Master Calculus 1 Practice Problems with Expert Help

Limit with a summation and sine: how to calculate $\underset{n\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{n}^2\underset{k=0}{\overset{n-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}\sin <!--\; \; -->\left(\frac{2\mathrm{\pi <!--\; \pi \; -->}k}{n}\right)$ ?

How do I solve ${\int <!--\; \int \; -->}_{-<!--\; -\; -->1}^{1}x{e}^{-<!--\; -\; -->x^4}dx$

How do you find the instantaneous rate of change of the function $f(x)=x^2+2x$ when x=0?

Prove that $\underset{n\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{n^{n-<!--\; -\; -->1}}{n!e^n}=0$

Operator on polynomials, antiderivative
We are given a linear map:
$\mathbb{R}[X]\ni <!--\; \ni \; -->p\to <!--\; \to \; -->q\in <!--\; \in \; -->\mathbb{R}[X],q^{\prime }=p,q(0)=0$ and two norms on $\mathbb{R}[X]$ : $||p|{|}_{\mathrm{\infty <!--\; \infty \; -->}}=\underset{t\in <!--\; \in \; -->[0,1]}{sup}|p(t)|$, $||p|{|}_1={\int <!--\; \int \; -->}_0^1|p(t)|dt$.
I want to check whether the map is continuous in these norms.
When it comes to the first norm, I get:
$||q|{|}_{\mathrm{\infty <!--\; \infty \; -->}}=\underset{t\in <!--\; \in \; -->[0,1]}{sup}|q(t)|$
By Lagrange mean value theorem we have that there exists $a\in <!--\; \in \; -->[0,1]$ such that $q^{\prime }(a)=\frac{q(t)-<!--\; -\; -->q(0)}{t}$, so $q(t)\le <!--\; \le \; -->\underset{[0,1]}{sup}|q^{\prime }(t)|$.
And $||p|{|}_{\mathrm{\infty <!--\; \infty \; -->}}=\underset{t\in <!--\; \in \; -->[0,1]}{sup}|p(t)|=\underset{t\in <!--\; \in \; -->[0,1]}{sup}|q^{\prime }(t)|\ge <!--\; \ge \; -->||q||$
Is that correct so far?

prove that ${\int <!--\; \int \; -->}_a^{b}f+{\int <!--\; \int \; -->}_{f(a)}^{f(b)}{f}^{-<!--\; -\; -->1}=bf(b)-<!--\; -\; -->af(a)$
Let $0<a<b$ let $f>0$ be continuous and strictly increasing function on [a,b]. Prove that ${\int <!--\; \int \; -->}_a^{b}f+{\int <!--\; \int \; -->}_{f(a)}^{f(b)}{f}^{-<!--\; -\; -->1}=bf(b)-<!--\; -\; -->af(a)$.
How to approach this problem . Any Hint? I am suppose to do it without using antiderivatives.

The integration is
${\displaystyle {\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}{\displaystyle \frac{e^x}{e^{x^{\mathrm{\lambda <!--\; \lambda \; -->}}}}}dx}$

How does one go about evaluating the limits of nonelementary antiderivatives of functions?
We obviously know the following to be equal:
$\underset{z\to <!--\; \to \; -->\pm <!--\; \pm \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\text{erf}(z)=\pm <!--\; \pm \; -->1$
But as far as I'm aware, the only way we know how to calculate any given value of the error function is to approximate it. How do we know the limits at infinities without any way to evaluate it algabraically?

How do you find the instantaneous rate of change for $f(x)=\frac{x^2-<!--\; -\; -->2}{x-<!--\; -\; -->1}$ for x=2?

Antiderivative of $\frac{1}{4}\sin <!--\; \; -->(2x)$
I am trying to find the antiderivative of $\frac{1}{4}\sin <!--\; \; -->(2x)$ but i'mn ot sure how to find this. Can someone maybe give me what formula to use when it involves sin?

$\underset{x\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\left(x^6[e^{-<!--\; -\; -->\frac{1}{2x^3}}-<!--\; -\; -->\cos <!--\; \; -->\left(\frac{1}{x\sqrt{x}}\right)]\right)$
Any tip on how to calculate it?

At first glance , I though that the series
$\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->z{)}^n$

Antiderivative fraction
I'm would like to find the antiderivative of $\frac{x^2}{(x-<!--\; -\; -->1{)}^5}$
I tried without using partial fractions but I did not manage, so I started a lesson on partial fractions but I do not succeed in proving
$\frac{x²}{(x-<!--\; -\; -->1{)}^5}=\frac{1}{(x-<!--\; -\; -->1{)}^3}+\frac{2}{(x-<!--\; -\; -->1{)}^4}+\frac{1}{(x-<!--\; -\; -->1{)}^5}$
Could you tell me how to proceed?

How do you find the instantaneous rate of change of the function $x^2+3x-<!--\; -\; -->4$ when x=2?

How to find instantaneous rate of change for $f(x)=x^2+3x+4$ when x=4?

Evaluate
${\displaystyle \underset{x\to <!--\; \backslash rightarrow\; -->0}{lim}(\frac{1}{\ln <!--\; \; -->(1+x)}+\frac{1}{\ln <!--\; \; -->(1-<!--\; -\; -->x)})}$

Are these two antiderivatives equivalent?
Regarding the anti derivative of: ${\cos }^3<!--\; \; -->\left(x\right){\sin }^3<!--\; \; -->\left(x\right)$
Is the answer $-<!--\; -\; -->{\displaystyle \frac{{\sin }^4<!--\; \; -->\left(x\right)(2{\sin }^2<!--\; \; -->\left(x\right)-<!--\; -\; -->3)}{12}}$ equivalent to ${\displaystyle \frac{{\sin }^4<!--\; \; -->\left(x\right)}{4}}$.
If so, how? Integral Calculator says they're equivalent, but I don't understand how.

Series expression of ${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{e^{-<!--\; -\; -->x}}{1+e^{-<!--\; -\; -->x}}dx$

Show that two primitives/ antiderivatives are related via a constant
Let $I\subset <!--\; \subset \; -->R$ be an interval. A differentiable function $F:I\to <!--\; \to \; -->R$ is called a primitive for the function $f:I\to <!--\; \to \; -->\mathbb{R}$ if $F^{\prime }(x)=f(x)$ for all $x\in <!--\; \in \; -->I$.
Show: If $F_1$ and $F_2$ are two primitives for f on I then there is a constant $C\in <!--\; \in \; -->\mathbb{R}$ such that $F_2\equiv <!--\; \equiv \; -->F_1+C$, i.e. $F_2(x)=F_1(x)+C$ for all $x\in <!--\; \in \; -->I$.
What we have covered so far is the formal definition of the derivative in terms of the limit:
$\underset{h\to <!--\; \to \; -->0}{lim}\frac{F_1(x+h)-<!--\; -\; -->F_1(x)}{h}=f(x)$
$\underset{h\to <!--\; \to \; -->0}{lim}\frac{F_2(x+h)-<!--\; -\; -->F_2(x)}{h}=f(x)$
I do not see how I can prove firsthand that these two functions differ by a constant. I know it to be true from my pre-calculus and calculus experience, but how would one make the argument from an analysis point of view, can someone give me a hint as to where this constant 'appears'.
We normally just define $F_2=F_1+C$ and then it would follow immediately that these two functions have the same derivative.