Help to Integral Calculus Solutions & Examples

Divide $(x-<!--\; -\; -->1{)}^6$ by (x+1)

Polar coordinates of a point are given. Find therectangular coordinates of the point. $(-<!--\; -\; -->3,-<!--\; -\; -->{135}^{\circ <!--\; \circ \; -->})$

Whats the integral of $x{e}^{-<!--\; -\; -->x}$

Find the percent of increase.
Original price:$4
New price:$4.50

In particular, if Euler's method is implemented on a computer, what's the minimum step size that can be used before rounding errors cause the Euler approximations to become completely unreliable?
I presume it's when step size reaches the machine epsilon? E.g. if machine epsilon is e-16, then once step size is roughly e-16, the Euler approximations are unreliable.

Consider the boundary-value problem
$\begin{array}{r}{y}^{″}=y^3+x,y(a)=\mathrm{\alpha <!--\; \alpha \; -->},y(b)=\mathrm{\beta <!--\; \beta \; -->},a\le <!--\; \le \; -->x\le <!--\; \le \; -->b\end{array}$
To use the shooting method to solve this problem, one needs a starting guess for the initial slope $y\prime (a)$. One way to obtain such a starting guess for the initial slope is, in effect, to do a ”preliminary shooting” in which we take a single step of Euler’s method with $h=b-a$.
(a) Using this approach, write out the resulting algebraic equation for the initial slope.
(b) What starting value for the initial slope results from this approach?
I'm really not sure how to begin here; my idea was to write the 2nd order ODE using a new variable, such as: $y^{\prime }=u$, $u^{\prime }=y^{″}=y^3+x$. Then, maybe I could use Euler's method to solve for $u=y^{\prime }$; however, I'm not sure how to go about doing this. Any help appreciated.

Find the asymptotes
Find the asymptotes of $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R},f(x)=\sqrt[3]{e^x-<!--\; -\; -->e^{2x}+e^{4x}{\ln }^2<!--\; \; -->(1+e^{-<!--\; -\; -->x})}.$
I found that y=0 is an asymptote when $x\to <!--\; \to \; -->-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$, but how do I calculate $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(x)$ ?

Finding f intervals
How would I find the f intervals for the following two functions.
$f(x)=(x-<!--\; -\; -->2{)}^2(x+1{)}^2$
using the chain rule I got $(x-<!--\; -\; -->2{)}^2(2)(x+1)(1)+(2)(x-<!--\; -\; -->2)(1)(x+1{)}^2$
then I got f decrease $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->1]$
and f increase $[2,\mathrm{\infty <!--\; \infty \; -->})$
but the area between -1 and 2 in confusing me.
My second function is $f(x)=x+\frac{4}{x^2}$
differentiating I got $\frac{x^3-<!--\; -\; -->8}{x^3}$
so I got f increase $(\mathrm{\infty <!--\; \infty \; -->},0)$ and $[2,\mathrm{\infty <!--\; \infty \; -->})$
f decrease (0, 2]
but did I do this correctly.

Find all asymptotes of given curve
$y^2(a+x)=x^2(a-<!--\; -\; -->x)$
origin i found will be a node
i'm trying to find asymptotes by equating highest degree term to 0Please help.

Asymptotes of function
How to find a if the function below doesn't have a vertical asymptotes.
$f(x)=\frac{x^2-<!--\; -\; -->ax+2}{x-<!--\; -\; -->2}$

Solve the following ODE by using the method of undetermined coefficients in which Euler's formula needs to be utilized:
$y^{″}-<!--\; -\; -->2y^{\prime }+y=\sin <!--\; \; -->(t)$
The way that I solved this doesn't involve Euler's formula, and I was wondering how I might use the formula here.

My approach:
The formula can be written as $y(t)=y_h(t)+y_p(t)$ where $y_h(t)$ is the "homogeneous version" of the ODE and $y_p(t)$ is the particular solution that we'll obtain via the basic rule of the method of undetermined coefficients.

$y_h(t)$:
Putting $r(t)=\sin <!--\; \; -->(t)=0$ in the original equation, the ODE we need to solve is:
$y^{″}-<!--\; -\; -->2y^{\prime }+y=0$
where we can set the general solution as $y=e^{\mathrm{\lambda <!--\; \lambda \; -->}t}$ and obtain the characteristic equation:
${\mathrm{\lambda <!--\; \lambda \; -->}}^2-<!--\; -\; -->2\mathrm{\lambda <!--\; \lambda \; -->}+1=0$
which has a real double root, hence giving us the solution:
$y_h(t)=(c_1+c_{2}t)e^t$

$y_p(t)$:
Judging by the fact that $r(t)$ is shape $k\sin <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t)$ and we know that $\mathrm{\omega <!--\; \omega \; -->}=1$ we can set the general solution to be of form:
$\begin{array}{rl}{y}_p(t)& =\phantom{-<!--\; -\; -->}K\cos <!--\; \; -->(t)+M\sin <!--\; \; -->(t)\\ y_p^{\prime }(t)& =-<!--\; -\; -->K\sin <!--\; \; -->(t)+M\cos <!--\; \; -->(t)\\ y_p^{″}(t)& =-<!--\; -\; -->K\cos <!--\; \; -->(t)-<!--\; -\; -->M\sin <!--\; \; -->(t)\end{array}$
substituting these equations into the original equation and then simplifying gives us:
$y_p(t)=\frac{1}{2}\cos <!--\; \; -->(t)$
And in conclusion, we can write that the solution to the given ODE is:
$\begin{array}{rl}y(t)& =y_h(t)+y_p(t)\\ & =(c_1+c_{2}t)e^t+\frac{1}{2}\cos <!--\; \; -->(t)\end{array}$
How would we be able to derive this conclusion via Euler's formula? Thanks in advance.

Find eccentricity when asymptotes are given
What is the Eccentricity of hyperbola whose asymptotes are $x+2y=3$ and $2x-<!--\; -\; -->y=1$.
I know for any hyperbola ${\displaystyle \frac{(x-<!--\; -\; -->h{)}^2}{a^2}}-<!--\; -\; -->{\displaystyle \frac{(y-<!--\; -\; -->k{)}^2}{b^2}}=1$
The asymptotes are $y-<!--\; -\; -->k=\pm <!--\; \pm \; -->{\displaystyle \frac{b}{a}}(x-<!--\; -\; -->h)$
But how to solve this problem

I have the following initial value problem on $t\in <!--\; \in \; -->[0,T]$:
$y^{\prime }=y^2(1-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}y)$
I wish to use the Trapezoidal method to solve this I.V.P., and I cannot use simply use the pblackictor-corrector method, i.e. improved Euler's method. The trapezoidal method is defined by:
$y_{n+1}=y_n+\frac{h}{2}(f_{n+1}+f_n)$
Plugging in the value from the O.D.E., we have after some simplification:
$y_{n+1}=y_n+\frac{h}{2}(y_{n+1}^2+y_n^2-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}{y}_{n+1}^3-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}{y}_n^3)$
This is an implicit method, so I must use a root finding method at each iteration to find the $y_{n+1}$ on the right side of the equation. I have read online that Newton's method is usually used for such purposes, but I cannot figure out how to implement it.
$y_{n+1}=y_n-<!--\; -\; -->\frac{f(y_n)}{f^{\prime }(y_n)}$
For example, what values would $f(y_n)$ and $f^{\prime }(y_n)$ take in this context? Thank you in advance!

The linear system $y^{\prime }=Ay,y(0)=y_0$, where $A$ is a symmetric matrix, is solved by the Euler method.
Let $e_n=y_n-<!--\; -\; -->y(nh)$
Where $y_n$ denotes the Euler approximation and $y(nh)$ the exact solution ($h$ is Euler step size).
Prove that $||e_n|{|}_2=||y_0|{|}_{2}max|(1+h\mathrm{\lambda <!--\; \lambda \; -->}{)}^n-<!--\; -\; -->e^{nh\mathrm{\lambda <!--\; \lambda \; -->}}|$
Where $\mathrm{\lambda <!--\; \lambda \; -->}\in <!--\; \in \; -->\mathrm{\sigma <!--\; \sigma \; -->}(A)$ where $\mathrm{\sigma <!--\; \sigma \; -->}(A)$ is the set of eigenvalues of $A$.
I have tried various approaches such as writing $e_n$ as the error bound of the Euler method and taking the norm, but I can't seem to get $||y_0|{|}_2$ in my answers.

Proving the existence of a non-monotone continuous function defined on [0,1]
Let $(I_n{)}_{n\in <!--\; \in \; -->\mathbb{N}}$ be the sequence of intervals of [0,1] with rational endpoints, and for every $n\in <!--\; \in \; -->\mathbb{N}$ let $E_n=\{f\in <!--\; \in \; -->C[0,1]:f\text{is monotone in}{I}_n\}$. Prove that for every $n\in <!--\; \in \; -->\mathbb{N}$, $E_n$ is closed and nowhere dense in $(C[0,1],d_{\mathrm{\infty <!--\; \infty \; -->}})$. Deduce that there are continuous functions in the interval [0,1] which aren't monotone in any subinterval.
For a given n, $E_n$ can be expressed as $E_n=E_{n\nearrow <!--\; \nearrow \; -->}\cup <!--\; \cup \; -->E_{n\swarrow <!--\; \swarrow \; -->}$ where $E_{n\nearrow <!--\; \nearrow \; -->}$ and $E_{n\swarrow <!--\; \swarrow \; -->}$ are the sets of monotonically increasing functions and monotonically decreasing functions in $E_n$ respectively. I am having problems trying to prove that these two sets are closed. I mean, take $f\in <!--\; \in \; -->(C[0,1],d_{\mathrm{\infty <!--\; \infty \; -->}})$ such that there is $\{f_k{\}}_{k\in <!--\; \in \; -->\mathbb{N}}\subset <!--\; \subset \; -->E_{n\nearrow <!--\; \nearrow \; -->}$ with $f_k\to <!--\; \to \; -->f$. How can I prove $f\in <!--\; \in \; -->E_{n\nearrow <!--\; \nearrow \; -->}$?. Suppose I could prove this, then I have to show that ${\stackrel{¯<!--\; ¯\; -->}{E_n}}^{\circ <!--\; \circ \; -->}=E_n^{\circ <!--\; \circ \; -->}=\mathrm{\varnothing <!--\; \varnothing \; -->}$. This means that for every $f\in <!--\; \in \; -->E_n$ and every $r>0$, there is $g\in <!--\; \in \; -->B(f,r)$ such that g is not monotone. Again, I am stuck. If I could solve this two points, it's not difficult to check the hypothesis and apply the Baire category theorem to prove the last statement.

Using the Euler's method (with $h={10}^{-<!--\; -\; -->n}$) to find $y(1)$
$y^{\prime }=\frac{\sin <!--\; \; -->(x)}{x}$
Since
$y^{\prime }(x)=\frac{\sin <!--\; \; -->(x)}{x}$
then
$f(x,y)=\frac{\sin <!--\; \; -->(x)}{x}$
I know
$y_{n+1}=y_n+h\cdot <!--\; \cdot \; -->f(x_n,y_n)$
and given $y(0)=0$, so
$x_0=y_0=0$
Therefore,
$x_1=x_0+h=0+{10}^{-<!--\; -\; -->0}=1\Rightarrow <!--\; \Rightarrow \; -->y_1=y(x_1)=y(1)$
Then,
$y_1=0+{10}^{-<!--\; -\; -->0}\cdot <!--\; \cdot \; -->f(x_0,y_0)$
$y_1=f(0,0)$
But
$f(0,0)=\frac{\sin <!--\; \; -->(0)}{0}=\frac{0}{0}$
How am I supposed to do it?

Intermediate value theorem on infinite interval $\mathbb{R}$
I have a continious function f that is strictly increasing. And a continious function g that is strictly decreasing. How to I rigorously prove that $f(x)=g(x)$ has a unique solution?
Intuitively, I understand that if I take limits to infinity, then f grows really large and g grows very small so the difference is less than zero. If I take the limits to negative infinity then the opposite happens. Using the intermediate value theorem, there must be an intersection, and since they are strictly increasing/decreasing, only one intersection happens.
My question is how do I apply the intermediate value theorem here? I don't have the interval to apply it on. I don't know when f crosses g and therefore can't take any interval. Or is there some sort of axiom applied here that I am missing?

Sequence defined by $u_0=1/2$ and the recurrence relation $u_{n+1}=1-<!--\; -\; -->u_n^2$.
I want to study the sequence defined by $u_0=1/2$ and the recurrence relation $u_{n+1}=1-<!--\; -\; -->u_n^{2}n\ge <!--\; \ge \; -->0.$.
I calculated sufficient terms to understand that this sequence does not converge because its odd and even subsequences converge to different limits.
In particular we have $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{a}_{2n+1}=0\text{ and}\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{a}_{2n+2}=1.$.
Now I can also prove that $u_n\le <!--\; \le \; -->1$ because the subsequent terms are related in this way
$f:<!--\; :\; -->x\mapsto <!--\; \mapsto \; -->1-<!--\; -\; -->x^2.$
In order to prove that the subsequences converge I need to prove either that they are bounded (in order to use Bolzano-Weierstrass) or that they are bounded and monotonic for $n>N$. Proved that the subsequences converge to two different limits, I will be able to prove that our $u_n$ diverges.
Can you help me? Have you any other idea to study this sequence?

Derivative of an integral from 0 to x when x is negative?
Given a function $F(x)={\int <!--\; \int \; -->}_0^x\frac{t+8}{t^3-<!--\; -\; -->9}dt,$, is F′(x) different when $x<0$, when $x=0$ and when $x>0$?
When $x<0$, is $F^{\prime }(x)=-<!--\; -\; -->\frac{x+8}{x^3-<!--\; -\; -->9}$.
... since you can't evaluate an integral going from a smaller number to a bigger number? That's what I initially thought, but when I graphed the antiderivative the intervals of increase and decrease were different from the ones in my calculation.
EDIT: Yes, I was talking about that identity. So is $F^{\prime }(x)=\frac{x+8}{x^3-<!--\; -\; -->9}$ or $F^{\prime }(x)=-<!--\; -\; -->\frac{x+8}{x^3-<!--\; -\; -->9}$.