Master Calculus 1 Practice Problems with Expert Help

Gamma function approximation
When $h\to <!--\; \to \; -->0$, ${\displaystyle \frac{\mathrm{\Gamma <!--\; \Gamma \; -->}(h+\mathrm{\alpha <!--\; \alpha \; -->})\cdot <!--\; \cdot \; -->h}{\mathrm{\Gamma <!--\; \Gamma \; -->}(h+1)}}\sim <!--\; \sim \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}(\mathrm{\alpha <!--\; \alpha \; -->})\cdot <!--\; \cdot \; -->h$

Function y-axis symmetry in (anti)derivatives
If for a function f the equation $f(x)=f(-<!--\; -\; -->x)$ is true, is it also automatically true for all its derivatives and antiderivatives?

Integrating a step function using antiderivatives
Let $f(x)=\{\begin{array}{l}\begin{array}{rl}1& \text{ if }0\le <!--\; \le \; -->x\le <!--\; \le \; -->1\\ 2& \text{ if }1<x\le <!--\; \le \; -->2\end{array}\end{array}$.
Then ${\int <!--\; \int \; -->}_0^{2}f(x)dx=3$ and an anti derivative of f(x) is
$F(x)=\{\begin{array}{l}\begin{array}{rl}x& \text{ if }0\le <!--\; \le \; -->x\le <!--\; \le \; -->1\\ 2x& \text{ if }1<x\le <!--\; \le \; -->2\end{array}\end{array}$
But, $F(2)-<!--\; -\; -->F(0)=4-<!--\; -\; -->0=4\ne <!--\; \ne \; -->3.$. Why has this happened?
Find an anti-derivative G(x) of f(x) such that $G(2)-<!--\; -\; -->G(0)=3$, the correct answer.
My attempt: $F(2)-<!--\; -\; -->F(0)=4\ne <!--\; \ne \; -->3$ because we need to break it up into two parts like this: ${\int <!--\; \int \; -->}_1^{2}2xdx+{\int <!--\; \int \; -->}_0^{1}xdx=x^2{|}_0^2+\frac{1}{2}{x}^2{|}_0^1=4-<!--\; -\; -->1+\frac{1}{2}-<!--\; -\; -->0=3.5$ but this is not right either.
For the "Find an anti-derivative G(x)..."I'm completely lost but perhaps if we solve the first part then I'll understand what it's asking for there.

Antiderivative of CDF
Let F(x) be the CDF of a random variable. Does its antiderivative ${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{x}F(x^{\prime })d{x}^{\prime }$ have an interpretation?

Antiderivative problem
What is the antiderivative of $(2x+7{)}^{1/2}$?
My understanding is that it would be $\frac{2}{3}\times <!--\; \times \; -->(2x+7{)}^{3/2}$ but according to the source I am working from the answer is $\frac{1}{3}\times <!--\; \times \; -->(2x+7{)}^{3/2}$.

Prove or disprove, limits of antiderivatives
If g(x), f(x) are real and continuous, and $g(x)\le <!--\; \le \; -->$ $f(x)\le <!--\; \le \; -->$ $g(x)+\frac{1}{x}$ for every $x>1$. The following limit is equal to 0:
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_x^{x+\sqrt{x}}(f(t)-<!--\; -\; -->g(t))dt=0$

For cosine approximation , how to arrive at $(9.11)$ from $(9.10)$ ?

Finding antiderivative by Maclaurin series
I want to find $\int <!--\; \int \; -->p(x)dx$ where:
$p(x)=\frac{e^{\frac{-<!--\; -\; -->x^2}{2}}}{\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}.$.
This function can not be manually computed so I am using Maclaurin series to find the antiderivative. I believe the Maclaurin series for this function would be:
$\frac{1}{\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{{\frac{-<!--\; -\; -->x^2}{2}}^n}{n!}.$
The bounds of integration would be from -1 to 1. How would I create a Maclaurin series for the antiderivative from this?

$\begin{array}{r}\underset{k=2}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{(1-<!--\; -\; -->e^{-<!--\; -\; -->n{s}^{k-<!--\; -\; -->1}})}^k\end{array}$
Compute an approximation of this series, for the regime where n is large.

Does the midpoint approximation rule over or under estimates a increasing and decreasing function?

Antiderivative of the greatest integer function
One of my homework problem wants me to prove that the greatest integer function $f(x)=\lfloor <!--\; \lfloor \; -->x\rfloor <!--\; \rfloor \; -->$ does not have an antiderivative. While thinking, I got to this expression, ${\int <!--\; \int \; -->}_0^x\lfloor <!--\; \lfloor \; -->t\rfloor <!--\; \rfloor \; -->dt=\frac{1}{2}\lfloor <!--\; \lfloor \; -->x\rfloor <!--\; \rfloor \; -->(2x-<!--\; -\; -->\lfloor <!--\; \lfloor \; -->x\rfloor <!--\; \rfloor \; -->-<!--\; -\; -->1)$
Is there an antiderivative for the greatest integer function?
The first part of the fundamental theorem of calculus implies the existence of an antiderivative for continuous functions. But what about discontinuous functions? Is there an antiderivative for any discontinuous functions?
And thank you for any hints or ideas for my homework problem.

Approximation to simplify an equation
${(1+\frac{{\mathrm{\Delta <!--\; \Delta \; -->}}^2}{d^2})}^{\frac{-<!--\; -\; -->3}{2}}+{(1-<!--\; -\; -->\frac{{\mathrm{\Delta <!--\; \Delta \; -->}}^2}{d^2})}^{-<!--\; -\; -->2}=(1-<!--\; -\; -->\frac{3}{2}\frac{{\mathrm{\Delta <!--\; \Delta \; -->}}^2}{d^2}+...)+(1+\frac{2{\mathrm{\Delta <!--\; \Delta \; -->}}^2}{d^2}+...)$

Antiderivative around inverse trigonometric function
I'm just learning about antiderivative that resulted some inverse trigonometric functions, since derivative of $\arccos <!--\; \; -->(x)$ is $-<!--\; -\; -->\frac{1}{\sqrt{1-<!--\; -\; -->x^2}}$. I tried put this value back into integral calculator, but I got $-<!--\; -\; -->\arcsin <!--\; \; -->(x)$ instead of $\arccos <!--\; \; -->(x)$. even though $-<!--\; -\; -->\arcsin <!--\; \; -->(x)\ne <!--\; \ne \; -->\arccos <!--\; \; -->(x)$ so why this happen? How should I fix my understanding?

Antiderivative of $(x^2+c{)}^{-<!--\; -\; -->3/2}$
What method should be used to determine the antiderivative of this expression?
Edit: I have $c>0$ in the problem I'm working on.

Integration by substitution for antiderivatives
Integration by substitution is an integration technique for definite integrals based upon this formula:
${\displaystyle {\int <!--\; \int \; -->}_a^{b}f(\mathrm{\varphi <!--\; \varphi \; -->}(x)){\mathrm{\varphi <!--\; \varphi \; -->}}^{\prime }(x)dx={\int <!--\; \int \; -->}_{\mathrm{\varphi <!--\; \varphi \; -->}(a)}^{\mathrm{\varphi <!--\; \varphi \; -->}(b)}f(u)du}$
I don't understand how it can be applied in the case of indefinite integrals, how it can be used to find antiderivatives, as in this example:
${\displaystyle \int <!--\; \int \; -->\frac{dx}{x^2+a^2}=\frac{1}{a}\int <!--\; \int \; -->\frac{du}{1+u^2}=\frac{1}{a}{\tan }^{-<!--\; -\; -->1}<!--\; \; -->(u)+C=\frac{1}{a}{\tan }^{-<!--\; -\; -->1}<!--\; \; -->\left(\frac{x}{a}\right)+C}$
The formula ${\displaystyle {\int <!--\; \int \; -->}_a^{b}f(\mathrm{\varphi <!--\; \varphi \; -->}(x)){\mathrm{\varphi <!--\; \varphi \; -->}}^{\prime }(x)dx={\int <!--\; \int \; -->}_{\mathrm{\varphi <!--\; \varphi \; -->}(a)}^{\mathrm{\varphi <!--\; \varphi \; -->}(b)}f(u)du}$ is for definite integrals. Why can it be applied to indefinite integrals?

Find the integral of $2t^5+t+3$ with respect to t.

Find the integral of $2x^2-<!--\; -\; -->2x-<!--\; -\; -->2$ with respect to x.

Find the integral of $-<!--\; -\; -->4x^2-<!--\; -\; -->3$ with respect to x.

Find the integral of $t^4$ with respect to t.

What is $\int <!--\; \int \; -->\frac{8}{9}x\cos <!--\; \; -->(3x)dx$?