Master Calculus 1 Practice Problems with Expert Help

If a function is holomorphic in a certain domain, does it mean it necessarily has an antiderivative
Say I have a holomrphic function f(z) in domain D, say $|z|>4$ and my function is not holomorphic for $|z|\le <!--\; \le \; -->4$ - it has some poles there. Does it mean there is necessarily an antiderivative for f(z)?
I know I can use Morera's theorem and calculate all the residues inside $|z|\le <!--\; \le \; -->4$ and check if their sum is zero. However what seemed more intuitive to me is to prove that since ${\int <!--\; \int \; -->}_{\mathrm{\gamma <!--\; \gamma \; -->}}f(z)dz$ for every $\mathrm{\gamma <!--\; \gamma \; -->}$ contained in D, depends only on the start and end points, there exists an antiderivative for f(z).
Is my prof correct? Can I conclude that in general, a holomorphic function in a domain D not only has infinitely many derivatives in D, but it has also infinitely many antiderivatives there?

Must a Function be Bounded for the Antiderivative to Exist over a Given Interval?
In my Calculus class, we were given the following definition of antiderivative:
Let f be a function defined on an interval.
An antiderivative of f is any function F such that $F^{\prime }=f$.
The collection of all antiderivatives of f is denoted ${\displaystyle \int <!--\; \int \; -->f(x)dx}$.
My question is, don't we have to say that f should be bounded in the definition?
If not, then f is not integrable by definition, so we can't say anything about ${\displaystyle \int <!--\; \int \; -->f(x)dx}$, right?

I used all the trig and log tricks but still can't compute this limit
$\underset{x\to <!--\; \backslash rightarrow\; -->0}{lim}\frac{\ln <!--\; \; -->(1+{\sin }^2<!--\; \; -->(2x))}{1-<!--\; -\; -->{\cos }^2<!--\; \; -->(x)}.$

I was trying to find asymptotes for a function $f(x)=x{e}^{\frac{1}{x-<!--\; -\; -->2}}$

Evaluate $\underset{n\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{e^{n^2}}{(2n)!}$

Use definition of limit to prove $lim(\ln <!--\; \; -->(\frac{ne+1}{n}))=1$

Antiderivative of $\arctan <!--\; \; -->(-<!--\; -\; -->x^2)$.
As I said in the title I'm trying to find an antiderivative of $f(x)=\arctan <!--\; \; -->(-<!--\; -\; -->x^2)$.
I am aware that e.g. WolframAlpha can find one, but I have no clue how to do it by hand. Can anyone give me a hint?

For $f:<!--\; :\; -->\mathbb{R}\to <!--\; \backslash rightarrow\; -->\mathbb{R}$ and $\mathrm{\forall <!--\; \forall \; -->}n\in <!--\; \in \; -->\mathbb{N}$ s. t. $n\ge <!--\; \ge \; -->2$
$\underset{x\to <!--\; \backslash rightarrow\; -->0}{lim}\{f(x)+f(nx)\}=0$
Prove/disprove :
$\underset{x\to <!--\; \backslash rightarrow\; -->0}{lim}f(x)=0$

We must to calculate
$\underset{x\to <!--\; \backslash rightarrow\; -->\mathrm{\pi <!--\; \pi \; -->}/2}{lim}\frac{\sin <!--\; \; -->x-<!--\; -\; -->1}{2x-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}.$

How would you go about solving this limit knowing it's result but no way of proving it.
$\underset{n\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(n-<!--\; -\; -->\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}\left(e^{\frac{k}{n^2}}\right))$

I need to solve the following problem:
$\underset{n\mapsto <!--\; \mapsto \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(\sqrt[n]{n}-<!--\; -\; -->1{)}^{\frac{1}{n}}.$

If
$M_{Z_n}(t)={\left(\frac{3e^{\frac{-<!--\; -\; -->t}{4\sqrt{n}}}+e^{\frac{3t}{4\sqrt{n}}}}{4}\right)}^{2n}$
Calculate ${\displaystyle \underset{n\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{M}_{Z_n}(t)}$ and interpret the result

If $y_n>0$, $lim\frac{x_n}{y_n}=a$ and $\sum <!--\; \sum \; -->y_n=+\mathrm{\infty <!--\; \infty \; -->}$, then $lim\frac{x_1+x_2+...+x_n}{y_1+y_2+...+y_n}=a$

How to find the general antiderivative of $f(x)=x(6-<!--\; -\; -->x{)}^2$?
I want to find the general antiderivative of $f(x)=x(6-<!--\; -\; -->x{)}^2$. However, I keep getting it wrong.
I am new to antiderivatives, but I think the first thing I should do is differentiate.
$\frac{d}{dx}(6-<!--\; -\; -->x{)}^2=2(6-<!--\; -\; -->x)\cdot <!--\; \cdot \; -->-<!--\; -\; -->1=-<!--\; -\; -->2(6-<!--\; -\; -->x)$
$f^{\prime }(x)=[(6-<!--\; -\; -->x{)}^2\cdot <!--\; \cdot \; -->1]+[-<!--\; -\; -->2(6-<!--\; -\; -->x)\cdot <!--\; \cdot \; -->x]=(6-<!--\; -\; -->x{)}^2-<!--\; -\; -->2x(6-<!--\; -\; -->x)$
According to the antiderivative power rule, when $n\ne <!--\; \ne \; -->-<!--\; -\; -->1$, $\int <!--\; \int \; -->x^{n}dx=\frac{x^{n+1}}{n+1}+C$.
So it seems like $\int <!--\; \int \; -->(6-<!--\; -\; -->x{)}^{2}dx=\frac{(6-<!--\; -\; -->x{)}^3}{3}+C$.
However, I can't find a rule that seems like it would work with $-<!--\; -\; -->2x(6-<!--\; -\; -->x)$. The closest thing that I can find is the "Multiplication by Constant Rule", $\int <!--\; \int \; -->cf(x)dx=c\cdot <!--\; \cdot \; -->\int <!--\; \int \; -->(f(x))dx$, but I'm not sure if I'm allowed to change $2x(6-<!--\; -\; -->x)$ into the form $2(6x-<!--\; -\; -->x^2)$.
$\int <!--\; \int \; -->2(6x-<!--\; -\; -->x^2)$
$=2\cdot <!--\; \cdot \; -->\int <!--\; \int \; -->(6x-<!--\; -\; -->x^2)$
$=2(6\int <!--\; \int \; -->x-<!--\; -\; -->\int <!--\; \int \; -->x^2)$
$=2(\frac{6x^2}{2}-<!--\; -\; -->\frac{x^3}{3})+C$
But $\frac{(6-<!--\; -\; -->x{)}^3}{3}-<!--\; -\; -->(6x^2-<!--\; -\; -->\frac{2x^3}{3})+C$ is incorrect.

Show that the sequence $\left\{b_n\right\}:b_n={\displaystyle \frac{5n^2+2n-<!--\; -\; -->1}{3n^2-<!--\; -\; -->14}}$ converges

Where is the absolute value when computing antiderivatives?
Here is a typical second-semester single-variable calculus question:
$\int <!--\; \int \; -->\frac{1}{\sqrt{1-<!--\; -\; -->x^2}}dx$
Students are probably taught to just memorize the result of this since the derivative of arcsin(x) is taught as a rule to memorize. However, if we were to actually try and find an antiderivative, we might let $x=\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}⟹<!--\; ⟹\; -->dx=\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}d\mathrm{\theta <!--\; \theta \; -->}$
so the integral may be rewritten as $\int <!--\; \int \; -->\frac{\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{\sqrt{1-<!--\; -\; -->{\sin }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}}d\mathrm{\theta <!--\; \theta \; -->}=\int <!--\; \int \; -->\frac{\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{\sqrt{{\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}}d\mathrm{\theta <!--\; \theta \; -->}$
At this point, students then simplify the denominator to just $\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$, which boils the integral down to
$\int <!--\; \int \; -->1d\mathrm{\theta <!--\; \theta \; -->}=\mathrm{\theta <!--\; \theta \; -->}+C=\arcsin <!--\; \; -->x+C$
which is the correct antiderivative. However, by definition, $\sqrt{x^2}=|x|$, implying that the integral above should really be simplified to $\int <!--\; \int \; -->\frac{\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{|\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}|}d\mathrm{\theta <!--\; \theta \; -->}=\int <!--\; \int \; -->\pm <!--\; \pm \; -->1d\mathrm{\theta <!--\; \theta \; -->}$ depending on the interval for $\mathrm{\theta <!--\; \theta \; -->}$. At this point, it looks like the answer that we will eventually arrive at is different from what we know the correct answer to be.

Existence of antiderivative without Cauchy-Goursat's theorem
I wonder if anybody has tried the following kind of direct proof for the existence of an antiderivative of an analytic function on a star-shaped domain.
Theorem: Let $f:D\to <!--\; \to \; -->\mathbb{C}$ be an analytic function on a star-shaped domain D. Then f has an antiderivative F on D.
"proof": For simplicity, assume that every point in D is connected to $0\in <!--\; \in \; -->\mathbb{C}$ by a line segment. Define
$F(z)={\int <!--\; \int \; -->}_0^{1}zf(zt)dt.$
Then $\underset{h\to <!--\; \to \; -->0}{lim}\left(\frac{F(z+h)-<!--\; -\; -->F(z)}{h}\right)=\underset{h\to <!--\; \to \; -->0}{lim}{\int <!--\; \int \; -->}_0^1\left(\frac{(z+h)f(zt+ht)-<!--\; -\; -->zf(t)}{h}\right)dt.$
It can be checked that as $h\to <!--\; \to \; -->0$, the integrand converges to $\frac{d}{dt}tf(zt).$.
Therefore, $F^{\prime }(z)=f(z)$, provided that the limit and integral are interchangeable. qed.
Of course, a limit and integral cannot always be interchanged. But I wonder if anybody seriously considered the above line of proof.
Thanks. As far as I can search from several textbooks in complex variables, the above theorem is proved by using Cauchy-Goursat's theorem. More concretely, they use the equality ${\int <!--\; \int \; -->}_{z_0}^{z+h}f(z)dz-<!--\; -\; -->{\int <!--\; \int \; -->}_{z_0}^{z}f(z)dz={\int <!--\; \int \; -->}_z^{z+h}f(z)dz,$, which can be justified by Cauchy-Goursat's theorem. The point of my question is: Is it possible to directly invoke to the computation as above, without using Cauchy-Goursat's theorem? If this is possible, we get another proof of Cauchy-Goursat's theorem, at least for star-convex domains.

Find an antiderivative of the function
Let $u:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be a function. Find an antiderivative of $u^2{u}^{‴}$ in terms of u and its derivatives.
Note: I know an antiderivative of uu′′′ is $u{u}^{″}-<!--\; -\; -->\frac{u^{\prime 2}}{2}$. However it seems for this problem one needs to evaualate antiderivative of either uu′u′′ or $u^{\prime 3}$ which can't be computed easily via integration by parts

Jump discontinuity in antiderivative
I am dealing with integral $I={\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}dx\frac{{\sin }^2<!--\; \; -->x}{a^2+{\sin }^2<!--\; \; -->x}.$
I know the answer is $I=\mathrm{\pi <!--\; \pi \; -->}(1-<!--\; -\; -->\frac{a}{\sqrt{1+a^2}}).$.
However I am having some uncertainty getting there. I know that before plugging in the integration limits the solution looks like $I={[x-<!--\; -\; -->\frac{a{\tan }^{-<!--\; -\; -->1}<!--\; \; -->(\frac{\sqrt{1+a^2}}{a}\tan <!--\; \; -->x)}{\sqrt{1+a^2}}]}_{x=0}^{x=\mathrm{\pi <!--\; \pi \; -->}}$.
but at first sight this would yield merely $\mathrm{\pi <!--\; \pi \; -->}$. I can see that the right step to recover the correct solution is to take ${\tan }^{-<!--\; -\; -->1}<!--\; \; -->(\frac{\sqrt{1+a^2}}{a}\tan <!--\; \; -->x){|}_{x=\mathrm{\pi <!--\; \pi \; -->}}=\mathrm{\pi <!--\; \pi \; -->}$ however how do I justify this (besides that it yields the correct answer)?

How do I show the following:
${\int <!--\; \int \; -->}_0^{\mathrm{\alpha <!--\; \alpha \; -->}}dx{\int <!--\; \int \; -->}_0^{\mathrm{\alpha <!--\; \alpha \; -->}}dy\frac{\ln <!--\; \; -->|x-<!--\; -\; -->y|}{\sqrt{xy}}=4\mathrm{\alpha <!--\; \alpha \; -->}(\ln <!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->})+2\ln <!--\; \; -->(2)-<!--\; -\; -->3)$
where $\mathrm{\alpha <!--\; \alpha \; -->}>0$.