).
I know that global error is defined as $e(x,h)=\mathrm{\eta <!--\; \eta \; -->}(x,h)-<!--\; -\; -->y(x)$ but I don't know how exactly should I use that formula. My guess it that I have to solve the differential equation and use the solution $y(x)$, but I don't know how to solve that system or what exactly is $\mathrm{\eta <!--\; \eta \; -->}(x,h)$.
Any help regarding to solving this problem will be highly appreciated.
Thanks!