Help to Integral Calculus Solutions & Examples

Solve the system:
$\{\begin{array}{l}{\displaystyle \frac{0.01-<!--\; -\; -->x}{0.02}}-<!--\; -\; -->0.5\le <!--\; \le \; -->{\displaystyle \frac{2-<!--\; -\; -->3x}{0.01}}\\ {\displaystyle \frac{3-<!--\; -\; -->x}{0.1}}\le <!--\; \le \; -->{\displaystyle \frac{6x+8}{0.4}}-<!--\; -\; -->5\end{array}$
Is it a good idea to multiply by 10?

Use Euler's method to approximate the solution for the following initial value problem. $y^{\prime }=t{e}^{3t}-<!--\; -\; -->2y,0\le <!--\; \le \; -->t\le <!--\; \le \; -->1,y(0)=0\text{ with }h=0.5$
h is the step size The initial condition is
$$\begin{gathered} y(0)=0\Rightarrow <!--\; \Rightarrow \; -->w_0=0 \\ {t}_0=0 \\ \text{thus} \\ {w}_1=w_0+h(t_0{e}^{3t_0}-<!--\; -\; -->2w_0) \\ 0+0.5[(0)e^{3(0)}-<!--\; -\; -->2(0)=0 \\ t=a+ih\text{ common distance} \\ {t}_1=.5 \end{gathered}$$
From this we can create this Table:
$\begin{array}{|cccc|}\hline i& t_i& w_i& y(t_i)\\ 0& 0& 0& 0\\ 1& .5& 0& \\ 2& 1& 1.2& \\ \hline\end{array}$
To find the the actual solutions requires one to solve the ordinary differential equation IVP problem.
A 1st order linear ODE has the form
$$\begin{gathered} y^{\prime }(x)+p(x)y=q(x) \\ y(x)=\frac{\int <!--\; \int \; -->e^{\int <!--\; \int \; -->p(x)dx}q(x)dx+C}{e^{\int <!--\; \int \; -->p(x)dt}} \\ \frac{\int <!--\; \int \; -->e^{\int <!--\; \int \; -->p(x)dx}q(x)dx+C}{e^{\int <!--\; \int \; -->p(x)dx}} \end{gathered}$$
$$\begin{gathered} \frac{\int <!--\; \int \; -->e^{\int <!--\; \int \; -->2dt}{e}^{3t}tdt+C}{e^{\int <!--\; \int \; -->2dt}} \\ \frac{\int <!--\; \int \; -->e^{5t}tdt}{e^{2t}}=\frac{t{e}^{3t}}{5}-<!--\; -\; -->\frac{e^{3t}}{25}+\frac{e^{-<!--\; -\; -->2t}C}{1} \end{gathered}$$
I was unable to reach the correct answer to the ODE which was
$\frac{1}{5}t{e}^{3t}-<!--\; -\; -->\frac{1}{25}{e}^{3t}+\frac{1}{25}{e}^{-<!--\; -\; -->2t}$
How do I correct my work and arrive to the correct conclussion?

$y(0)=1$
How can we solve this equation with operator splitting methods in matlab by using forward Euler method?

Find all horizontal, vertical, or slant asymptotes
I have been instructed to find all the horizontal, vertical or slant asymptotes of the following function:
$y=x^3-<!--\; -\; -->4x^2$
I know there are no horizontal or vertical asymptotes, but I'm having a hard time understanding slant asymptotes. How do I figure out if there are any of those?

The repeating decimals $0.abab\stackrel{¯<!--\; ¯\; -->}{ab}$ and $.abcabc\stackrel{¯<!--\; ¯\; -->}{abc}$ satisfy $0.abab\stackrel{¯<!--\; ¯\; -->}{ab}+0.abcabc\stackrel{¯<!--\; ¯\; -->}{abc}=\frac{33}{37}$where a,b, and c are (not necessarily distinct) digits. Find the three-digit number abc.

Find intervals decrease and increase of a function $f(x)=\frac{\mathrm{\pi <!--\; \pi \; -->}x}{2}-<!--\; -\; -->x\arctan <!--\; \; -->x$.
Find intervals decrease and increase of a function.
$f(x)=\frac{\mathrm{\pi <!--\; \pi \; -->}x}{2}-<!--\; -\; -->x\arctan <!--\; \; -->x$
$f^{\prime }(x)=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}-<!--\; -\; -->\arctan <!--\; \; -->x-<!--\; -\; -->\frac{x}{1+x^2}$
$f^{″}(x)=\frac{-<!--\; -\; -->2}{(1+x^2{)}^2}$
which is negative. What can I say from here?

Fourth solution of $y^{\prime }={\displaystyle \frac{10}{3}}x{y}^{2/5}$, $y(0)=0$?
By inspection, we know that $y=0$ is a solution. If we separate variables, we get another solution, $y=x^{10/3}$. The book tells me this much, and asks me to find the other 2 solutions that differ on every open interval containing $x=0$ and are defined on $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},\mathrm{\infty <!--\; \infty \; -->})$.
I found the third solution (by guessing) to be $y=-<!--\; -\; -->x^{10/3}$. The only other possibility i could think about is $x=0$, but I'm pretty sure that's not valid since y′ would be undefined. What is the last solution? How do we know there are only 4 possible solutions (and not more)?

Let $f(x)=\frac{\ln <!--\; \; -->(x)}{\sqrt{x}}$. Find the critical values of f(x)
How would you find the critical value of this equation? So far... I have gotten that you find the derivative of f(x) which is $f^{\prime }(x)=\frac{1}{x^{3/2}}-<!--\; -\; -->\frac{\ln <!--\; \; -->(x)}{2x^{3/2}}.$.
To find the critical values set $f^{\prime }(x)=0$? But how would you factor out the derivative to find the critical values?
Also, after that how would you find the intervals of increase and decrease and relative extreme values of f(x).
And lastly the intervals of concavity and inflection points.

How to find asymptotes
I'm supposed to find asymptotes for x⋅arctan(x) for a homework assignment.
Are there any theorems regarding this I can utilize to find the equation for the asymptotes, or is this one of those solve case by case things, where this particular equation can be solved in some neat way, but there is no easy universal way to find asymptotes like this?
I know it should be:
$\pm <!--\; \pm \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}x-<!--\; -\; -->1$
but "how do I show this?" is essentially my question.
Thanks in advance

Can asymptotes be any straight line?
I was thinking if graphs can have asymptotes as x=y or any other random line . I couldn't think of a function . please help

I am trying to show that, for the equation
$y^{\prime }+\mathrm{\alpha <!--\; \alpha \; -->}y=0$
alternating between a forward Euler method step for $y_{2n}$ and a backward Euler step for $y_{2n+1}$ with time-step $h$ is equivalent to
$y_{n+1}=\frac{1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}h}{1+\mathrm{\alpha <!--\; \alpha \; -->}h}{y}_n$
I have that
$y_1=y_0+h(-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}{y}_0)=(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}h)y_0$
How exactly $(1+\mathrm{\alpha <!--\; \alpha \; -->}h)$ works into this I don't see. If I just press forward anyway, using the trapezoid rule for the backward Euler step next, and call ${\stackrel{^<!--\; ^\; -->}{y}}_n$ the approximation of $y_n$ using the forward Euler method, I get
$y_2=y_1+h\left(\frac{y_1+{\stackrel{^<!--\; ^\; -->}{y}}_2}{2}\right)$
I'm kind of doing this a priori but that seems right to me: the trapezoid rule is to multiply the change in the $x$-axis by the average of the bases which are $y_1$ and $y_2$. This becomes
$y_2=y_1+h\left(\frac{y_1+(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}h)y_1}{2}\right)=(1+h\left(\frac{2-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}h}{2}\right))y_1$
This doesn't seem to be shaping up to the form I'm supposed to be getting. Am I doing something wrong?

Finding the critical points of a function and the interval where it increases and decreases.
I am having trouble finding the critical points of
$f(x)=(x+1)/x-<!--\; -\; -->3$
I found the derivative to be $f^{\prime }(x)=-<!--\; -\; -->4/(x-<!--\; -\; -->3{)}^2$.
My next step was to equate my derivative to zero, but that does not seem to work as my x cancels out. Usually I would take the x-value(worked out by equating the derivative with zero) and substitute it into the original equation to get a y-value. This would then be the critical points. Is there anyone who could maybe help me out (maybe with an example or so) as I also have to find the intervals where the function is increasing and decreasing?

Convert decimal fraction to decimal: $\frac{16}{1,000}$

Find the general indefinite integrals integral of x^(2) square root of (x^(3)+1)dx integral of pi sin x cos^(3)x dx

Find the interval in which f(x) increases and decreases.
Let $f(x)=2x^3-<!--\; -\; -->9x^2+12x+6$ so $f^{\prime }(x)=6x^2-<!--\; -\; -->18x+12=6(x-<!--\; -\; -->1)(x-<!--\; -\; -->2)$.
I need the intervals in which f(x) strictly increases, $f^{\prime }(x)>0$ when $x<1$ and $x>2$ and thus f(x) strictly increases in these intervals and $f^{\prime }(x)<0$ when $1<x<2$ so f(x) strictly decreases in this interval.
My Question:
What about at points $x=1,2$.
If $f^{\prime }(x)=0$ at points (not intervals) then f(x) can still be considered strictly monotone. And it also seems reasonable (I'll add the reason below) to include the points 1,2 in the intervals of increase (IOI, for short) and decreases (IOD).
Eg: Take $x=1$. Let's say I include this point in both IOI and IOD. So IOI is now, $x\in <!--\; \in \; -->(-<!--\; -\; -->\infty ,1]$ and you can see that it doesn't contradict the definition of "strictly increasing function in interval" either. Take any $p,q\in <!--\; \in \; -->(-<!--\; -\; -->\infty ,1],p>q\Rightarrow <!--\; \Rightarrow \; -->f(p)>f(q)$ similarly my IOD, now would be $x\in <!--\; \in \; -->[1,2]$ (notice I included 2) and it still follows the definition.
As I understand that, definition of monotonicity functions at a point, would now get in the way.
I can't include $x=1$ because there exists no $h>0$ such that taking $p,q\in <!--\; \in \; -->(1-<!--\; -\; -->h,1+h)\nRightarrow <!--\; \nRightarrow \; -->f(p)>f(q)$ similarly, I can't add $x=1$ in interval of decrease either.
If I'd have to pick, my intuition would lead me to pick the second one, but I can't see why I should reject the first one either since it actually follows the definition of strictly increasing function in interval.

Is f(x) increasing or decreasing? on $(-<!--\; -\; -->1,0]$
Let the function f(x) be defined by $f(x)=\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{(-<!--\; -\; -->1{)}^{n-<!--\; -\; -->1}}{(2n+1)!}{x}^n$. Since f(x) is a function, then it can increase and decrease. So, in the interval $(-<!--\; -\; -->1,0]$, is f(x) increasing or decreasing? Here we restrict x to any real number between this interval. Fortunately, I found from a previous question I asked that when $x<0$, $f(x)=\frac{\sinh <!--\; \; -->(\sqrt{x})}{\sqrt{x}}$. Credits to the people that answered that question. Graphing this function, I found that it is decreasing throughout. Of course, at $x=0$, the entire function is 0, and it is neither decreasing or increasing.

How can one convert radicals to decimals(approximate value) when the number is not perfect such as $\sqrt{2}$ , $\sqrt{3}$ , $\sqrt{5}$ , etc. Without the use of calculators.

Is $f(x)=x^{5/3}-5x^{2/3}$ defined over $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0]$?
I've encountered this question: Find and describe all local extrema of $f(x)=x^{5/3}-5x^{2/3}.$.
Also indicate on which regions the function is increasing and decreasing.
I've managed to find the extrema, but I am not sure whether the function is defined on $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0]$. To make sure I looked on the internet at some graphing calculators and some of them graphed the function on that interval while others did not. Is it defined on that interval?

Let
$x^{\prime }(t)=f(t,x(t)),t\in <!--\; \in \; -->(0,T)$ with $x(0)=x_0$
$f$ satifies the Lipschitz-condition $f(t,x)-<!--\; -\; -->f(t,y)\le <!--\; \le \; -->L|x-<!--\; -\; -->y|$
$h\in <!--\; \in \; -->(0,\frac{1}{L})$ is the step size and the approximation $x_k$ for $x(t_k)=hk$ is given by $x_k=x_{k-<!--\; -\; -->1}+hf(t_k,x_k)$.
Now I would be very interested how to derive the error
$|x_k-<!--\; -\; -->x(t_k)|\le <!--\; \le \; -->\frac{1}{1-<!--\; -\; -->Lh}(|x_{k-<!--\; -\; -->1}-<!--\; -\; -->x(t_{k-<!--\; -\; -->1})|+\frac{h^2}{2}\underset{s\in <!--\; \in \; -->[0,T]}{max}|x^{″}(s)|)$
I tried to look up it up in some numerical analysis books but it is always different

Find all vertical and horizontal asymptotes.
Find all vertical and horizontal asymptotes:
$f(x)=\frac{4x}{x^2+16}$
My Work: (Vertical)
1) $x^2+16=0$
2) $x^2=-<!--\; -\; -->16$
3) not possible so there is no vertical asymptote
My Work: (Horizontal)
1) $4x/x^2$
2) 4/x
3) What do I do from here?