Master Calculus 1 Practice Problems with Expert Help

How to find instantaneous rate of change for $F(x)=e^x$ when x=0?

How to find instantaneous rate of change for $f(x)=e^x$ at x=1?

Is there a way to find "polynomial rational functions" having point of inflections, in their graphs?
For example if the derivative of f(x) is
$f^{\prime }(x)=\frac{(x-<!--\; -\; -->1{)}^2(x-<!--\; -\; -->3)}{(x-<!--\; -\; -->2)}$
then f(x) has a point of inflection at $x=1$.
But f(x) is not a polynomial rational !
Is there a way to determine which antiderivatives are the ratio of to polynomials?

Looking for antiderivative
Find $\int <!--\; \int \; -->\log <!--\; \; -->(\sqrt{x+1}+\sqrt{x})dx$
I used the subtitution $\sqrt{x}=\tan <!--\; \; -->y$.
but i the integral became too long and involves $\int <!--\; \int \; -->{\sec }^3<!--\; \; -->ydy$
and also $\int <!--\; \int \; -->{\sec }^5<!--\; \; -->ydy$
but it solvable so i search for a simple solution if exist? Note log is the natural logarithm

Two apparently different antiderivatives of $\frac{1}{2x}$
What is right way to calculate this integral and why?
$\int <!--\; \int \; -->\frac{1}{2x}\text{d}x$
I thought, that this substitution is right:
$t=2x$
$\text{d}t=2\text{d}x$
$\frac{\text{d}t}{2}=\text{d}x$
$\int <!--\; \int \; -->\frac{1}{2x}\text{d}x=\int <!--\; \int \; -->\frac{1}{t}\frac{\text{d}t}{2}=\frac{1}{2}\ln <!--\; \; -->|2x|+C.$
But it's not right, because this is the correct answer:
$\int <!--\; \int \; -->\frac{1}{2x}\text{d}x=\frac{1}{2}\int <!--\; \int \; -->\frac{1}{x}\text{d}x=\frac{1}{2}\ln <!--\; \; -->|x|+C.$
Can someone explain me, why is the first way wrong?

I need to determine the convergence of the series whose general term is given by:
$a_n=\frac{1}{\sqrt{1}\cdot <!--\; \cdot \; -->n^2}+\frac{1}{\sqrt{2}\cdot <!--\; \cdot \; -->(n-<!--\; -\; -->1{)}^2}+\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->\cdot <!--\; \cdot \; -->+\frac{1}{\sqrt{n}\cdot <!--\; \cdot \; -->1^2}$

I have to evaluate
$\underset{n\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(-<!--\; -\; -->1{)}^{n-<!--\; -\; -->1}\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}\sqrt{n^2+0.5n+1})wheren\mathrm{ϵ<!--\; ϵ\; -->}N$

Given a multivariate rational function $p(\stackrel{\to <!--\; \to \; -->}{x})=\frac{f(\stackrel{\to <!--\; \to \; -->}{x})}{g(\stackrel{\to <!--\; \to \; -->}{x})}$ over $[0,1{]}^n$ with $p(\stackrel{\to <!--\; \to \; -->}{x})\in <!--\; \in \; -->[0,1]$, how can we come up with a polynomial approximation of $p$, say $q(\stackrel{\to <!--\; \to \; -->}{x})$ such that $|p(\stackrel{\to <!--\; \to \; -->}{x})-<!--\; -\; -->q(\stackrel{\to <!--\; \to \; -->}{x})|\le <!--\; \le \; -->\mathrm{ϵ<!--\; ϵ\; -->}$ for all $\stackrel{\to <!--\; \to \; -->}{x}$ ?
For the univariate case, we might use chebyshev approximation, however, what are the results for the multivariate case?

Definition of setting the integration constant equal to zero.
In Calculus books, we often hear of setting the integration constant equal to zero. But what does that actually mean? For example, the set of antiderivatives of 2x is $x^2+C$. If we set the constant equal to zero, we get simply $x^2$. However, I can also write the set of antiderivatives as $x^2+\mathrm{\pi <!--\; \pi \; -->}+C$. In that case, setting the constant equal to zero would give us $x^2+\mathrm{\pi <!--\; \pi \; -->}$. So, my question is, what does setting the constant equal to zero in an antiderivative actually mean?

Is there a relationship between $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\left(\frac{1}{2}\right)}^n=1$ and ${\int <!--\; \int \; -->}_1^{\mathrm{\infty <!--\; \infty \; -->}}\frac{1}{x^2}dx=1$ ?

How to prove this function has Darboux's property?
$f(x)=\{\begin{array}{ll}\cos <!--\; \; -->(\frac{1}{x})& \text{if }x\ne <!--\; \ne \; -->0\\ 0& \text{if }x=0\end{array}$
How do I prove this function has Darboux's property? I know it has it because it has antiderivatives, but how do I prove it otherwise, with intervals maybe?

How to find instantaneous rate of change for $f(x)=\ln <!--\; \; -->(x)$ when x=0?

Can we allow f to be undefined at finitely many points in (a,b) when formulating ${\int <!--\; \int \; -->}_a^{b}f(x)dx=F(b)-<!--\; -\; -->F(a)$, (F is the antiderivative of f)
Let f be a real-valued function on a closed interval [a,b] undefined only at finite points in (a,b). Let F be antiderivative of f. Then:
${\int <!--\; \int \; -->}_a^{b}f(x)dx=F(b)-<!--\; -\; -->F(a)$
Is the theorem true? How shall we prove it?
I ask this because I have read that function undefined at finite points in the interval (a,b) doesn't effect integration i.e. we can take the antiderivatives and apply the limits as usual to get the definite integral.

Antiderivative of $|\sin <!--\; \; -->(x)|$
Because ${\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}\sin <!--\; \; -->(x)\mathrm{d}x=2,$, then ${\int <!--\; \int \; -->}_0^{16\mathrm{\pi <!--\; \pi \; -->}}|\sin <!--\; \; -->(x)|\mathrm{d}x=32.$.
And Wolfram Alpha agrees to this, but when I ask for the indefinite integral $\int <!--\; \int \; -->|\sin <!--\; \; -->(x)|\mathrm{d}x,$,
Wolfram gives me $-<!--\; -\; -->\cos <!--\; \; -->(x)\mathrm{s}\mathrm{g}\mathrm{n}(\sin <!--\; \; -->(x))+c.$.
However, $[-<!--\; -\; -->\cos <!--\; \; -->(x)\mathrm{s}\mathrm{g}\mathrm{n}(\sin <!--\; \; -->(x)){]}_0^{16\mathrm{\pi <!--\; \pi \; -->}}=0,$,
So what's going on here? What is the antiderivative of $|\sin <!--\; \; -->(x)|$?

How to tackle the integral ${\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}\frac{3\mathrm{\pi <!--\; \pi \; -->}{x}^2-<!--\; -\; -->2x^3}{(1+\sin <!--\; \; -->x{)}^n}dx,\text{ where }n\in <!--\; \in \; -->N$

Integrate:
$\int <!--\; \int \; -->\sqrt{x^2-<!--\; -\; -->x+1}dx$

Antiderivative of a polynomial fraction
I am trying to solve this problem and I can't seem to understand where I am going wrong. My understanding of antiderivatives is this formula:
$\int <!--\; \int \; -->x^n\mathrm{d}x=\frac{x^{n+1}}{n+1}+c$
The problem I am trying to solve is:
$f(x)=\frac{x^5-<!--\; -\; -->x^3+8x}{x^4}$
I think there are two possible parts that I am getting wrong, I started the problem by separating them into individual fractions and solving each fraction at a time. I think I get it wrong at either the ln(x) or $\frac{8x^{-<!--\; -\; -->2}}{-<!--\; -\; -->2}$. Here is my attempt: $f(x)=\frac{x^5}{x^4}-<!--\; -\; -->\frac{x^3}{x^4}+\frac{8x}{x^4}=x-<!--\; -\; -->\frac{1}{x}+\frac{8}{x^3}$
Then from here I get the antiderivative:
$F(x)=\frac{x^2}{2}-<!--\; -\; -->\ln <!--\; \; -->(x)+\frac{8x^{-<!--\; -\; -->2}}{-<!--\; -\; -->2}+c=\frac{x^2}{2}-<!--\; -\; -->\ln <!--\; \; -->(x)-<!--\; -\; -->\frac{4}{x^2}+c$
I keep getting the wrong answer when I plug it in, am I missing some fundamental piece of information about antiderivatives which is causing me to mess up the ln(x) or is my algebra wrong and I mess up around $\frac{8x^{-<!--\; -\; -->2}}{-<!--\; -\; -->2}$?

How to calculate $\underset{x\to <!--\; \backslash rightarrow\; -->1}{lim}\frac{(x^2-<!--\; -\; -->3x+2)}{(x-<!--\; -\; -->1)(x^3-<!--\; -\; -->1)}$

Complex antiderivative
I am confused on a couple things:
1.) Why is it that an integral of a complex valued function of a complex variable exists if f(z(t)) is piecewise continuous (and/or piecewise continuous on $\mathbb{C}$) and not continuous, like a real?
2.) Why is it that one cannot make use of an antiderivative to evaluate an integral of a function like 1/z, on a contour of something like $z=2e^{i\mathrm{\theta <!--\; \theta \; -->}}$ positively oriented with $-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->\mathrm{\pi <!--\; \pi \; -->}$? That is, because $F^{\prime }(z)=1/z$ is undefined at 0, it is disqualified (I think). But why?