Help to Integral Calculus Solutions & Examples

Finding $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_1^a\frac{n}{1+x^n}dx$

Non-decreasing sequences
I have the following sequence
$\{n^4-<!--\; -\; -->6n^2\}$
I have to determine if the sequence is non-decreasing, increasing or decreasing.
In my opinion, the sequence is neither, decreasing, non-decreasing nor increasing because it seems to be increasing for all the terms except for the first and second term. Am I right or I am right?
By the way,is there category for such a sequence? where you have an interval of increasing terms and one of decreasing terms or is such a sequence not even possible?

$\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}t}+c\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}x}=0,$
$y=cos(x),t=0,$
$\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}t}=csin(x),t=0$
has a solution
$y=cos(x-ct).$
I wanted to expand both the derivatives as centeblack differences. So, in order to expand my derivatives, I did as shown below;
$\frac{y(i+1)-<!--\; -\; -->y(i-<!--\; -\; -->1)}{2\Delta t}+c(\frac{y(i+1)-<!--\; -\; -->y(i-<!--\; -\; -->1)}{2\Delta x})=0$
$\frac{y}{\Delta t}+c(\frac{y}{\Delta x})=0$
Now, I also intend to prove that the algebraic solution is an exact solution of the difference formula if I choose $\Delta x=c\Delta t$. How do I achieve this goal?

Calculus, solving for increasing/decreasing and concavity
Determine the intervals where the function is increasing and where it is decreasing. (Enter your answers using interval notation.)
$f(x)=\frac{\ln <!--\; \; -->(x)}{x}$
Find intervals of concavity for the graph of the function $f(x)=\frac{\ln <!--\; \; -->(x)}{x}$.
I have already found the first and second derivative but I am confused on how to solve for it $f^{\prime }(x)=-<!--\; -\; -->\frac{\ln <!--\; \; -->(x)-<!--\; -\; -->1}{x^2}$
$f^{″}(x)=\frac{2\ln <!--\; \; -->(x)-<!--\; -\; -->3}{x^3}$

Intervals of increase and decrease
If you have a function and there's an asymptote at say -7, then when doing the intervals for increase decrease, would you do something like increasing from $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->7)\cup <!--\; \cup \; -->(-<!--\; -\; -->7,\text{wherever increase stops})$ and not include the -7, or would the -7 be included

Solve:
$${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}[x]e^{-<!--\; -\; -->x}dx$$

Why do I get different results when testing for increasing/decreasing intervals of a function here?
I have the function $f(x)=6+\frac{6}{x}+\frac{6}{x^2}$ and I want to find the intervals where it increases or decreases. The problem is that when I find $f^{\prime }(x)=0$, which becomes $x=-<!--\; -\; -->2$. Once I put -2 on a number line, and find whether the numbers higher and lower than -2 produce positive or negative numbers when plugged into f′(x), I get both positives and negatives for different numbers when $x>-<!--\; -\; -->2$. For example, $f^{\prime }(-<!--\; -\; -->1)>0$ and $f^{\prime }(5)<0$.
I am not sure if this is because the function has a vertical asymptote at $x=0$ and a horizontal attribute at $y=6$. If so, I'd like to know what I need to do to handle them.
How I found the first derivative:
$\frac{d}{dx}\frac{6}{x}=\frac{[0]-<!--\; -\; -->[6\cdot <!--\; \cdot \; -->1]}{x^2}=\frac{-<!--\; -\; -->6}{x^2}$
$\frac{d}{dx}\frac{6}{x^2}=\frac{[0]-<!--\; -\; -->[6\cdot <!--\; \cdot \; -->2x]}{(x^2{)}^2}=\frac{-<!--\; -\; -->12}{x^3}$
$f^{\prime }(x)=\frac{-<!--\; -\; -->6}{x^2}-<!--\; -\; -->\frac{-<!--\; -\; -->12}{x^3}$
How I found the asymptotes:
When you combine the fractions of f(x), $f(x)=\frac{6x^2+6x+6}{x^2}$. When set equal to zero, $x^2$ has an x-value of zero. Therefore, f(x) has a vertical asymptote at $x=0$.
Once the fractions are combined, and when everything but the coefficients of the leading terms on the numerator and denominator are removed, $f(x)=\frac{6}{1}$. So f(x) has a horizontal asymptote at $y=6$.
How I found the value of $f^{\prime }(x)=0$:
$\frac{-<!--\; -\; -->6}{x^2}-<!--\; -\; -->\frac{12}{x^3}=0$
Add $\frac{12}{x^3}$ to both sides
$\frac{6}{x^2}=\frac{-<!--\; -\; -->12}{x^3}$
Multiply both sides by $x^3$.
$6x=-<!--\; -\; -->12$
Divide both sides by 6.
$x=\frac{-<!--\; -\; -->12}{6}=-<!--\; -\; -->2$

The original functional was $J[x]={\int <!--\; \int \; -->}_0^{1}x\cdot <!--\; \cdot \; -->f(x)\cdot <!--\; \cdot \; -->f^{\prime }(x)dx$ with $h(x)=x$. Solved for the Gateau derivative: $\mathrm{\partial <!--\; \partial \; -->}J[f,h]={\int <!--\; \int \; -->}_0^{1}x\cdot <!--\; \cdot \; -->f(x)\cdot <!--\; \cdot \; -->h^{\prime }(x)dx+{\int <!--\; \int \; -->}_0^{1}x\cdot <!--\; \cdot \; -->f^{\prime }(x)\cdot <!--\; \cdot \; -->h(x)dx$. To use the Euler method, $L=x\cdot <!--\; \cdot \; -->f\cdot <!--\; \cdot \; -->x+x\cdot <!--\; \cdot \; -->f^{\prime }\cdot <!--\; \cdot \; -->x$ such that $L_f=x^2,L_{f^{\prime }}=x^2,\frac{d}{dx}{L}_{f^{\prime }}=2x$. However, I must have made a mistake that I can't see with that because it yields $x^2-<!--\; -\; -->2x=0$ which doesn't tell me anything about what $f$ must be.

How to compute ${\int <!--\; \int \; -->}_a^b(b-<!--\; -\; -->x{)}^{\frac{n-<!--\; -\; -->1}{2}}(x-<!--\; -\; -->a{)}^{-<!--\; -\; -->1/2}dx$

Increasing and decreasing intervals of a function
$f(x)=x^3-<!--\; -\; -->4x^2+2$, which of the following statements are true:
(1) Increasing in $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0)$, decreasing in $(\frac{8}{3},+\mathrm{\infty <!--\; \infty \; -->})$.
(2) Increasing in both $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0)$, decreasing in $(\frac{8}{3},+\mathrm{\infty <!--\; \infty \; -->})$.
(3) decreasing in both $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0)$, and $(\frac{8}{3},+\mathrm{\infty <!--\; \infty \; -->})$.
(4) Decreasing in $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0)$, Increasing in $(\frac{8}{3},+\mathrm{\infty <!--\; \infty \; -->})$.
(5) None of the above.
$f^{\prime }(x)=0=3x^2-<!--\; -\; -->8x=0\Rightarrow <!--\; \Rightarrow \; -->x=\frac{8}{3},x=0$ are the singular point/point of inflection.Could anyone tell me what next?

Difference between first and second fundamental theorem of calculus
In first fundamental theorem of calculus,it states if $A(x)={\int <!--\; \int \; -->}_a^{x}f(t)dt$ then $A^{\prime }(x)=f(x)$.But in second they say ${\int <!--\; \int \; -->}_a^{b}f(t)dt=F(b)-<!--\; -\; -->F(a)$,But if we put x=b in the first one we get A(b).Then what is the difference between these two and how do we prove A(b)=F(b)−F(a)?

Interval related to increasing/decreasing and concavity/convexity
Why do some people use closed intervals when describing the intervals where a function is increasing/decreasing or concave/convex?
For example, given the function $f(x)=x^2-<!--\; -\; -->5x+6$, it says the interval of increase is $[5/2,\mathrm{\infty <!--\; \infty \; -->})$. Why is this written as a closed interval, and not an open one?
Concavity, on the other hand, uses open intervals.

One step of Forward Euler method: $u_{j+1}=u_j+h{f}_j$
Find the region of absolute statibility for the Forward Euler method?
I'm struggling to solve this problem, could I please get a hint?

Obliques asymptotes of a function
$f:[0,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->R,f(x)=\sqrt{x^2+x\ln <!--\; \; -->(e^x+1)}$
I have this function and i need to find out the asymptotes to $+\mathrm{\infty <!--\; \infty \; -->}$ (+infinity)
i calculate the horizontal ones, and they are $+\mathrm{\infty <!--\; \infty \; -->}$
i can't calculate the limits at the obliques asymptotes

I need some help solving the differential equation
$$\begin{gathered} y^{‴}=x+y \\ y(1)=3 \\ {y}^{\prime }(1)=2 \\ {y}^{″}(1)=1 \end{gathered}$$
and h=0.5 with Euler's method

I don't know how to rewrite the equation to a system of equations of the first order..

${\displaystyle \frac{dA}{dt}}=0.5\times <!--\; \times \; -->A\times <!--\; \times \; -->(1-<!--\; -\; -->{\displaystyle \frac{A}{100}})-<!--\; -\; -->10$
with $A(0)=70$ and we want to use Euler's method to get an approximate value for $A(10)$, with a step size of 1.
So the answer sheet says you basically have to use $\text{Ans}+0.5\times <!--\; \times \; -->\text{Ans}\times <!--\; \times \; -->(1-<!--\; -\; -->{\displaystyle \frac{\text{Ans}}{100}})-<!--\; -\; -->10$ with the first Ans being 70, and then of course repeat 10 times.
But I'm wondering, doesn't this actually give you ${\displaystyle \frac{dA(10)}{dt}}$? How is this a correct method?

Use the Linear Approximation to estimate $\mathrm{\Delta <!--\; \Delta \; -->}f=f(3.5)-<!--\; -\; -->f(3)\text{for}f(x)=\frac{3}{1+x^2}$
a) $\mathrm{\Delta <!--\; \Delta \; -->}f\approx <!--\; \approx \; -->$
b) $\mathrm{\Delta <!--\; \Delta \; -->}f=$

Do all functions with vertical asymptotes also have oblique asymptotes?
I just started learning about asymptotes in my Advanced Functions class, and as I was taking a look at all this stuff, a question came up. Do all rational functions that have vertical asymptotes also have an oblique asymptote? Or is an oblique asymptote only formed when the degree of the numerator is 1 higher than the degree of the denominator, and so only functions with a vertical asymptote with a degree of 1 can also have an oblique asymptote?

The notation of the second fundamental theorem of Calculus
I am self studying calculus, and just finished the lesson on the second fundamental theorem of calculus.
the way the theorem is described is:
$\frac{d}{dx}({\int <!--\; \int \; -->}_a^{x}f(t)dt)=f(x)$
and it was told that the meaning is that the derivative of an integral of a function is the function itself.
I don't get how you can get that from this. the expression that I would think suggests this is:
$\frac{d}{dx}(\int <!--\; \int \; -->f(x)dt)=f(x)$
so the derivative of an indefinite integral (as oppose to integrating over a range) of a function is the function itself.
another interpretation of the FToC2 I read here, is that it means that the derivative of the functions that gives the area under the curve of a different function is the different function. this is also something I don't understand how the FToC2 suggests of?
to me, it seems like what this means:
$\frac{d}{dx}({\int <!--\; \int \; -->}_a^{x}f(t)dt)=f(x)$
is how a very small change in x affects that area under f(t) between a (a constant) and x. how do I get from that to the right interpretation?

$f(x)=-<!--\; -\; -->x$ and initial condition $x(0)=1$
Using the Euler Method with the step size $\mathrm{\Delta <!--\; \Delta \; -->}t=1$, estimate $x(1)$ numerically.
I so far did:
$X_{n+1}=X_n+f(x_n)(1)$
$X_1=0$
$X_2=0$
I have a similar question on my test tomorrow. Any help will be appreciated