Help to Integral Calculus Solutions & Examples

Second Fundamental Theorem of Calculus...
Let f have a continuous second derivative. Prove that
$f(x)=f(a)+(x-<!--\; -\; -->a)f^{\prime }(a)+{\int <!--\; \int \; -->}_a^x(x-<!--\; -\; -->t)f^{″}(t)dt.$
Here is my attempt at the problem.
Since f has a continuous second derivative, then the first derivative is also continuous. Therefore, by the first fundamental theorem of calculus, we have that
$f(x)=f(a)+{\int <!--\; \int \; -->}_a^x{f}^{\prime }(t)dt.$
Expanding out the right-hand side of the above using integration by parts, we see that
$f(x)=f(a)+f^{\prime }(t)t-<!--\; -\; -->{\int <!--\; \int \; -->}_a^{x}t{f}^{″}(t)dt.$
This is where I am confused.

If a rock is thrown upward o the planet Mars with a velocity of 10m/s, its height in meters t seconds later is given by $y=10t-<!--\; -\; -->1.86t^2$
a) Find the average velocity over the given time intervals:
i)[1,2] ii)[1,1.5] iii)[1,1.1] iv)[1,1.01] v) [1,1.001]
b) Estimate the instantaneous velocity when t=1.

Does xcos(x) have oblique asymptotes?
Looking at the graph of xcos(x) or xsin(x) etc., it looks like the magnitude of the waves are following a line. Are they oblique asymptotes or something else?
I am familiar with finding the oblique asymptotes of a rational function like $\frac{P(x)}{Q(x)}$ by dividing Q(x) into P(x); however, it doesn't seem like I can do that with xcos(x) et al. what is going on here? and how would you find the equation of the function that is 'controlling' the original function's behavior?

Winding number is locally-constant for general curves (not $C^1$) using variation of argument definition
I've been searching for this proof, here and on the web too, but it seems like the answer is harder to find than expected. There are many similar questions about this but they all (implicitly or explicitly) suppose continuous differentiability.
First, let me state the definition of winding number:
Let $\mathrm{\gamma <!--\; \gamma \; -->}:[0,1]\to <!--\; \to \; -->{\mathbb{C}}^{\mathbb{\ast <!--\; \ast \; -->}}$ be a continuos function. We know that for such a path, there exists a continuous argument function, that is a function $\mathrm{\varphi <!--\; \varphi \; -->}:[0,1]\to <!--\; \to \; -->\mathbb{R}$ such that $\mathrm{\gamma <!--\; \gamma \; -->}(t)=|\mathrm{\gamma <!--\; \gamma \; -->}(t)|e^{i\mathrm{\varphi <!--\; \varphi \; -->}(t)}$, for all $t\in <!--\; \in \; -->[0,1]$. We define the variation of argument along $\mathrm{\gamma <!--\; \gamma \; -->}$:
$\mathrm{v}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{r}\mathrm{g}(\mathrm{\gamma <!--\; \gamma \; -->})=\mathrm{\varphi <!--\; \varphi \; -->}(1)-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->}(0)$
(We can also prove that all such argument functions are of the form $\mathrm{\varphi <!--\; \varphi \; -->}+2n\mathrm{\pi <!--\; \pi \; -->}$, for any integer n, so the variation of argument is well-defined. Also, if $\mathrm{\gamma <!--\; \gamma \; -->}$ is closed, the variation of argument is a multiple of $2\mathrm{\pi <!--\; \pi \; -->}$)
The winding number (index) of the closed curve $\mathrm{\gamma <!--\; \gamma \; -->}$ around 0 is defined as:
$\mathrm{I}(\mathrm{\gamma <!--\; \gamma \; -->},0)=\frac{\mathrm{v}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{r}\mathrm{g}(\mathrm{\gamma <!--\; \gamma \; -->})}{2\mathrm{\pi <!--\; \pi \; -->}}$
For a closed path $\mathrm{\gamma <!--\; \gamma \; -->}:[0,1]\to <!--\; \to \; -->\mathbb{C}\setminus <!--\; \setminus \; -->\{z_0\}$ for some $z_0\in <!--\; \in \; -->\mathbb{C}$, the winding number of $\mathrm{\gamma <!--\; \gamma \; -->}$ around $z_0$ is defined by translation as:
$\mathrm{I}(\mathrm{\gamma <!--\; \gamma \; -->},z_0)=\mathrm{I}(\mathrm{\gamma <!--\; \gamma \; -->}-<!--\; -\; -->z_0,0)$
I consider this as the best definition of winding number because it is intuitive, general and everything follows from it.
Now I want to prove that the winding number is constant on the connected components of the complement of the curve. This is not very hard to prove using the integral expression of the winding number, but that assumes that $\mathrm{\gamma <!--\; \gamma \; -->}$ is $C^1$.
I know that I should somehow prove that $\mathrm{\varphi <!--\; \varphi \; -->}$ is continuous or just that it is constant in an open ball around any point included in the connected component.
Intuitively, I can prove it like this: consider a curve γ that doesn't go through 0, a point $z_0\ne <!--\; \ne \; -->0$ inside a ball around 0 that is contained in the connected component of 0. Imagine two segments from 0 to the curve and from $z_0$ to the curve, as we move along it. Both segments follow the same point on the circle around 0 and their movements along little circles around their respective start points have the same intervals of increase/decrease in angle (going counter-/clockwise). Therefore, whenever one completes a circle in one direction, the other does so too. Hence, the variation of the argument of $\mathrm{\gamma <!--\; \gamma \; -->}$ around 0 and around $z_0$ is the same. By translation, we generalize the result for other points than 0.
This argument is hard to write rigorously and gets a little too geometric, I think. I wonder if an easier take on this can be carried out.

Absolute value f interval?
How would I find where the interval in which f increases and decreases in the following function.
$f(x)=\mid <!--\; \mid \; -->4-<!--\; -\; -->x^2\mid <!--\; \mid \; -->$
What would I do find the zero
I know $f(x)=\{\begin{array}{r}4-<!--\; -\; -->x^2\text{for}|x|<2\\ x^2-<!--\; -\; -->4\text{for}|x|>2\end{array}$
I suspect there are zeroes at 2 and -2 looking at the graph of the function. But how would find the zeroes.So that I can find the f interval.

Say we have a function $y(t)$, that satisfies the ordinary differential equation $\frac{\mathrm{d}y}{\mathrm{d}t}=f(t,y)\text{for }t\in <!--\; \in \; -->(t_0,t_{max}],$, where $t$ takes discrete values, $t_n$, with a constant step size $h=t_{n+1}-<!--\; -\; -->t_n$. We also are given the initial condition, for example, $y(0)=1$.

If we were to use the Backward Euler Method:
$y_{n+1}\approx <!--\; \approx \; -->y_n+hf(t_{n+1},y_{n+1}),$
to approximate a solution for a function, say, $f(y)=-<!--\; -\; -->y$, I am told that it is possible to use this information to show that the Backward Euler Scheme is first order, by expanding the global error as a Taylor series in powers of $h$. (The global error at fixed $t$ is the difference between the approximate solution and the exact solution, $y(t)$)

Can anyone figure out how this might be possible? I cannot find any resources online that dmeonstrate this idea.

Even or Odd symmetry
What type of symmetry does the function $y=\frac{1}{|x|}$ have? Specify the intervals over which the function is increasing and the intervals where it is decreasing.

Equation of a hyperbola given its asymptotes
Find the equation of the hyperbola whose asymptotes are $3x-<!--\; -\; -->4y+7$ and $4x+3y+1=0$ and which pass through the origin.
The equation of the hyperbola is obtained in my reference as
$(3x-<!--\; -\; -->4y+7)(4x+3y+1)=K=7$
So it make use of the statement, the equation of the hyperbola = equation of pair of asymptotes + constant
I understand that the pair of straight lines is the limiting case of hyperbola.
Why does the equation to the hyperbola differ from the equation of pair of asymptotes only by a constant ?

Conjecture: If $a_k-<!--\; -\; -->k\mathrm{\pi <!--\; \pi \; -->}(a_k)=0$ then $a_{k+1}/a_k\to <!--\; \to \; -->e$.
A natural question to ask is for what n does $\mathrm{\pi <!--\; \pi \; -->}(n)\mid <!--\; \mid \; -->n$ where $\mathrm{\pi <!--\; \pi \; -->}(n)$ is the prime counting function. It can be shown with a discrete variant of the intermediate value theorem that such an n exists for any $k=n/\mathrm{\pi <!--\; \pi \; -->}(n)$. (Intuition: $f(n)=n-<!--\; -\; -->k\mathrm{\pi <!--\; \pi \; -->}(k)$ can only increase by 1 and decrease by $k-<!--\; -\; -->1$. Since $f(2)<0$ for $k>3$ and $f(n)>0$ for large n, there must be a zero).
Computations suggest that all solutions to $n-<!--\; -\; -->k\mathrm{\pi <!--\; \pi \; -->}(n)=0$ are grouped at some $n_k$ in an interval of around $\sqrt{n_k}$ width. But more surprisingly to me, for any pair of solutions $a_k$ and $a_{k+1}$, with the former being in the k group and the latter in the $k+1$ group, the ratio $a_{k+1}/a_k$ seems to be about e.
My conjecture: If $a_k-<!--\; -\; -->k\mathrm{\pi <!--\; \pi \; -->}(a_k)=0$ then $a_{k+1}/a_k\to <!--\; \to \; -->e$. In other words, the solutions of $f(n)=n-<!--\; -\; -->k\mathrm{\pi <!--\; \pi \; -->}(k)$ grow exponentially as $e^k$.
I am at a loss why this is the case. My guess is the gaps between primes $[n!,n!+n]$ comes into play, giving us Stiring's formula and hence our e factor, but I really have no clue.
Can someone either (1) provide a proof or disproof of this conjecture or (2) give an extremely plausible method of attack? I would also appreciate suggestions in the comments for ideas/theorems/techniques to investigate to help me make my own crack at the problem.

"Whats the result of one step of Backward Euler Method with h = 0.1 applied on the IVP:
$y^{\prime }(t)=5y(t)+10$
$y(0)=1$"?
So, been trying to understand the Backward Euler Method for a while now and almost get it. However, I don't know exactly what to do and would really appreciate to see how someone who knows what they're doing tackle this problem.

Is the floor function increasing or decreasing or both in the interval [0,1)
I was attempting a quiz in which I encountered this question, Now the graph of the floor function is constant in the interval [0,1) so technically the function should be neither increasing nor decreasing, but the answer to the question is its both increasing and decreasing. So is the answer wrong or am I unaware of a concept?
Similarly, there was another question about the function $f(x)=x^2-<!--\; -\; -->3x+2$ in the interval $[0,inf]$ but the graph of the function comes down from 0 and there is a critical point at 1 and then goes up, so is decreasing and increasing in the interval $[0,inf)$ but the answer is its neither increasing nor decreasing, now one of my classmates gave me a line of reasoning that $f^{\prime }(x)<=0and{f}^{\prime }(x)>=0$ in the inerval $[0,inf)$ which implies $f^{\prime }(x)=0$ but I'm not really convinced with the argument because the graph clearly shows that the function is decreasing then increasing. So am I missing a concept here or the answers are wrong?

Solve the first-order system that satisfies the given initial conditions using the Euler Method for $y(0.5)$ and $z(0.5)$, using a mesh size of $h=0.1$:
1. $y^{″}-<!--\; -\; -->6z^2{z}^{\prime }-<!--\; -\; -->y^{\prime }-<!--\; -\; -->x^{3}y=0;y(0)=1,y^{\prime }(0)=1.5$
2. $z^{‴}+3y^2(z^{″}{)}^2-<!--\; -\; -->5z^{\prime }-<!--\; -\; -->x^{2}z=0;z(0)=1.25,z^{\prime }(0)=1.5,z^{″}(0)=2$
Please help. I just can’t figure this problem out. We’re supposed to use $u=y$, $v=y^{\prime }$, $w=z$, $g=z^{\prime }$, $k=z^{″}$ when defining first-order system of ODE’s.

I am asked to solve the following problem. Use Euler's modified method to solve the initial value problem $\frac{dx}{dt}=\frac{1+x^2}{t},1\le <!--\; \le \; -->t\le <!--\; \le \; -->4,x(1)=0$
The step size is not given here. Now, I can solve this using Euler's method but I am not able to find the formula for modified one in the mentioned textbook. So, I need the method here. I need to use the following as textbook: Brain Bradie, A friendly Introduction to Numerical Analysis (Pearson)

How do you check which intervals a cubic function will increase and in which intervals it will decrease?
I was trying to find the intervals in which the cubic function $4x^3-<!--\; -\; -->6x^2-<!--\; -\; -->72x+30$ would be strictly increasing and strictly decreasing.
I managed to get the fact that at the values {-2,3} the differential of the function is zero. However this divides the function in three intervals, how can i know which intervals the function increases in and which intervals the function decreases in?
Note: I know you could simply plot the function. I was hoping for a more analytical method.

Some doubts with the sign of a derivative
Good evening to everyone. The derivative is defined in the following order:
$\frac{d}{dx}f(x)=\frac{-<!--\; -\; -->x^2-<!--\; -\; -->x+11}{{(x+3)}^2}{e}^{2-<!--\; -\; -->x}$ for $x<-<!--\; -\; -->3$
$\frac{d}{dx}f(x)=\frac{x^2+x-<!--\; -\; -->11}{{(x+3)}^2}{e}^{2-<!--\; -\; -->x}$ for $-<!--\; -\; -->3<x<2$ and $\frac{d}{dx}f(x)=\frac{x^2+x+1}{{(x+3)}^2}{e}^{x-<!--\; -\; -->2}$ for $x>2$. If you'll do the sign of the first one it'll be $x^2+x-<!--\; -\; -->11<0$ therefore only $x_1=\frac{-<!--\; -\; -->1+\sqrt{45}}{2}$ verifies the conditions, for the second one it is $x^2+x-<!--\; -\; -->11>0$ therefore $x_1=\frac{-<!--\; -\; -->1+\sqrt{45}}{2}$ and $x_1=\frac{-<!--\; -\; -->1-<!--\; -\; -->\sqrt{45}}{2}$, the last one $x^2+x+1>0$ has the solutions $x_1=\frac{-<!--\; -\; -->1-<!--\; -\; -->i\sqrt{3}}{2}$ and $x_2=\frac{-<!--\; -\; -->1+i\sqrt{3}}{2}$ (here I think that I did a mistake). Therefore our function should increase on the interval $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},\frac{-<!--\; -\; -->1-<!--\; -\; -->\sqrt{45}}{2})$ then decrease on the interval $(\frac{-<!--\; -\; -->1-<!--\; -\; -->\sqrt{45}}{2},\frac{-<!--\; -\; -->1+\sqrt{45}}{2})$ then increase again on the interval $(\frac{-<!--\; -\; -->1+\sqrt{45}}{2},\frac{-<!--\; -\; -->1-<!--\; -\; -->i\sqrt{3}}{2})$ and decrease again on $(\frac{-<!--\; -\; -->1+i\sqrt{3}}{2},\mathrm{\infty <!--\; \infty \; -->})$. But Instead the solutions of the third equation are $x_1=\frac{-<!--\; -\; -->1-<!--\; -\; -->\sqrt{5}}{2}$ and $x_2=\frac{-<!--\; -\; -->1+\sqrt{5}}{2}$. And it is decreasing on $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},\frac{-<!--\; -\; -->1-<!--\; -\; -->\sqrt{45}}{2})$, then increasing on $(\frac{-<!--\; -\; -->1-<!--\; -\; -->\sqrt{45}}{2},-<!--\; -\; -->3)$ then decreasing on (-3,2) then increasing again on $(2,\mathrm{\infty <!--\; \infty \; -->})$. I don't get where I'm wrong.. neither with the results of quadratic equation nor with the intervals of monotony(sign) of the derivative.

I have a question from my book and it says essentially, consider the IVP $x^{\bullet <!--\; \bullet \; -->}=-<!--\; -\; -->x$ with $x(0)=1$, what is the exact value of $x(1)$, then using Eulers method with step size1 , estimate $x(1)$ call this $x^{\ast <!--\; \ast \; -->}(1)$ , then repeat for step sizes of ${10}^{-<!--\; -\; -->n}$ for $n=1,2,3,4$ then finally plot $E=|x^{\ast <!--\; \ast \; -->}(1)-<!--\; -\; -->x(1)|$ as a function of step size and then as $lnE$ vs $lnt$.
Now I am having some issues and ill explain,
for the first part I get that $x(1)=e^{-<!--\; -\; -->1}$
We have $f(x)=-<!--\; -\; -->x$ and $x_0=1$
so I have that by Euler method
$x_1=x_0+f(x_0)t$
which would imply that for $t=1,x_1=0$
and then for the second part it would imply $x_1=0.9$, then 0.99, then 0.999 and finally 0.9999.
But this doesn't seem to make any sense to me. Seeing as none of these are close to $e^{-<!--\; -\; -->1}$
So I am confused in regard to where I am making mistakes, or where everything is going wrong. For the plotting part, I am also stuck because of this. I am looking for any help and advice.

Now I consider the harmonic oscillator problem.
The ordinal differential equation is
$\begin{array}{rl}\stackrel{\dot{}<!--\; \dot{}\; -->}{q}& =p\\ \stackrel{\dot{}<!--\; \dot{}\; -->}{p}& =-<!--\; -\; -->q\end{array}$
In symplectic Euler method, where
$\begin{array}{rl}\frac{p^{(m+1)}-<!--\; -\; -->p^{(m)}}{\mathrm{\Delta <!--\; \Delta \; -->}t}& =-<!--\; -\; -->q^{(m)}\\ \frac{q^{(m+1)}-<!--\; -\; -->q^{(m)}}{\mathrm{\Delta <!--\; \Delta \; -->}t}& =p^{(m+1)}\end{array}$
the flow operator ${\mathrm{\psi <!--\; \psi \; -->}}_{\mathrm{d},\mathrm{\Delta <!--\; \Delta \; -->}}=((1-<!--\; -\; -->\mathrm{\Delta <!--\; \Delta \; -->}{t}^2)q+\mathrm{\Delta <!--\; \Delta \; -->}t\cdot <!--\; \cdot \; -->p,-<!--\; -\; -->\mathrm{\Delta <!--\; \Delta \; -->}t\cdot <!--\; \cdot \; -->q+p)$ is symplectic.
Here, the textbook suggests that this flow operator stricktly integrates the modified Hamiltonian and this Hamiltonian is invariant. This modified hamiltonian is
$\begin{array}{r}\stackrel{~<!--\; ~\; -->}{H}=\frac{q^2+p^2}{2}-<!--\; -\; -->\frac{qp}{2}\mathrm{\Delta <!--\; \Delta \; -->}t\end{array}$
I cannot understand how to derive $\stackrel{~<!--\; ~\; -->}{H}$.

Having trouble wrapping my brain around this question.. don't know where to start.
A. Consider the differential equation $f^{\prime }(x)=(x+1)f(x)$ with $f(0)=1$. What is the exact formula for $f(x)$?
I tried solving for $f(x)$ but that just gives me $(x+1)f^{\prime }(x)$. I feel like that's not sufficient here.
B. Solve for $f(.2)$ using Eulers approximation method with increment $h=.01$ for $x\in <!--\; \in \; -->[0,0.2]$.
Edit- my work for part A.
integral f'(x) = integral(x+1) *f(x)+(x+1) * integral (f(x)
so f(x) = ((x^2+x)/2)*f(x) + (x+1) * (f(x)^2)/2.
I don't believe this is the right answer.

Consider the function $f(x)=sup\{x^2,\cos <!--\; \; -->x\}$, $x\in <!--\; \in \; -->[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}]$. Show that there exists a point $x_0\in <!--\; \in \; -->[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}]$ such that $f(x_0)=\underset{x\in <!--\; \in \; -->[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}]}{min}f(x)$. Also show that $\cos <!--\; \; -->x_0=x_0^2$.
The first part is obvious due to continuity of f in $[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}]$. Second part is also obvious from the graph of f. But how to mathematically prove that $\cos <!--\; \; -->x_0=x_0^2$?
I was trying to show it using the expression of f $f(x)=\frac{1}{2}\{x^2+\cos <!--\; \; -->x+|x^2-<!--\; -\; -->\cos <!--\; \; -->x|\}$. Can anybody please help me out?