Help to Integral Calculus Solutions & Examples

Show that :
$${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{{\text{e}}^{-<!--\; -\; -->x}}{x^2}{(\frac{1}{1-<!--\; -\; -->{\text{e}}^{-<!--\; -\; -->x}}-<!--\; -\; -->\frac{1}{x}-<!--\; -\; -->\frac{1}{2})}^2\text{d}x=\frac{7}{36}-<!--\; -\; -->\ln <!--\; \; -->A+\frac{\mathrm{\zeta <!--\; \zeta \; -->}\left(3\right)}{2{\mathrm{\pi <!--\; \pi \; -->}}^2}$$

How do you show that
$${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{\log <!--\; \; -->|\tan <!--\; \; -->x|}{1+x^2}dx=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}\log <!--\; \; -->(\tanh <!--\; \; -->1)?$$

Evaluate the intergral using the fundamental theorem of calculus.
$${\int <!--\; \int \; -->}_{-<!--\; -\; -->10}^{10}(8x^9+5x^5)dx$$

Consider the ODE
$f(u)\equiv <!--\; \equiv \; -->u^{\prime }=au+b,u(t_0=0)\equiv <!--\; \equiv \; -->u^0=u_0$
For a timestep $k$, the forward-difference approximator for $u^{\prime }$ gives the Euler update as
$u^{n+1}=u^n+kf(u^n)=u^n+k(a{u}^n+b)$
I am trying to come up with a closed form expression for $u^n$, that is, one that involves only the constants $a,b,u_0,k$.
I've tried to do this by first writing $u^n$ as a cumulative sum
$u^n=u^n+k(a{u}^n+b)=u_0+\underset{i=0}{\overset{n-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}k(a{u}^i+b)$
I then tried to expand the sum on the right and see if there was any easily identifiable pattern to write more generally, or if it would perhaps resemble an expansion of some other known function.After some ugly algebra, this is something like
$\begin{array}{rl}{u}^n& =u_0+k(a{u}_0+b)+k(a{u}_1+b)+k(a{u}_2+b)+...\\ & =u_0+k(a{u}_0+b)+(k^{2}a+k)(a{u}_0+b)+(k^3{a}^2+2k^{2}a+k)(a{u}_0+b)+...\\ & =u_0+(k^3{a}^2+3k^{2}a+3k+...)(a{u}_0+b)\end{array}$
Nothing jumps out to me, and I'm not sure this approach is promising. Any ideas?
I've massaged it some more and found that it can be written as
$u^n=(1+ka{)}^n{u}_0+C_n$
Where C is a function of a,b,k,n, which I've yet to deduce...

Oblique asymptotes?
A rational function, $\frac{p(x)}{q(x)}$ has an oblique asymptote only when the degree of p(x)= degree of q(x)−1.
What "causes" the "slant" of the asymptote? Most asymptotes are caused by a function approaching an undefined value - I assume this is the same, but why (unlike others) would these asymptotes be slanted?
Why does this only work with a difference of 1 between degrees?

An upper bound for the global error in Euler's method to solve a first order ODE numerically is given by the equation
$\frac{Mh}{2L}(e^{L(t_{i+1}-<!--\; -\; -->a)}-<!--\; -\; -->1),$
where $t_{i+1}$ is the $i+1$ cell in the abcsisa and $a$ is the first abcisa. M is a bound for the second derivative of the ODE unknown $y$ and $L$ is the Lipschitz continuity property bound. This is derived in most of numerical analysis books.

My question is the following: Having $L$ small is good for convergence and stability (right?) in the sense that $L$ small implies little change. If $L=0$ then the function $f(t,y)$ (on the right hand side of the ODE) is constant on y. If $L<1$ the function is contracting. So why is $L$ in the denominator of the bound? My intuition (which is not working good) says that the error bound should diminish as $L$ diminshes.

What is wrong with my intuition here?

Does improved Euler method fail when we increase step size?meaning does it give worse answer than standard Euler method when our step size is big?

Suppose f(x) is such that
$${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}{e}^{tx}f(x)dx=\arcsin <!--\; \; -->(t-<!--\; -\; -->\sqrt{\frac{1}{2}})$$

Round 8.647 to one decimal place.

Given that x' = tx, x(0) = 1, find x(1), and use Euler’s method with n steps to find an approximation to x(1). (You will end up with a product of n terms whose limit (as $n\to <!--\; \to \; -->$$\mathrm{\infty <!--\; \infty \; -->}$) is sqrt(e))
I calculated x(1) using the standard way of solving the ODE by integration and managed to find sqrt(e) but I'm having trouble finding this using Euler. What I did is:
Xn+1 = Xn + X'h where h=$\frac{1}{N}$
substituting the X' by tx I have
Xn+1 = Xn + tx($\frac{1}{N}$) = Xn(1+$\frac{t}{N}$)
So Xn = (1+$(1+\frac{t}{N}{)}^N$ and simplifying this I have Xn = $((1+\frac{1}{N}{)}^N{)}^t$ and taking the limit as $n\to <!--\; \to \; -->$$\mathrm{\infty <!--\; \infty \; -->}$ I only get $e^t$. Where did I make a mistake? I don't know where does the square root comes from.

I have
$y\mathrm{\prime <!--\; \prime \; -->}=9x,h=\frac{1}{2},y(0)=1$
And I want to find y(2) using the improved euler's method. And I know that it goes:
$y_{n+1}=y_n+h\frac{f(x_n,y_n)+f(x_{n+1},u_{n+1})}{2}$
What I don't really "get" is the $u$. Is it a sort of an replacement for $y$? In which case it should not mean anything considering the equation I'm given.

For ${\displaystyle f(x)={\int <!--\; \int \; -->}_x^{x+1}\sin <!--\; \; -->({\text{e}}^t)\text{d}t}$
Prove that : ${\displaystyle {\text{e}}^x|f(x)|\le <!--\; \le \; -->2}$

Find the interval of increase and decrease ,local extremes , inflection points of $f(x)=x^{\frac{4}{3}}-<!--\; -\; -->x^{\frac{1}{3}}$
Since $f(x)=x^{\frac{4}{3}}-<!--\; -\; -->x^{\frac{1}{3}}\Rightarrow <!--\; \Rightarrow \; -->f^{\prime }(x)=\frac{4}{3}{x}^{\frac{1}{3}}-<!--\; -\; -->\frac{1}{3}{x}^{\frac{-<!--\; -\; -->1}{3}}$

Intervals in which f(x) is Strictly Increasing/Decreasing
Find the intervals in which $f(x)=\sin <!--\; \; -->x+\cos <!--\; \; -->x,0\le <!--\; \le \; -->x\le <!--\; \le \; -->2\mathrm{\pi <!--\; \pi \; -->}$ is strictly increasing/decreasing.
First I find the derivative $f^{\prime }(x)=\cos <!--\; \; -->x-<!--\; -\; -->\sin <!--\; \; -->x$, then put $f^{\prime }(x)=0$, getting $\tan <!--\; \; -->x=1$. The principal solutions of $\tan <!--\; \; -->x=1$ are $x=\mathrm{\pi <!--\; \pi \; -->}/4$ and $x=5\mathrm{\pi <!--\; \pi \; -->}/4$, which gives the intervals $[0,\mathrm{\pi <!--\; \pi \; -->}/4)$, $(\mathrm{\pi <!--\; \pi \; -->}/4,5\mathrm{\pi <!--\; \pi \; -->}/4)$, and $(5\mathrm{\pi <!--\; \pi \; -->}/4,2\mathrm{\pi <!--\; \pi \; -->})$.
After this I am stuck. The book just solves the question by making a table showing the interval, sign of the derivative as positive or negative, and strictly increasing for positive and vice versa.
I just want to know how do they get it positive or negative.It would really help if someone did one for the interval $(\mathrm{\pi <!--\; \pi \; -->}/4,5\mathrm{\pi <!--\; \pi \; -->}/4)$.

Intervals on which function is increasing and decreasing
Let $p(x)=x^5-<!--\; -\; -->q^{2}x-<!--\; -\; -->q$, where q is a prime number. I want to understand how to determine when the function will be decreasing and increasing on the intervals given below.
We compute $p^{\mathrm{\prime <!--\; \prime \; -->}}(x)=5x^4-<!--\; -\; -->q^2$ and look for the critical points.
$5x^4-<!--\; -\; -->q^2=0⟺<!--\; ⟺\; -->x=\pm <!--\; \pm \; -->\frac{\sqrt{q}}{\sqrt[4]{5}}$
Hence we have to investigate the behavior of p′(x) for each of these intervals $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->\frac{\sqrt{q}}{\sqrt[4]{5}})$, $(-<!--\; -\; -->\frac{\sqrt{q}}{\sqrt[4]{5}},\frac{\sqrt{q}}{\sqrt[4]{5}})$ and $(\frac{\sqrt{q}}{\sqrt[4]{5}},\mathrm{\infty <!--\; \infty \; -->})$ this will indicate when the function will be increasing and decreasing. How can this be determined when the expression $\frac{\sqrt{q}}{\sqrt[4]{5}}$ contains a prime number???
The answer should be : the function will be increasing for $x<\frac{\sqrt{q}}{\sqrt[4]{5}}$ and strictly decreasing for $-<!--\; -\; -->\frac{\sqrt{q}}{\sqrt[4]{5}}<x<\frac{\sqrt{q}}{\sqrt[4]{5}}$ and strictly increasing again for $x>\frac{\sqrt{q}}{\sqrt[4]{5}}$.

Identifying Asymptotes of a Hyperbola
basic hyperbolic functionHow would I find the vertical and horizontal asymptotes of a $y=\frac{1}{x}$ function algebraically? For example, $y=-<!--\; -\; -->\frac{2}{x+3}-<!--\; -\; -->1$ (as you would type into a calculator). Simply, how do I find the x and y values by looking at this equation? In other words, the middle point where the asymptotes in the picture has moved and the whole graph has been vertically stretched, where are the asymptotes now?

Finding the horizontal asymptotes of a function
Here is the function I am trying to find the horizontal asymptotes for:
$y=\frac{1-<!--\; -\; -->2^x}{1+2^x}$
Could you please explain how the horizontal asymptotes can be found for this function?

Evaluating the definite integral ${\int <!--\; \int \; -->}_{-<!--\; -\; -->4}^4(10x^9+7x^5)dx$

Finding the asymptotes of a hyperbola
Given the hyperbola $y^2/9$ what would the equation of the asymptotes be?

Prove the integral inequality
$\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}-<!--\; -\; -->\frac{1}{2}\le <!--\; \le \; -->{\int <!--\; \int \; -->}_0^1\frac{\arcsin <!--\; \; -->x}{1+x^8}\le <!--\; \le \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}-<!--\; -\; -->1$
$\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}-<!--\; -\; -->\frac{1}{2}\le <!--\; \le \; -->{\int <!--\; \int \; -->}_0^1\frac{\arcsin <!--\; \; -->x}{1+x^8}\le <!--\; \le \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}-<!--\; -\; -->1$
I can say that I have close to zero experience with integral inequalities. I've searched for couple of theorems most of them involving monotonous function on [0,1]. Arcsin is increasing in this case and the other one is decreasing. You could say we have a product of increasing and decreasing function on this interval however I didn't find any theorems that cover this case. Also this integral doesn't seem to have elementary function.
Another thing I noticed is that most of the theorems cover only 1 part of this inequality; either $\le <!--\; \le \; -->or\ge <!--\; \ge \; -->$ so my guess is we need some type of combination or perhaps find a new integral that is equal to $\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}-<!--\; -\; -->\frac{1}{2}$ and another that is equal to $\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}-<!--\; -\; -->1$.
What's the correct approach when we need to prove that the value of certain integral is in a provided interval?